Telescoping Series

Introduction

Consider the series \(1/(1\cdot2) + 1/(2\cdot3) + 1/(3\cdot4) + \ldots\) up to \(n\) terms.

The \(k\)-th term of this series can be expressed as \(1/(k(k+1))\). This term can be rewritten as:

\[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \]

Let's see how the terms cancel out when we add them:

\[ \begin{align*} S &= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \\ &= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \ldots - \frac{1}{n+1} \\ &= 1 - \frac{1}{n+1} \end{align*} \]

Here, all intermediate terms \(- \frac{1}{2} + \frac{1}{2}\), \(- \frac{1}{3} + \frac{1}{3}\), ..., cancel out, showcasing the telescoping nature where the sum \(S\) effectively collapses to the first term minus the last term of the sequence \(\{t_k\}\). The term "telescoping" comes from the series' ability to collapse, much like a telescope, into a much simpler form.

Definition:

The essence of a telescoping series lies in its unique structure, where each term is cleverly split into two or sometimes more components, leading to a systematic cancellation of terms as the series progresses. This splitting is designed such that, when the series is summed, most intermediate terms cancel each other out, leaving only a few terms at the beginning and end of the series. This phenomenon greatly simplifies the calculation of the series sum, reducing a potentially complex series to a simple expression involving only the first and last terms of the sequence. The ability to transform a series in this way hinges on identifying an appropriate form for each term, often through partial fraction decomposition or similar techniques, which reveals the series' telescoping nature. This approach not only streamlines calculations but also unveils the underlying simplicity masked by the series' initial complex appearance.

Example

In general, for a telescoping series, the strategy involves splitting the general term \(t_k\) into the difference of two terms, such that \(t_k = a_k - a_{k-1}\). This manipulation sets the stage for a sequence where, upon summing, most terms cancel out, leading to a simplified expression for the sum.

Consider the Series \(1/(1\cdot3) + 1/(3\cdot5) + 1/(5\cdot7) + \ldots\) up to \(n\) terms:

Step 1: Identify the General Term
The general term of the given series is \(t_k = 1/((2k-1)(2k+1))\), corresponding to each term in the series.

Step 2: Decompose the General Term

• We multiply the numerator and the denominator by 2, acknowledging that multiplying by a form of 1 (2/2) does not change the value:

  \[ \frac{1}{(2k-1)(2k+1)} = \frac{2}{2 \cdot (2k-1)(2k+1)} \]

• We express 2 in the numerator as a difference between \(2k+1\) and \(2k-1\), exploiting the structure of the denominator:

  \[ \frac{2}{2 \cdot (2k-1)(2k+1)} = \frac{(2k+1) - (2k-1)}{2 \cdot (2k-1)(2k+1)} \]

• Now split this into two terms:

  \[ \frac{1}{2} \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) \]

This step effectively splits each term into two parts \(t_k = a_k - a_{k-1}\), which sets up the series for telescoping.

Step 4: Telescoping the Series
Now, applying this form to our series and adding the terms from \(k=1\) to \(n\), we get:

\[ S = \frac{1}{2} \left( \left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \ldots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) \right) \]

Step 5: Observe the Cancellation
Most terms within the parentheses cancel out, leaving only the first and the last term:

\[ S = \frac{1}{2} \left( 1 - \frac{1}{2n+1} \right) \]

Conclusion:
By decomposing the general term \(t_k\) into the difference \(a_k - a_{k-1}\) and leveraging the telescoping effect, we've simplified the series \(1/(1\cdot3) + 1/(3\cdot5) + 1/(5\cdot7) + \ldots\) up to \(n\) terms to a more manageable expression. This illustrates the power of the telescoping series to turn a complex summation problem into a straightforward calculation.

How to telescope?

