Skip to page content Home     Posts     CV     Contact     People

Telescoping Sums

2017 June 9

Find the values of each of the following infinite sums: $$\sum_{x=1}^\infty \frac{1}{x(x+1)}$$ $$\sum_{x=2}^\infty \frac{1}{x^2-1}$$ $$\sum_{x=1}^\infty \frac{1}{x(x+1)(x+2)}$$ $$\sum_{x=1}^\infty \frac{x}{2^x}$$

Okay, let's start with the first problem: $$\sum_{x=1}^\infty \frac{1}{x(x+1)}$$ This is a simple telescoping sum problem. Note that $$\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$$ So, when expanded, the infinite sum looks like $$\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...$$ Look! Each term other than the very first one forms a zero pair with some other term. They all cancel out, and we are left with $$\sum_{x=1}^\infty \frac{1}{x(x+1)}=1$$ as our answer. But wait a minute... couldn't we do an infinite sum like $$\sum_{x=1}^\infty \frac{1}{x(x+2)}$$ the same way? Since $$\frac{1}{x(x+2)}=\frac{1}{2}\bigg(\frac{1}{x}-\frac{1}{x+2}\bigg)$$ then the entire sum is $$\frac{1}{2}\bigg(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...\bigg)$$ Now the only terms that don't get cancelled are $1$ and $\frac{1}{2}$, so the sum is $$\sum_{x=1}^\infty \frac{1}{x(x+2)}=\frac{1}{2}\bigg(1+\frac{1}{2}\bigg)=\frac{3}{4}$$ In fact, we can generalize this method of solving sums of this sort. If we have a sum of the form $$\sum_{x=1}^\infty \frac{1}{x(x+k)}$$ We can use partial fractions to set $$\frac{1}{x(x+k)}=\frac{1}{k}\bigg(\frac{1}{x}-\frac{1}{x+k}\bigg)$$ And so, by expanding, we can see that the sum becomes $$\frac{1}{k}\bigg(1-\frac{1}{k+1}+\frac{1}{2}-\frac{1}{k+2}+...+\frac{1}{k}+\frac{1}{2k}+\frac{1}{k+1}+\frac{1}{2k+1}+...\bigg)$$ Which can be rearranged to form $$\frac{1}{k}\bigg(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k}+\frac{1}{k+1}-\frac{1}{k+1}+\frac{1}{k+2}-\frac{1}{k+2}+...\bigg)$$ All of the terms cancel except for those from $1$ to $k$, so we get $$\frac{1}{k}\bigg(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k}\bigg)$$ and we have the formula $$\sum_{x=1}^\infty \frac{1}{x(x+k)}=\frac{1}{k}\sum_{x=1}^k \frac{1}{x}$$ or $$\sum_{x=1}^\infty \frac{1}{x(x+k)}=\frac{1}{k}H_k$$ where $H_n$ denotes the nth harmonic number. Okay! Onto the next sum: $$\sum_{x=2}^\infty \frac{1}{x^2-1}$$ We can split up the fraction into $$\frac{1}{(x+1)(x-1)}$$ $$\frac{1}{2}\bigg(\frac{1}{x-1}-\frac{1}{x+1}\bigg)$$ And so, when we expand the sum, we get $$\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...\bigg)$$ $$\frac{1}{2}\bigg(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...\bigg)$$ $$\frac{1}{2}\bigg(1+\frac{1}{2}\bigg)$$ $$\frac{3}{4}$$ and so we have our answer: $$\sum_{x=2}^\infty \frac{1}{x^2-1}=\frac{3}{4}$$

Are you noticing a pattern in these summation problems?

