Telescoping

Today we will learn about a new type of series that can be evaluated using the a method called $telescoping$

Let us begin with an example problem.

Find the sum of the series:

$\dfrac{1}{1\times2} + \dfrac{1}{2 \times 3} + \dfrac{1}{3 \times 4} + ... + \dfrac{1}{99 \times 100}$

The aim of telescoping is to split up each term into two parts such that one part cancels out with a corresponding part in the next term.

We can observe that for each term of the given series, the denominator is a product of two number whose difference is $1$. So, we can express the $1$s in each numerator as the difference of the two numbers in the denominator.

Our series becomes:

$\dfrac{2 - 1}{1\times2} + \dfrac{3 - 2}{2 \times 3} + \dfrac{4 - 3}{3 \times 4} + ... + \dfrac{100 - 99}{99 \times 100}$

$= \dfrac{2 - 1}{1\times2} + \dfrac{3 - 2}{2 \times 3} + \dfrac{4 - 3}{3 \times 4} + ... + \dfrac{100 - 99}{99 \times 100}$

$= \left(\dfrac{1}{1} - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{3} - \dfrac{1}{4}\right) + ... + \left(\dfrac{1}{99} - \dfrac{1}{100}\right)$

Now we have achieved our goal. The $-\dfrac{1}{2}$ of the first term cancels out with the $\dfrac{1}{2}$

Likewise the second part of every term will cancel out with the first part of the next term.

-----------book page break-----------

Thus we will be left with only the first part of the first term, which is $1$ and the second part of the last term, which is $-\dfrac{1}{100}$

Therefore our result will be $1 - \dfrac{1}{100} = \dfrac{99}{100}$

We will look at another similar problem.

Find the sum of the series:

$\dfrac{1}{1.2.3} + \dfrac{1}{2.3.4} + \dfrac{1}{3.4.5} + ... + \dfrac{1}{48.49.50}$

Let us observe that the $k\xasuper{th}$ term $\dfrac{1}{k(k+1)(k+2)}$ can be split as:

$\dfrac{1}{2}\left\{\dfrac{1}{k(k+1)} - \dfrac{1}{(k+1)(k+2)}\right\}$

Therefore, the second part of the $k\xasuper{th}$ term will cancel out with the first term of the $(k+1)\xasuper{th}$ term.

We can rewrite our series as:

$\dfrac{1}{2}\left(\dfrac{1}{1.2} - \dfrac{1}{2.3}\right) + \dfrac{1}{2}\left(\dfrac{1}{2.3} - \dfrac{1}{3.4}\right) + ... + \dfrac{1}{2}\left(\dfrac{1}{48.49} - \dfrac{1}{49.50}\right)$

$= \dfrac{1}{2}\left(\dfrac{1}{1.2} - \dfrac{1}{2.3} + \dfrac{1}{2.3} - \dfrac{1}{3.4} + ... + \dfrac{1}{48.49} - \dfrac{1}{49.50}\right)$

$= \dfrac{1}{2}\left(\dfrac{1}{1.2} - \dfrac{1}{49.50}\right)$

-----------book page break-----------

$= \dfrac{1}{2}\left(\dfrac{49.25 - 1}{49.50}\right)$

$= \dfrac{1}{2}\left(\dfrac{1224}{49.50}\right)$

$= \dfrac{308}{49.25}$

$= \dfrac{308}{1225} = \dfrac{44}{175}$

Now we will see a more generic for of this type of series. In this case each term of the denominator contains $t$ consecutive terms, and looks like:

$S_n = \dfrac{1}{1.2.3...\ t\ terms} + \dfrac{1}{2.3.4...\ t\ terms} + ... + \dfrac{1}{n(n+1)(n+2)...\ t\ terms}$

The $k\xasuper{th}$ term would be:

$\dfrac{1}{k(k+1)(k+2)...(k+t-1)}$

This term can be split as follows:

$\dfrac{1}{k-1} \times \left\{\dfrac{1}{k(k+1)(k+2)...(k+t-2)} - \dfrac{1}{(k+1)(k+2)...(k+t-1)}\right\}$

-----------book page break-----------

Therefore, the sum, $S_n$, would be:

$\dfrac{1}{k-1} \times \left\{\dfrac{1}{1.2.3...\ (t-1)(t)} - \dfrac{1}{(n+1)(n+2)...\ (n + t -2)} \right\}$

We can take a look at another variant of this series, where the multipliers of the denominator are not consecutive integers, but are integers in $A.P$ series. For example:

$S_n = \dfrac{1}{1.3.5} + \dfrac{1}{3.5.7} + \dfrac{1}{5.7.9} + ... + \dfrac{1}{99.101.103}$

First we will generalise this with the multipliers denominator being the terms of any $A.P.$ series, starting with $a$ and has a common difference $d$ and each denominator contain $t$ consecutive terms of this series, and the overall series contains $n$ terms, as follows:

$S_n = \dfrac{1}{a(a+d)(a+2d)..\{a+(t-1)d\}} + \dfrac{1}{(a+d)(a+2d)..\{a+(t)d\}} + $ $ ... + \dfrac{1}{\{a+(n-1)d\}\{a+(n)d\}...\{a+(n + t -2)d\}}$

The $k\xasuper{th}$ term of this series will be:

$\dfrac{1}{\{a+(k-1)d\}\{a+(k)d\}...\{a+(k + t -2)d\}}$

This term can be split as:

$\dfrac{1}{2d} \left\{\dfrac{1}{\{a+(k-1)d\}\{a+(k)d\}...\{a+(k + t - 3)d\}} \right.$ $\left. \ \ \ - \dfrac{1}{\{a+(k)d\}\{a+(k+1)d\}...\{a+(k+t-2)d\}}\right\}$

-----------book page break-----------

After cancellation of the telescoped terms, we will be left with:

$S_n = \dfrac{1}{2d} \left\{\dfrac{1}{a(a+d)(a+2d)..\{a+(t-2)d\}} \right.$ $\left. \ \ \ - \dfrac{1}{\{a+(n)d\}\{a+(n+1)d\}...\{a+(n + t -2)d\}}\right\}$

For our example problem above, $a = 1$, $d = 2$, $t = 3$, and the number of terms $n = \dfrac{99 - 1}{2} + 1 = 50$

Directly substituting these values in the our previous formula we get:

$S_{50} = \dfrac{1}{4}\left\{\dfrac{1}{1\times 3} - \dfrac{1}{101\times 103}\right\}$

$S_{50} = \dfrac{101\times 103 - 3}{4 \times 3 \times 101 \times 103} = \dfrac{2600}{31209}$