To solve telescoping series well, start by finding the general term of the series, which we call \(t_r\). Then, try to split this term into a format that looks like \(t_r = a_r - a_{r-1}\). This step might need some guessing and testing, plus using what you've learned from solving similar problems before. The key to getting better at telescoping series is to practice a lot. By working through many different examples, you'll learn various ways to handle new problems you come across. So, the best way to improve is by solving as many telescoping series as you can, using what you learn each time to tackle new challenges.

Some illustrating examples of telesoping Series

Example

Problem: Find the sum of \(\sum_{r=1}^{n} \frac{1}{r^2 + 2r}\)

Solution: Given the series \(\sum_{r=1}^{n} \frac{1}{r^2 + 2r}\), we aim to simplify and identify the telescoping nature of the series. The general term can be expressed as \(\frac{1}{r(r+2)}\). To manipulate this into a telescoping form, we observe that the difference in the denominator, \(r+2-r\), is 2. Multiplying both the numerator and denominator by 2, we aim to utilize this difference effectively.

By multiplying, we have \(\frac{2}{2r(r+2)}\), which, considering the denominator's difference, allows us to represent 2 in the numerator as \((r+2) - r\). Thus, the general term becomes:

\[ \frac{1}{2} \left( \frac{1}{r} - \frac{1}{r+2} \right) \]

Applying this to the series from \(r=1\) to \(n\), we get:

\[ \frac{1}{2} \left( \frac{1}{1} - \frac{1}{3} \right) + \frac{1}{2} \left( \frac{1}{2} - \frac{1}{4} \right) + \ldots + \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+2} \right) \]

To illustrate the cancellation, write in a vertical form with each term in seperate line

\[ \begin{align*} &\frac{1}{2} \left( \frac{1}{1} - \cancel{\frac{1}{3}} \right) \\ &\frac{1}{2} \left( \frac{1}{2} - \cancel{\frac{1}{4}} \right) \\ &\frac{1}{2} \left( \cancel{\frac{1}{3}} - \cancel{\frac{1}{5}} \right) \\ &\frac{1}{2} \left( \cancel{\frac{1}{4}} - \cancel{\frac{1}{6}} \right) \\ &\vdots \\ &\frac{1}{2} \left( \cancel{\frac{1}{n-2}} - \cancel{\frac{1}{n}} \right)\\ &\frac{1}{2} \left( \cancel{\frac{1}{n-1}} - \frac{1}{n+1} \right)\\ &\frac{1}{2} \left( \cancel{\frac{1}{n}} - \frac{1}{n+2} \right) \end{align*} \]

In this arrangement, terms cancel in a pattern that skips one line, revealing the telescoping nature. Specifically, \(\frac{1}{3}\) from the first term cancels with \(\frac{1}{3}\) in a subsequent term, and so on, leaving uncancelled the last part of the final term and the first part of the initial term.

Thus, the sum simplifies to the non-cancelled portions:

\[ S = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right) \]

This expression provides the simplified sum of the series by leveraging the telescoping principle, where intermediate terms cancel out, leaving only the initial and final terms of the sequence.

Example

Problem: Evaluate the sum of the series \( \frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7} + \cdots + \frac{n^2}{(2n-1)(2n+1)} \) up to \( n \) terms.

Solution:

The \( k \)-th term of the series is given by:

\[ t_k = \frac{k^2}{(2k-1)(2k+1)} \]

To find the sum of the series, we manipulate the \( k \)-th term to make the series telescopic. Multiply and divide \( t_k \) by 4:

\[ t_k = \frac{4k^2}{4 \cdot (2k-1)(2k+1)} \]

Now, add and subtract 1 in the numerator to split into two terms:

\[ t_k = \frac{4k^2 - 1 + 1}{4(2k-1)(2k+1)} \]

\[ t_k = \frac{(2k-1)(2k+1) + 1}{4(2k-1)(2k+1)} \]

\[ t_k = \frac{1}{4} + \frac{1}{4(2k-1)(2k+1)} \]

