Contest math problem: $\sum_{n=1}^\infty \frac{\{H_n\}}{n^2}$

Question

$$\sum_{n=1}^\infty \frac{\{H_n\}}{n^2}$$

I have managed to prove that it converges, but am having trouble with a closed form. This came from a school contest from last year, but can't really figure it out.

I came up with a numeric solution, but am having lots of trouble with the closed form. Thanks in advance.

Sorry about that other problem, everybody. I had issues with the composition of the problem itself — , Dec 28 '15 at 02:13
@JimmyK4542 Yes, and $H_n$ denotes the $n$ th harmonic number — , Dec 28 '15 at 02:13
Note that $\sum_{n=1}^\infty \frac{H_n}{n^2}$ already converges, to $2\zeta(3)$. — Andrew Dudzik, Dec 28 '15 at 02:16
@Slade I actually tried using that and the definition of "{n}" as in n-floor(n) but note that $\sum_{n=1}^\infty \frac{1}{n}$ diverges — , Dec 28 '15 at 02:18
@Apple Cuz $\zeta(2)=\pi^2/6$ but $\zeta(3)$ is just $\sum\frac1{n^3}$ which is well known to be unknown. So I always took it as $\zeta(3)$ has no known closed form, while $\zeta(2)$ does. But it's just terminology, I take it you're happy with a solution in terms of $\zeta$. — Gregory Grant, Dec 28 '15 at 02:31
$\zeta(3)$ is known as "Apéry's constant" - see https://en.wikipedia.org/wiki/Apéry%27s_theorem — David Holden, Dec 28 '15 at 03:31
The numerical value for the sum (to an accuracy of $10^{-5}$) is $0.32830$ — Winther, Dec 28 '15 at 03:35
@JohnVine Really? What country? This looks pretty hard for a local contest. I almost wonder if a different problem was intended instead (like the one answered below). — Potato, Dec 28 '15 at 20:28
@Patrick Yes this was local. But it was too hard so the person running the contest disbanded the problem. I was still curious so I asked it here. — , Dec 29 '15 at 00:28

yhhuang · Answer 1 · 2015-12-28T09:59:11.300

Note: this is the answer to calculate $\sum_{n=1}^\infty \frac{H_n}{n^2}$ instead of the one $\sum_{n=1}^\infty \frac{\{H_n\}}{n^2}$ in the question.

First the $n$-the harmonic number can be written as $$ H_n = 1+\frac{1}{2}+\cdots+\frac{1}{n}=\int_{0}^1 \left(1+x+\cdots+x^{n-1}\right)dx =\int_0^1 \frac{1-x^n}{1-x}dx. $$ Then $$ \sum_{n=1}^\infty \frac{H_n}{n^2} =\sum_{n=1}^\infty \int_0^1 \frac{1}{n^2}\frac{1-x^n}{1-x}dx =\int_0^1 \frac{1}{1-x}\left(\sum_{n=1}^\infty\frac{1}{n^2}-\sum_{n=1}^\infty\frac{x^n}{n^2}\right)dx.$$ By the Taylor expansion of $\log(1-t)$, $$ \log(1-t) =-\left(t+\frac{t^2}{2}+\cdots+\frac{t^n}{n}+\cdots\right),$$ we can write $$ \sum_{n=1}^\infty \frac{x^2}{n^2} = -\int_0^x \frac{\log(1-t)}{t}dt $$ which is exactly the dilogarithm function $\mbox{Li}_2(x)$.

