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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:

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

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.

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  • $\begingroup$ 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
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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?

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  • $\begingroup$ 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
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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.

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

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