Prove convergence by considering the partial sums

Question

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.

Answer 1

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

$$\sum_{n=1}^{2p}\frac1{k(k+p)}=\frac1p\left(1-\color{red}{\frac1{p+1}}+\ldots+\frac1p-\color{green}{\frac1{2p}}+\color{red}{\frac1{p+1}}-\frac1{2p+1}+\ldots+\color{green}{\frac1{2p}}-\frac1{3p}\right)$$

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

$$\frac1p\left(1+\frac12+\ldots+\frac1p\right)$$

Answer 2

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.