Separate the second term into partial fractions:

\[ t_k = \frac{1}{4} + \frac{1}{8} \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) \]

The series up to \( n \) terms is:

\[ S_n = \sum_{k=1}^{n} \left[ \frac{1}{4} + \frac{1}{8(2k-1)} - \frac{1}{8(2k+1)} \right] \]

Upon summing, the middle terms cancel out, leaving us with the first and the last terms:

\[ S_n = \frac{n}{4} + \frac{1}{8} \left( 1 - \frac{1}{2n+1} \right) \]

Simplify to find the sum:

\[ S_n = \frac{n}{4} + \frac{1}{8} - \frac{1}{8(2n+1)} \]

\[ S_n = \frac{2n(2n+1) + 2n + 1 - 1}{8(2n+1)} \]

\[ S_n = \frac{4n^2 + 4n}{8(2n+1)} \]

\[ S_n = \frac{n(n + 1)}{2(2n+1)} \]

The sum of the series \( \frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7} + \cdots + \frac{n^2}{(2n-1)(2n+1)} \) up to \( n \) terms is therefore \( \frac{n(n + 1)}{4(2n+1)} \). \(\blacksquare\)

Example

Problem: Determine the sum of the series

\[ \frac{1}{1+1^2+1^4} + \frac{2}{1+2^2+2^4} + \frac{3}{1+3^2+3^4} + \cdots + \frac{n}{1+n^2+n^4} \]

up to \( n \) terms.

Solution:

Consider the \( k \)-th term of the series:

\[ t_k = \frac{k}{1+k^2+k^4} \]

The denominator can be factored as:

\[ t_k = \frac{k}{(1+k^2)^2 - k^2} \]

This is a difference of squares, which we can express as:

\[ t_k = \frac{k}{[(1+k^2)+k][(1+k^2)-k]} \]

Simplify the expression:

\[ t_k = \frac{k}{(1+k+k^2)(1-k+k^2)} \]

Write the \( k \)-th term as a difference of two fractions:

\[ t_k = \frac{1}{2} \left[ \frac{1}{1-k+k^2} - \frac{1}{1+k+k^2} \right] \]

Now, let's consider the sum of the first \( n \) terms of the series:

\[ S_n = \frac{1}{2} \sum_{k=1}^{n} \left[ \frac{1}{1-k+k^2} - \frac{1}{1+k+k^2} \right] \]

As we sum from \( k=1 \) to \( n \), most terms will cancel due to the telescoping nature of the series:

\[\begin{align*} S_n &= \frac{1}{2} \left[ \left(\frac{1}{1-1+1^2} - \frac{1}{1+1+1^2}\right) + \left(\frac{1}{1-2+2^2} - \frac{1}{1+2+2^2}\right) + \left(\frac{1}{1-3+3^2} - \frac{1}{1+3+3^2}\right) + \ldots + \left(\frac{1}{1-n+n^2} - \frac{1}{1+n+n^2}\right) \right] \\ &= \frac{1}{2} \left[ \left(\frac{1}{1} - \cancel{\color{red}\frac{1}{3}}\right) + \left(\cancel{\color{red}\frac{1}{3}} - \cancel{\color{red}\frac{1}{7}}\right) + \left(\cancel{\color{red}\frac{1}{7}} - \cancel{\color{red}\frac{1}{13}}\right) + \ldots + \left(\cancel{\color{red}\frac{1}{1+(n-1)+(n-1)^2}} - \frac{1}{1+n+n^2}\right) \right] \\ &= \frac{1}{2} \left[ 1 - \frac{1}{1+n+n^2} \right] \end{align*}\]

Hence, the sum of the series is:

\[ S_n = \frac{1}{2} \left( 1 - \frac{1}{1+n+n^2} \right) \]