Suppose we have any summation in the form $$\sum_{x=1}^\infty f(x)-f(x+k)$$ or $$\lim_{n\to \infty}\sum_{x=1}^n f(x)-f(x+k)$$ we can start by finding a formula for $$\sum_{x=1}^n f(x)-f(x+k)$$ which can be found by expanding the sum and telescoping it: $$f(1)-f(k+1)+f(2)-f(k+2)+...+f(n)-f(k+n)$$ If we assume that $n \gt k$, then we have $$f(1)-f(k+1)+f(2)-f(k+2)+...+f(k)-f(2k)+f(k+1)-f(2k+1)+...+f(n)-f(k+n)$$ Which collapses to $$f(1)+f(2)+...+f(k)-f(n+1)-f(n+2)-...-f(n+k)$$ and this is equal to $$\sum_{x=1}^k f(x)-\sum_{x=n+1}^{n+k} f(x)$$ And our original infinite sum is $$\lim_{n\to \infty}\Bigg(\sum_{x=1}^k f(x)-\sum_{x=n+1}^{n+k} f(x)\Bigg)$$ If $\lim_{x\to\infty}f(x)$ exists, then we can say that this is equal to $$\sum_{x=1}^k f(x)-\lim_{x\to\infty}kf(x)$$ This is the formula for a telescoping sum: $$\sum_{x=1}^\infty f(x)-f(x+k)=\sum_{x=1}^k f(x)-\lim_{x\to\infty}kf(x)$$ Given that $k$ is a natural number.

Moving on to the next problem: $$\sum_{x=1}^\infty \frac{1}{x(x+1)(x+2)}$$ Using partial fractions, we can split up the fraction to get $$\frac{1}{x(x+1)(x+2)}=\frac{1}{2}\frac{1}{x}-\frac{1}{x+1}+\frac{1}{2}\frac{1}{x+2}$$ $$\frac{1}{x(x+1)(x+2)}=\frac{1}{2}\bigg(\frac{1}{x}-\frac{1}{x+1}\bigg)-\frac{1}{2}\bigg(\frac{1}{x+1}-\frac{1}{x+2}\bigg)$$ and so our sum is equal to $$\frac{1}{2}\sum_{x=1}^\infty \bigg(\frac{1}{x}-\frac{1}{x+1}\bigg)-\frac{1}{2}\sum_{x=1}^\infty \bigg(\frac{1}{x+1}-\frac{1}{x+2}\bigg)$$ $$\frac{1}{2}\bigg(1-\frac{1}{2}\bigg)$$ $$\frac{1}{4}$$ And so the answer to this problem is $$\sum_{x=1}^\infty \frac{1}{x(x+1)(x+2)}=\frac{1}{4}$$

Next up is a different type of sum: $$\sum_{x=1}^\infty \frac{x}{2^x}$$ Yay! Something that can't be done using partial fractions! This one telescopes in a different way. Look what happens if we multiply and divide the sum by $\frac{1}{2}$: $$\frac{\frac{1}{2}}{\frac{1}{2}}\sum_{x=1}^\infty \frac{x}{2^x}$$ $$\frac{(1-\frac{1}{2})}{\frac{1}{2}}\sum_{x=1}^\infty \frac{x}{2^x}$$ $$\frac{1}{\frac{1}{2}}\sum_{x=1}^\infty (1-\frac{1}{2})\frac{x}{2^x}$$ $$2\sum_{x=1}^\infty \bigg(\frac{x}{2^x}-\frac{x}{2^{x+1}}\bigg)$$ Now look what happens when we write this out: $$2\bigg(\frac{1}{2}-\frac{1}{4}+\frac{2}{4}-\frac{2}{8}+\frac{3}{8}-\frac{3}{16}+...\bigg)$$ $$2\bigg(\frac{1}{2}+\frac{-1+2}{4}+\frac{-2+3}{8}+...\bigg)$$ $$2\bigg(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\bigg)$$ Now we have a geometric series inside of the parentheses. That geometric series can be easily calculated to converge to $1$ using the formula, and we have $$2(1)$$ $$2$$ and so $$\sum_{x=1}^\infty \frac{x}{2^x}=2$$

Great. All of these sums were telescoping sums, but in a future post I hope to tackle a few infinite sums that cannot be solved in this way.