# Sum of Telescoping Series

Another fascinating series is telescoping series. This series is quite different from a regular progression. If you came across to a problem that ask for the sum and you can’t find it by using arithmetic or geometric series. It could be the solve using telescoping technique. This series requires our knowledge to express fractional expression to its partial fraction form. I won’t be surprise if this is your first time to hear such series because the application of this topic is usually for Olympiad Mathematicians. Ingenious way to express telescopic sum $\displaystyle\frac{1}{n(n+1)}=\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}$ $\displaystyle\frac{k}{n(n+k)}=\displaystyle\frac{1}{k}[\frac{1}{n}-\frac{1}{n+k}]$ $\displaystyle\frac{k}{n(n+1)(n+k)}= \displaystyle\frac{1}{k}[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+k)}]$

Sample Problem 1:

Find the sum of the series $\frac{1}{1(2)}+\frac{1}{2(3)}+\frac{1}{2(3)}+\ldots +\frac{1}{2013(2014)}$

Solution:

We can express the nth term as $\displaystyle\frac{1}{n(n+1)}$

As mentioned above $\displaystyle\frac{1}{n(n+1)}=\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}$

Let n=1 $\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}=\displaystyle\frac{1}{1}-\displaystyle\frac{1}{2}$

Let n=2 $\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}=\displaystyle\frac{1}{2}-\displaystyle\frac{1}{3}$

Let n=3 $\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}=\displaystyle\frac{1}{3}-\displaystyle\frac{1}{4}$

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Let n=2013 $\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}=\displaystyle\frac{1}{2013}-\displaystyle\frac{1}{2014}$

Expressing $\frac{1}{1(2)}+\frac{1}{2(3)}+\frac{1}{2(3)}+\ldots +\frac{1}{2013(2014)}$ to its equivalent partial fractions we have $(\displaystyle\frac{1}{1}-\displaystyle\frac{1}{2})+(\displaystyle\frac{1}{2}-\displaystyle\frac{1}{3})+(\displaystyle\frac{1}{3}-\displaystyle\frac{1}{4})+\ldots +(\displaystyle\frac{1}{2013}-\displaystyle\frac{1}{2014})$

Cancel out terms leaving $1-\displaystyle\frac{1}{2014}=\displaystyle\frac{2013}{2014}$

Therefore we can conclude that $\sum_{n=1}^k \displaystyle\frac{1}{n(n+1)}=\displaystyle\frac{k}{k+1}$

Sample Problem 2:

Find the sum of $\sum_{k=0}^{99} \log(\displaystyle\frac{k}{k+1})$

Solution:

Using the law of logarithm, $\log(\displaystyle\frac{k}{k+1})=\log(k)-\log(k+1)$

Let k=1 $\log 1-\log 2$

Let k=2 $\log 2-\log 3$

Let k=3 $\log 3-\log 4$

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Let k=99 $\log 99-\log 100$ $\sum_{k=0}^{99} \log(\displaystyle\frac{k}{k+1})=\sum_{k=0}^{99} \log(k)-log(k+1)$ $(\log 1-\log 2)+(\log 2-\log 3)+ (\log 3-\log 4)+\ldots +(\log 99-\log 100)$

Cancel out terms from log2 to log99 we have $\log 1-log 100$ but log1=0 $-\log 100=-2$

Therefore we can conclude that the formula for Find the sum of $\sum_{k=0}^{n} \log(\displaystyle\frac{k}{k+1})=-\log n$

Sample Problem 3:

Find the sum of $\displaystyle\frac{1}{1\sqrt{2}+2\sqrt{1}}+\displaystyle\frac{1}{2\sqrt{3}+3\sqrt{2}}+\displaystyle\frac{1}{3\sqrt{4}+4\sqrt{3}}+\ldots +\displaystyle\frac{1}{840\sqrt{841}+841\sqrt{840}}$

Solution:

The nth term of this series is $\displaystyle\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}$

By rationalizing the denominator, $\displaystyle\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}\cdot \displaystyle\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{(n+1)\sqrt{n}-n\sqrt{n+1}}$ $=\displaystyle\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{((n+1)\sqrt{n})^2-(n\sqrt{n+1})^2}$ $=\displaystyle\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)^2-n^2(n+1)}$

Factor the common terms $=\displaystyle\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)[(n+1)-n]}$ $=\displaystyle\frac{n\sqrt{n+1}-(n+1)\sqrt{n}}{n(n+1)(1)}$ $=\displaystyle\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)}$ $=\displaystyle\frac{(n+1)\sqrt{n}}{n(n+1)}-\displaystyle\frac{n\sqrt{n+1}}{n(n+1)}$ $=\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}$

This means that $\displaystyle\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}$

Let n=1 $\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}=1-\displaystyle\frac{\sqrt{2}}{2}$

Let n=2 $\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}=\displaystyle\frac{\sqrt{2}}{2}-\displaystyle\frac{\sqrt{3}}{3}$

Let n=3 $\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}=\displaystyle\frac{\sqrt{3}}{3}-\displaystyle\frac{\sqrt{4}}{4}$

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Let n=840 $\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}=\displaystyle\frac{\sqrt{840}}{840}-\displaystyle\frac{\sqrt{841}}{841}$

We can express $\displaystyle\frac{1}{1\sqrt{2}+2\sqrt{1}}+\displaystyle\frac{1}{2\sqrt{3}+3\sqrt{2}}+\displaystyle\frac{1}{3\sqrt{4}+4\sqrt{3}}+\ldots +\displaystyle\frac{1}{840\sqrt{841}+841\sqrt{840}}$

As $(1-\displaystyle\frac{\sqrt{2}}{2})+(\displaystyle\frac{\sqrt{2}}{2}-\displaystyle\frac{\sqrt{3}}{3})+(\displaystyle\frac{\sqrt{3}}{3}-\displaystyle\frac{\sqrt{4}}{4})+\ldots +(\displaystyle\frac{\sqrt{840}}{840}-\displaystyle\frac{\sqrt{841}}{841})$

Cancel out terms leaving $1-\displaystyle\frac{\sqrt{841}}{841}=\displaystyle\frac{28}{29}$