# Two harmonic subseries

Let $S_0=1$, and let $S_{n+1}$ be the sum of $1/k+1/(k+1)+1/(k+2)\cdots 1/(k+n)$, for the integer $k$ such that $S_{n+1}$ is maximal while $S_{n+1}<S_n$.

Does $\sum_{k=1}^{\infty}S_k(-1)^{k}$ converge?

Minor addition:
What is $\lim_{n\to\infty} S_n ?$
And is it possible to obtain a formula for $k$ as a function of $n$?

And the series $1-(1/2+1/3)+(1/4+1/5+1/6)-(1/7+1/8+1/9+1/10)+\cdots$ converges by the alternating series test, does it have a closed form for its sum?

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For the first series, the alternating series test will work, if only we can show $S_k \to 0$ as $k$ grows. (Otherwise, of course, the series diverges anyway.) – Srivatsan Nov 2 '11 at 6:48

## 1 Answer

The first series does not converge, because $S_n$ does not tend to zero. I don't have a full proof of this, but I do have an outline.

Note that the $k$ in the definition of $S_{n+1}$ is between $(n+1)/S_n-n$ and $(n+1)/S_n$. Note also that increasing $k$ by one decreases the sum $1/k + 1/(k+1) + \cdots + 1/(k+n)$ by $1/k-1/(k+1)+1/(k+n)-1/(k+n+1)$, which is something on the order of $1/n^2$.

Therefore $k$ can be tweaked until the sum is within something like $1/n^2$ of $S_n$; it follows that $S_n-S_{n+1}$ is bounded above by something like $1/n^2$.

When that "something on the order of $1/n^2$" is made explicit, one should be able to compute the first several values of $S_n$ and then sum the convergent telescoping series of differences $S_{n+1}-S_n$ to prove that $\lim_{n\to\infty} S_n > 0$. Calculated data strongly suggests that $\lim_{n\to\infty} S_n$ equals a number slightly larger than $0.405$.

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Increasing $k$ decreases the sum by $1/k-1/(k+n+1)$, which isn't of order $1/n^2$. – joriki Nov 2 '11 at 7:59
I'm finding convergence of $S_n$ to $\log2$. Your value seems to be $\log (3/2)$. – joriki Nov 2 '11 at 8:34
@joriki He started with S_1=1/2, with s_1=1/3 i get log(4/3) – user1708 Nov 2 '11 at 13:52
@joriki: yeah, you're right about the decrease formula. Hm. Empirically the decrease is like $1/n^2$; how could we prove that? – Greg Martin Nov 2 '11 at 18:06