Simplify to find the final expression for the sum:

\[ S_n = \frac{1}{2} - \frac{1}{2(1+n+n^2)} \]

\[ S_n = \frac{1+n+n^2 - 1}{2(1+n+n^2)} \]

\[ S_n = \frac{n(n+1)}{2(1+n+n^2)} \]

Thus, the sum of the series

\[ \frac{1}{1+1^2+1^4} + \frac{2}{1+2^2+2^4} + \frac{3}{1+3^2+3^4} + \cdots + \frac{n}{1+n^2+n^4} \]

up to \( n \) terms is

\[ \frac{n(n+1)}{2(1+n+n^2)} \]

Infinite telescoping Series

When finding the sum of an infinite series, the process typically involves first determining the sum up to \(n\) terms and then considering what happens as \(n\) approaches infinity (\(n \rightarrow \infty\)).

1. Identify the General Term: Start by identifying the general term of the series, which is usually given by \(a_n\), where \(n\) is the term number.

2. Find the Sum up to \(n\) Terms: Derive a formula for \(S_n\), the sum of the first \(n\) terms. This might involve techniques like telescoping, or recognizing a pattern that allows for simplification.

3. Apply the Limit as \(n\) Approaches Infinity: Examine the limit of \(S_n\) as \(n \rightarrow \infty\). This step is crucial for understanding the behavior of the series over an infinite number of terms.

4. If the Limit Exists and Is Finite: If \(\lim_{n \rightarrow \infty} S_n = L\), where \(L\) is a finite number, then the sum of the infinite series converges to \(L\). This means that as you add more and more terms, the total sum approaches \(L\) more closely.

5. If the Limit Does Not Exist or Is Infinite: If \(\lim_{n \rightarrow \infty} S_n\) does not converge to a finite value, then the series diverges. This indicates that as you add more terms, the sum does not settle to a specific value but instead grows without bound or oscillates.

Example

Problem: Find the sum of the series \(\frac{3}{1^2 \cdot 2^2} + \frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \cdots\) upto infinity.

Solution:

Step 1: Express the series in sigma notation

Given the series \(\frac{3}{1^2 \cdot 3^2} + \frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \cdots\), we can see that the numerator follows the arithmetic progression \(3, 5, 7, \ldots\), which can be written as \(2k+1\) for the \(k\)-th term. Therefore, the general term of the series is \(\frac{2k+1}{k^2(k+1)^2}\).

Step 2: Rewrite the \(k\)-th term using a telescoping series

The numerator \(2k+1\) can be rewritten as \((k+1)^2 - k^2\), allowing us to express the \(k\)-th term as:

\[ \frac{(k+1)^2 - k^2}{k^2(k+1)^2} = \frac{(k+1)^2}{k^2(k+1)^2} - \frac{k^2}{k^2(k+1)^2} \]

This simplifies to:

\[ \frac{1}{k^2} - \frac{1}{(k+1)^2} \]

Step 3: Apply the telescoping property

The series now takes the form:

\[ \sum_{k=1}^{n} \left( \frac{1}{k^2} - \frac{1}{(k+1)^2} \right) \]

When we sum from \(k=1\) to \(n\), most terms cancel out due to the telescoping nature, leaving us with:

\[ 1 - \frac{1}{(n+1)^2} \]

Step 4: Find the limit as \(n\) approaches infinity

To find the sum to infinity, we take the limit of the partial sum as \(n\) approaches infinity:

\[ \lim_{n \to \infty} \left(1 - \frac{1}{(n+1)^2}\right) = 1 - \lim_{n \to \infty} \frac{1}{(n+1)^2} = 1 - 0 = 1 \]

Conclusion: The sum of the given series to infinity is 1.

Example

Problem: Find the sum of the series \(3/1^2 + 5/(1^2+2^2) + 7/(1^2+2^2+3^2) + \cdots\) up to infinity.

