1
$\begingroup$

I was trying to generalize an integral I found yesterday on this website and ran into the following interesting sum:

$S_k=\sum_{n=1}^\infty \frac{1}{n+k}-\frac{1}{n}$. I have seen this sum come up a lot in my calculus classes but today I realized it converges for all $k \in \mathbb{R}_{>0}$. I was only able to prove that for $k \in \mathbb{N}$ that $S_k = -H_k$ where $H_k$ is the $k^{th}$ harmonic number. My proof is by induction on k and is as follows:

The base case is $S_1=\sum_{n=1}^\infty \frac{1}{n+1}-\frac{1}{n}$ which if one writes out the terms clearly telescope to -1 while $-H_1=-1$. Now I assume $S_k = -H_k$ and note that $S_{k+1}=\sum_{n=1}^\infty \frac{1}{n+k+1}-\frac{1}{n}=\sum_{n=1}^\infty \frac{1}{n+k+1}-\sum_{n=1}^\infty \frac{1}{n}$. Reindexing the first sum to $n=2$ and throwing out the first term in the harmonic sum gives, $-1+\sum_{n=2}^\infty \frac{1}{n+k}-\sum_{n=2}^\infty \frac{1}{n}=-1+\sum_{n=2}^\infty \frac{1}{n+k}-\frac{1}{n}$. Now the remaining the sum is just $S_k$ but without the first term so we may rewrite this as $-1+(S_k-(\frac{1}{k+1}-1))$ simplifying and using the inductive assumption that $S_k = -H_k$ we have that $S_{k+1}=-H_k-\frac{1}{k+1}=-H_{k+1}$ so the result holds by induction.

So now my question is does anyone have a result for $k \in \mathbb{Q}$ or $k \in \mathbb{I}$? It looks as though Mathematica is cabable of finding a closed form for any $k \in \mathbb{R}_{>0}$.

$\endgroup$
2
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Marco Cantarini
    Commented Jun 18, 2016 at 17:24
  • $\begingroup$ Wow! I really need to start looking into special functions before I post questions! So this is really just a special case of the digamma function is what I'm gathering? $\endgroup$
    – mike van der naald
    Commented Jun 18, 2016 at 17:27

1 Answer 1

Reset to default
1
$\begingroup$

You cannot split the sums as you want in your inductive step, because the harmonic series diverges.

You can do it like this, involving two convergent sums: \begin{align} S_{k+1}&=\sum_{n=1}^\infty \frac{1}{n+k+1}-\frac{1}{n}=\sum_{n=1}^\infty \frac{1}{n+k+1}-\frac1{n+k}+\sum_{n=1}^\infty \frac1{n+k}-\frac{1}{n}\\ \ \\ &=-\frac1{n+1}+\sum_{n=1}^\infty \frac1{n+k}-\frac{1}{n}\\ \ \\ &=-\frac1{k+1}-H_k=-H_{k+1}. \end{align}

$\endgroup$
3
  • $\begingroup$ Wow that is slick. So by adding a clever 0 you can avoid ever having a divergent sum in your calculations. $\endgroup$
    – mike van der naald
    Commented Jun 18, 2016 at 17:34
  • 1
    $\begingroup$ It is worth noting that the proper proof for telescopic sums involves looking at finite sums and then taking the limit (e.g., $$\sum_{n=1}^Na_{n+1}-a_n=\sum_{n=1}^Na_{n+1}-\sum_{n=1}^Na_n=-a_1+a_{N+1}$$ and now take the limit as $N\to\infty$). The generalization is then $$\sum_{n=1}^Na_{n+k}-a_n=-\left(\sum_{n=1}^ka_n\right)+\sum_{n=N+1}^{N+k}a_{n}.$$ In particular, although TC's writing is not spot on, it is the usual approach. $\endgroup$
    – Clayton
    Commented Jun 18, 2016 at 17:36
  • $\begingroup$ @Clayton Thank you for your comment. I have not had much analysis but I remember this sort of technique for proving that the geometric sum converges to $\frac{1}{1-x}$ for $-1<x<1$. $\endgroup$
    – mike van der naald
    Commented Jun 18, 2016 at 18:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .