The question I've been posed is:

Show that the following series converges, and compute its value

$$\sum_{k=1}^\infty \frac{1}{k(k+2)}$$

From this I decided to use partial fractions to put into the form:


And from this I noticed that this is in the form of a telescoping series which I think would cancel down to:

$$\frac12\cdot\left(1+\frac12\right)= \frac{3}{4}$$

So I've got to this point, but I don't think what I've worked out is substantial enough to prove what I've been asked.

Would anyone mind giving any tips to make my working more thorough.

  • $\begingroup$ Concerning the convergence alone you can use the fact that $$\frac1{k(k+2)}<\frac1{k^2}$$ $\endgroup$ – mrtaurho Feb 27 '19 at 21:30

Note that$$\frac1k-\frac1{k+2}=\left(\frac1k-\frac1{k+1}\right)+\left(\frac1{k+1}-\frac1{k+2}\right).$$This will give you two telescoping series. Can you take it form here?

  • $\begingroup$ Sorry but I don't quite understand where I should go from there, would you mind giving another tip? $\endgroup$ – king Feb 28 '19 at 17:05
  • $\begingroup$ \begin{align}\sum_{n=1}^\infty\frac1k-\frac1{k+2}&=\sum_{k=0}^\infty\frac1k-\frac1{k+1}+\sum_{n=1}^\infty\frac1{k+1}-\frac1{k+2}\\&=1-\lim_{n\to\infty}\frac1{k+1}+\frac12-\lim_{k\to\infty}\frac1{k+2}\\&=\frac32.\end{align} $\endgroup$ – José Carlos Santos Feb 28 '19 at 17:33

Let's write, as you have done, the following :

$$S_n =\sum_{k=1}^n \frac{1}{k(k+2)} = \frac{1}{2} \sum_{k=1}^n \left(\frac{1}{k} - \frac{1}{k+2} \right) = \frac{1}{2} \left(\sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n\frac{1}{k+2} \right) = \frac{1}{2} \left(\sum_{k=1}^n \frac{1}{k} - \sum_{k=3}^{n+2}\frac{1}{k} \right) $$

So you see that $$S_n = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right)$$

Now this is obvious that the limit of $S_n$ is equal to $\frac{3}{4}$, i.e.

$$\sum_{k=1}^{+\infty} \frac{1}{k(k+2)} = \frac{3}{4}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.