Solution:

Step 1: Identify the pattern and express the series in sigma notation

Observing the given series, we notice that the numerator follows an arithmetic progression \(3, 5, 7, \ldots\), which can be represented as \(2r+1\) for the \(r\)-th term. The denominator accumulates the squares of natural numbers up to \(r\), which can be represented by the sum \(1^2 + 2^2 + \cdots + r^2\). The sum of the squares of the first \(r\) natural numbers is given by the formula \(\frac{r(r+1)(2r+1)}{6}\). Therefore, the \(r\)-th term of the series can be written as:

\[ \frac{2r+1}{\frac{r(r+1)(2r+1)}{6}} = \frac{6}{r(r+1)} \]

Step 2: Rewrite the series in sigma notation

The series can now be expressed in sigma notation as:

\[ \sum_{r=1}^{\infty} \frac{6}{r(r+1)} \]

Step 3: Split the rth term**

The term \(\frac{6}{r(r+1)}\) can be decomposed into partial fractions as:

\[ \frac{6}{r(r+1)} = \frac{6(r+1-r)}{r(r+1)} \]

\[ =\frac{6}{r} - \frac{6}{r+1} \]

Step 4: Apply the telescoping property

The series now takes the form:

\[ \sum_{r=1}^{n} \left( \frac{6}{r} - \frac{6}{r+1} \right) \]

Applying the telescoping property, most terms cancel out, leaving us with:

\[ 6\left(1 - \frac{1}{n+1}\right) \]

Step 5: Find the limit as \(n\) approaches infinity

To find the sum to infinity, take the limit of the partial sum as \(n\) approaches infinity:

\[ \lim_{n \to \infty} 6\left(1 - \frac{1}{n+1}\right) = 6(1-0) = 6 \]

Conclusion: The sum of the given series up to infinity is 6.

vn method

To understand this method which is nothing but telescoping applied on a special kind of series, consider the following example. The series has a speical form. Consider the sequence \(1,2, 3, 4, 5, \ldots \) which is clearly an AP. The series \(1.2.3 + 2.3.4 + 3.4.5 + \cdots\) is created from the arithmetic progression (AP) \(1, 2, 3, 4, 5, 6, \ldots\) by "shifting" the starting point of each product by one term sequentially. Each term in the series is a product of three consecutive numbers from the AP, starting with the first term as \(1 \cdot 2 \cdot 3\), then shifting the group one term forward to get the second term as \(2 \cdot 3 \cdot 4\), and continuing this process of shifting to generate subsequent terms.

We apply this method in general on series formed with this process. Taking any AP, grouping a fixed number of terms, multiplying, shiting by one term and adding to get a new series.

Example

Problem: Evaluate the sum of the series \(1.2.3 + 2.3.4 + 3.4.5 + \cdots\) up to \(n\) terms. Solution:

Step 1: Define the Series

The given series is represented by the \(n^{th}\) term \(t_n = n(n + 1)(n + 2)\).

Step 2: Introduce a new sequence \(v_n\) for Telescoping

Define \(v_n = n(n + 1)(n + 2)(n + 3)\), which includes an extra term \((n + 3)\) compared to \(t_n\).

Step 3: Calculate \(v_n - v_{n-1}\)

\[ v_n - v_{n-1} = n(n + 1)(n + 2)(n + 3) - (n-1)n(n + 1)(n + 2) \]

Taking common factors from both terms:

\[ = (n + 1)(n + 2) [n(n + 3) - (n - 1)n] \]

\[ = n(n + 1)(n + 2)(n+3 - n+1) \]

\[ = n(n + 1)(n + 2)(4) \]

\[ = 4n(n + 1)(n + 2) \]

Observe that this is exactly \(4\) times the \(n^{th}\) term \(t_n\) of the original series.

Thus, \(v_n - v_{n-1} = 4t_n\).

Step 4: Express \(t_n\) in Terms of \(v_n\) and \(v_{n-1}\)

Given \(v_n - v_{n-1} = 4t_n\), we find that \(t_n = \frac{1}{4}(v_n - v_{n-1})\).