Therefore, $$\sum_{n=1}^\infty \frac{H_n}{n^2} =-\int_0^1 \frac{1}{1-x}\int_x^1 \frac{\log(1-t)}{t}dt dx =-\int_0^1 \frac{\log(1-t)}{t}\int_0^t \frac{1}{1-x}dxdt =\int_0^1 \frac{(\log(1-t))^2}{t}dt $$ The final step is to convert the above integral into some integral representation of the Riemann Zeta function. For example, using the change of variable $s=-\log(1-t)$, $$ \sum_{n=1}^\infty \frac{H_n}{n^2}=\int\frac{s^2}{e^s-1}ds=\Gamma(3)\zeta(3)=2\zeta(3). $$

Hi. You seem to be answering a different question. This one takes the fractional part of each $H_n$ before summing. — Potato, Dec 28 '15 at 09:50
@Patrick you are right. I didn't read all the comments carefully following the question. — yhhuang, Dec 28 '15 at 09:54

score 3 · Answer 2 · edited Apr 13 '17 at 12:21

Not an answer to the actual question, but some thoughts, some useful references and a derivation of a good approximation to the sum.

We can rewrite the sum as (see e.g. this answer or the answer above for the $2\zeta(3)$ evaluation) $$\sum_{n=1}^\infty \frac{\{ H_n \}}{n^2}=\sum_{n=1}^\infty \frac{H_n}{n^2}-\sum_{n=1}^\infty \frac{\lfloor H_n \rfloor}{n^2} = 2\zeta(3) - \sum_{n=1}^\infty \frac{\lfloor H_n \rfloor}{n^2}$$ and we can further rewrite $$ \begin{array}{cll}\tag{1} \sum_{n=1}^\infty \frac{\lfloor H_n \rfloor}{n^2} &=& 1\left(\frac{1}{1^2} + \ldots + \frac{1}{3^2}\right) \\&+& 2\left(\frac{1}{4^2} + \ldots + \frac{1}{10^2}\right) \\&+& 3\left(\frac{1}{11^2} + \ldots + \frac{1}{30^2}\right) \\&+& 4\left(\frac{1}{31^2} + \ldots + \frac{1}{82^2}\right) \\&+& \ldots\\ &=& \sum_{k=1}^\infty k\sum_{j=n_k}^{n_{k+1}-1}\frac{1}{j^2} \end{array} $$ where $n_k$ are defined as the smallest integer where $\lfloor H_n\rfloor = k$. Since $H_n = \log(n) + \gamma + \frac{1}{2n} + \mathcal{O}(n^{-2})$ we expect $n_k = \lfloor e^{k-\gamma} + \frac{1}{2}\rfloor$. This formula holds for $k\leq 100$ and heuristics suggest that it could hold for all $k$, see comments in OEIS A002387 and R. P. Boas, Jr. and J. W. Wrench, Jr., Partial Sums of the Harmonic Series, The American Mathematical Monthly.

Evaluating the sum requires knowledge of $n_k$ and since this is not known in the litterature (as far as I have chekced) it seems unlikely that the problem has a (known) closed form solution. Even if the simple formula for $n_k$ holds it still seems pretty hopeless to evaluate the sums, however it gives us a good starting point for an approximation. We can approximate

$$\sum_{j=n_k}^{n_{k+1}-1}\frac{1}{j^2} \approx \int_{n_k}^{n_{k+1}}\frac{{\rm d}x}{x^2} = \frac{(n_{k+1}-n_k)}{n_kn_{k+1}}$$

Now if we take $n_k = e^{k-\gamma}$ in the expression above then the integral is $(e-1)e^{\gamma-1-k}$ and the resulting sum can be evaluated analytically. A good approximation can be found by explicitly summing the first $K-1$ brackets in $(1)$ and using the approximation above to estimate the rest of the terms. This leads to

$$\sum_{n=1}^\infty \frac{\{H_n\}}{n^2} \approx 2\zeta(3)-\sum_{k=1}^{K-1} k\sum_{j=n_k}^{n_{k+1}-1}\frac{1}{j^2} -\frac{K(e-1)+1}{(e-1)e^{K-\gamma}}$$

Taking $K=5$ gives us $$2\zeta(3)-2.00822-\frac{5 (e-1) + 1}{(e-1)e^{5-\gamma}} \approx 0.32890$$

which is within $0.2\%$ of the exact result (found by summing numerically up to $n=10^7$).

Contest math problem: $\sum_{n=1}^\infty \frac{\{H_n\}}{n^2}$

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