Let $p$ be a non-zero natural number. Prove by considering the partial sums that

$\sum \frac{1}{k(k+p)}$

converges. What is $\sum\limits_{k=1}^{\infty} \frac{1}{k(k+p)}$

No idea. Obviously, it looks like a telescoping series. Sure doesn't act like one. I have tried to treat it like I would a telescoping series to see if it would get me any where--it did not.

Alongside advice on how to do this one. If people could also offer general advice on taking partial sums that would be greatly appreciated.

  • 1
    $\begingroup$ it looks like a telescoping series - That's because it is. $\endgroup$ – Lucian Nov 29 '14 at 21:56

The series converges by the comparison test, for example. Now:$$\frac1{k(k+p)}=\frac1p\left(\frac1k-\frac1{k+p}\right)\implies$$


Let us write the summands in columns, the plus sign column and the minus sign column:

$$\begin{align*}&\frac11&-\frac1{p+1}\\ &\frac12&-\frac1{2+p}\\ &\frac13&-\frac1{3+p}\\&\ldots&\ldots\\ &\frac1p&-\frac1{2p}\\ &\frac1{p+1}&-\frac1{2p+1}\\ &\ldots&\ldots\\ &\frac1{2p}&-\frac1{3p}\end{align*}$$

We can see the first $\;p\;$ plus summands remain, whereas the first minus summands cancel with the last $\;p\;$ plus summands, and this is so no matter what multiple $\;mp\;$ we take, so passing to the limit when $\;m\to\infty\;$, we get the sum


  • 1
    $\begingroup$ Sorry, but does this tell me the value to which this converges? Because that is the main reason that I am struggling. I know how to show convergence. I need to show convergence using partial sums and then state the value of the sum. $\endgroup$ – Bob the Builds Nov 29 '14 at 22:52
  • $\begingroup$ Didn't you see the very last figure? That's the sum! $\endgroup$ – Timbuc Nov 30 '14 at 0:22
  • $\begingroup$ Yeah. Was having trouble at first understanding how you arrived there, but I got it, and gave this answer the checkmark $\endgroup$ – Bob the Builds Nov 30 '14 at 1:40

Hint: Write $$\frac1{k(k+p)}=\frac{A}{k}+\frac{B}{k+p}.$$ Solve for $A$ and $B,$ and you'll find that the series does telescope.

Added: Now that I'm back at my computer and have a little time, and now that you've accepted an answer, I'll expand on my own. You should readily find that $A=\frac1p$ and $B=-\frac1p.$ Consequently, we can rewrite the series as $$\frac1p\sum_{k=1}^\infty\left(\frac1k-\frac1{k+p}\right).$$ To prove convergence and find the limit, it will suffice to consider the partial sums $$S_n:=\frac1p\sum_{k=1}^n\left(\frac1k-\frac1{k+p}\right).$$ We'd like to get this into a more convenient form, as a difference of sums, rather than a sum of differences. That is, we'll rewrite it as $$S_n=\frac1p\sum_{k=1}^n\frac1k-\frac1p\sum_{k=1}^n\frac1{k+p},\tag{1}$$ which identity is readily proved by arithmetic properties.

Now, take any integer $m\ge1$ and note that $$\sum_{k=1}^{p+m}\frac1k=\sum_{k=1}^p\frac1k+\sum_{k=p+1}^{p+m}\frac1k.\tag{2}$$ On the other hand, $$\sum_{k=1}^{p+m}\frac1{k+p}=\sum_{k=1}^{p+m}\frac1{p+k}=\sum_{k=1}^m\frac1{p+k}+\sum_{k=m+1}^{p+m}\frac1{p+k}=\sum_{k=p+1}^{p+m}\frac1k+\sum_{k=m+1}^{p+m}\frac1{p+k}.\tag{3}$$

Consequently, by $(1)$ through $(3),$ we have for any integer $m\ge1$ that $$S_{p+m}=\frac1p\sum_{k=1}^p\frac1k-\frac1p\sum_{k=m+1}^{p+m}\frac1{p+k}=\frac1p\sum_{k=1}^p\frac1k-\frac1p\sum_{k=1}^p\frac1{p+m+k}.\tag{$\star$}$$ Now, for any such integer $m,$ we have $$0<\sum_{k=1}^p\frac1{p+m+k}<\sum_{k=1}^p\frac1{p+m+1}=\frac{p}{p+m+1},$$ so $$-\frac1{p+m+1}<-\frac1p\sum_{k=1}^p\frac1{p+m+1}<0,$$ and so by $(\star),$ we have $$-\frac1{p+m+1}<S_{p+m}-\frac1p\sum_{k=1}^p\frac1k<0$$ for all integers $m\ge1.$ A quick application of the Squeeze Theorem shows that the sequence of partial sums converges to $\frac1p\sum\limits_{k=1}^p\frac1k,$ proving series convergence and giving us its sum.

  • $\begingroup$ Not sure why this was downvoted. Nice hint, +1 $\endgroup$ – Zubin Mukerjee Nov 29 '14 at 22:15
  • $\begingroup$ I'm sorry, but you are going to have to give me a little more than that. When I say that I tried to treat it like a telescoping series, I mean to say that I tried to place the summation in the form of a telescoping series. $\endgroup$ – Bob the Builds Nov 29 '14 at 22:50
  • 1
    $\begingroup$ @Bob: The thing to notice here is that the telescoping begins on the $p$th term--that is, when $k=p.$ Moreover, $\frac1{k+p}$ takes on every value that $\frac1k$ does as $k$ ranges over the positive integers, except for $1,\frac12,...,\frac1p.$ Hence, noting the values of $A$ and $B,$ the series's sum is simply a multiple of $\sum\limits_{k=1}^p\frac1k.$ $\endgroup$ – Cameron Buie Nov 29 '14 at 23:03

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.