Step 5: Apply the Telescoping Method

The sum of the first \(n\) terms of the series can be expressed as:

\[ S_n = \sum_{k=1}^{n} t_k = \frac{1}{4} \sum_{k=1}^{n} (v_k - v_{k-1}) \]

This series telescopes, as consecutive terms cancel out, leaving:

\[ S_n = \frac{1}{4} (v_n - v_0) \]

Given \(v_0 = 0\), the final expression for the sum of the first \(n\) terms is:

\[ S_n = \frac{1}{4} v_n = \frac{1}{4} n(n + 1)(n + 2)(n + 3) \]

Conclusion:

The sum of the series \(1.2.3 + 2.3.4 + 3.4.5 + \cdots\) up to \(n\) terms is given by \( \frac{1}{4} n(n + 1)(n + 2)(n + 3) \).

For an arithmetic progression (AP) \(a_1, a_2, a_3, \ldots\) with a common difference \(d\), a series is formed by multiplying \(p\) consecutive terms, shifting one term forward with each step, and summing these products. Each term of the series, for a given \(k\), is \(T_k = a_k \cdot a_{k+1} \cdot \ldots \cdot a_{k+p-1}\), creating a series by this repetitive shifting and multiplying process.

\[ T_1 = a_1 \cdot a_2 \cdot \ldots \cdot a_p \]

\[ T_2 = a_2 \cdot a_3 \cdot \ldots \cdot a_{p+1} \]

\[ T_3 = a_3 \cdot a_4 \cdot \ldots \cdot a_{p+2} \]

So, the resulting series is:

\[ S = a_1 \cdot a_2 \cdot \ldots \cdot a_p + a_2 \cdot a_3 \cdot \ldots \cdot a_{p+1} + a_3 \cdot a_4 \cdot \ldots \cdot a_{p+2} + \ldots \]

We can sum such series using v-n method.

Example

Problem: Find the sum of the series \(1\cdot3\cdot5\cdot7 + 3\cdot5\cdot7\cdot9 + 5\cdot7\cdot9\cdot11 + \cdots\) up to \(n\) terms using the \(v_n\) method.

Solution:

The \(n\)-th term of the series is given by \(t_n = (2n-1)(2n+1)(2n+3)(2n+5)\). We define \(v_n = (2n-1)(2n+1)(2n+3)(2n+5)(2n+7)\) to apply the \(v_n\) method.

To find the sum, we calculate \(v_n - v_{n-1}\):

\[ v_n - v_{n-1} = (2n-1)(2n+1)(2n+3)(2n+5)(2n+7) - (2(n-1)-1)(2(n-1)+1)(2(n-1)+3)(2(n-1)+5)(2(n-1)+7) \]

Simplifying the above expression by taking common factors:

\[ = (2n-1)(2n+1)(2n+3)(2n+5)\left[(2n+7) - (2n-3)\right] \]

\[ = (2n-1)(2n+1)(2n+3)(2n+5)\cdot10 = 10t_n \]

Thus, \(v_n - v_{n-1} = 10t_n\).

The sum of the series up to \(n\) terms can be found by telescoping the series:

\[ S_n = \sum_{k=1}^{n} t_k = \frac{1}{10} \sum_{k=1}^{n} (v_k - v_{k-1}) \]

Since the series telescopes, all terms cancel except the first and last, resulting in:

\[ S_n = \frac{1}{10} (v_n - v_0) \]

\[ S_n = \frac{1}{10} \left((2n-1)(2n+1)(2n+3)(2n+5)(2n+7) + 1\cdot1\cdot3\cdot5\right) \]

This formula provides the sum of the series for \(n\) terms using the \(v_n\) method.

vn method for reciprocal series

Example

Problem: Determine the sum of the series

\[ S_n = \frac{1}{1\cdot2\cdot3} + \frac{1}{2\cdot3\cdot4} + \frac{1}{3\cdot4\cdot5} + \ldots + \frac{1}{n(n+1)(n+2)} \]

up to \( n \) terms.

Solution:

Let's denote the \( n \)-th term of the series by \( t_n \), which is given by:

\[ t_n = \frac{1}{n(n+1)(n+2)} \]

To apply a version of the \( v_n \) method for telescoping series, we define a new sequence \( v_n \) as:

\[ v_n = \frac{1}{(n+1)(n+2)} \]

Note that this is \( t_n \) without its first term in the denominator.

Next, we compute \( v_n - v_{n-1} \):

\[ v_n - v_{n-1} = \frac{1}{(n+1)(n+2)} - \frac{1}{n(n+1)} \]

Taking the least common multiple (LCM) to combine the fractions, we get:

\[ v_n - v_{n-1} = \frac{n - (n+2)}{n(n+1)(n+2)} \]

\[ v_n - v_{n-1} = \frac{-2}{n(n+1)(n+2)} \]

Notice that this expression is negative two times the \( n \)-th term of the original series \( S_n \), hence:

\[ t_n = -\frac{1}{2}(v_n - v_{n-1}) \]

Now let's consider the sum of the first \( n \) terms of the series using the telescoping nature:

\[ S_n = \sum_{k=1}^{n} t_k = -\frac{1}{2} \sum_{k=1}^{n} (v_k - v_{k-1}) \]

As the series telescopes, we see that for consecutive terms, the \( v_n \) and \( v_{n-1} \) terms will cancel each other:

\[ S_n = -\frac{1}{2} \left( v_n - v_0 \right) \]

The series starts at \( k=1 \), so \( v_0 \) corresponds to \( k=0 \) in our sequence:

\[ v_0 = \frac{1}{(0+1)(0+2)} = \frac{1}{2} \]

Therefore, the sum is:

\[ S_n = -\frac{1}{2} \left( \frac{1}{(n+1)(n+2)} - \frac{1}{2} \right) \]

\[ S_n = -\frac{1}{2} \left( \frac{1}{(n+1)(n+2)} - \frac{n(n+1)(n+2)}{2n(n+1)(n+2)} \right) \]

\[ S_n = -\frac{1}{2} \left( \frac{2 - (n+1)(n+2)}{2(n+1)(n+2)} \right) \]

\[ S_n = \frac{(n+1)(n+2) - 2}{4(n+1)(n+2)} \]

\[ S_n = \frac{n^2 + 3n}{4(n+1)(n+2)} \]

The sum of the series \( \frac{1}{1\cdot2\cdot3} + \frac{1}{2\cdot3\cdot4} + \frac{1}{3\cdot4\cdot5} + \ldots + \frac{1}{n(n+1)(n+2)} \) up to \( n \) terms is \( \frac{n^2 + 3n}{4(n+1)(n+2)} \).

Example

Problem: Evaluate the sum of the series

\[ S = \frac{1}{1\cdot3\cdot5\cdot7} + \frac{1}{3\cdot5\cdot7\cdot9} + \frac{1}{5\cdot7\cdot9\cdot11} + \ldots \]

up to infinity.

Solution:

To find the sum up to \( n \) terms, first identify the \( n \)-th term of the series (\( t_n \)):

\[ t_n = \frac{1}{(2n-1)(2n+1)(2n+3)(2n+5)} \]

Following the \( v_n \) method, define \( v_n \) as:

\[ v_n = \frac{1}{(2n+1)(2n+3)(2n+5)} \]

Now compute the difference \( v_n - v_{n-1} \):

\[ v_n - v_{n-1} = \frac{1}{(2n+1)(2n+3)(2n+5)} - \frac{1}{(2n-1)(2n+1)(2n+3)} \]

Taking the least common denominator (LCD) to combine the fractions:

\[ v_n - v_{n-1} = \frac{(2n-1) - (2n+5)}{(2n-1)(2n+1)(2n+3)(2n+5)} \]

Simplify the numerator:

\[ v_n - v_{n-1} = \frac{-6}{(2n-1)(2n+1)(2n+3)(2n+5)} \]

Now we have that \( v_n - v_{n-1} \) is negative six times the \( n \)-th term of the original series (\( t_n \)). Hence, we can express \( t_n \) as:

\[ t_n = -\frac{1}{6}(v_n - v_{n-1}) \]

The sum of the first \( n \) terms of the series can be written as:

\[ S_n = \sum_{k=1}^{n} t_k = -\frac{1}{6} \sum_{k=1}^{n} (v_k - v_{k-1}) \]

The series is telescopic, so most terms will cancel out when we sum from \( k=1 \) to \( n \), leaving us with:

\[ S_n = -\frac{1}{6} (v_n - v_0) \]

To find the sum as \( n \) approaches infinity, we first observe that:

\[ \lim_{n \to \infty} v_n = \lim_{n \to \infty} \frac{1}{(2n+1)(2n+3)(2n+5)} = 0 \]

Since \( v_0 \) corresponds to \( k=0 \) in our sequence, we find that:

\[ v_0 = \frac{1}{1\cdot3\cdot5} \]

Thus, the sum to infinity \( S \) is:

\[ S = \lim_{n \to \infty} S_n = -\frac{1}{6} \left(0 - \frac{1}{1\cdot3\cdot5}\right) \]

\[ S = -\frac{1}{6} \left(-\frac{1}{15}\right) \]

\[ S = \frac{1}{90} \]

Therefore, the sum of the series

\[ \frac{1}{1\cdot3\cdot5\cdot7} + \frac{1}{3\cdot5\cdot7\cdot9} + \frac{1}{5\cdot7\cdot9\cdot11} + \ldots \]

up to infinity is \( \frac{1}{90} \).

More illustrations on telescopic series

Example

Problem:

We aim to find the sum up to \( n \) terms of the series:

\[ S_n = \frac{1}{1 \cdot 3} + \frac{2}{1 \cdot 3 \cdot 5} + \frac{3}{1 \cdot 3 \cdot 5 \cdot 7} + \ldots \]

Solution:

Let's denote the \( n \)-th term of the series by \( t_n \):

\[ t_n = \frac{n}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} \]

Now, multiply both the numerator and the denominator of \( t_n \) by 2:

\[ t_n = \frac{2n}{2 \cdot 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} \]

We then add and subtract 1 to the numerator:

\[ t_n = \frac{(2n+1) - 1}{2 \cdot 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} \]

This allows us to split \( t_n \) into two fractions:

\[ t_n = \frac{1}{2} \left( \frac{2n+1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} - \frac{1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} \right) \]

The term \( 2n+1 \) in the first fraction cancels with the last term of the denominator:

\[ t_n = \frac{1}{2} \left( \frac{1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)} - \frac{1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} \right) \]

This is in the form of \( \frac{1}{2} \left( a_{n-1} - a_n \right) \), where:

\[ a_n = \frac{1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} \]

By summing the series from \( n = 1 \) to \( n \), we notice the telescoping nature:

\[ S_n = \frac{1}{2} \left( a_0-a_1+a_1 - a_2 + a_2 - a_3 + a_3 - a_4 + \ldots + a_{n-1} - a_n \right) \]

Everything cancels except the first term of the first fraction \( a_0\) and the last term of the last fraction \( -a_n \):

\[ S_n = \frac{1}{2} \left( a_0 - a_n \right) \]

Since \( a_0 = 1 \) (when \( n = 0 \)) and \( a_n = \frac{1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} \), we can write:

\[ S_n = \frac{1}{2} \left( 1 - \frac{1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)} \right) \]

This expression provides us with the sum of the first \( n \) terms of the given series.