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I need some ideas to exploit for finding the closed form of $$\sum_{n=1}^{\infty} \left(\frac{1}{\lfloor\sqrt{3n}\rfloor^2}-\frac{1}{3n}\right)$$

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  • $\begingroup$ The sum seems to be $$\frac13( \log 3 + \gamma ) + \frac{\pi^2}{54} + \frac{1}{27}\left(\psi'\left(\frac13\right) - \psi'\left(\frac23\right)\right) \sim 1.00181390736073477589... $$ where $\psi$ is the digamma function. The convergence of the original sum is pretty slow. I cannot verify the correctness of the assertion numerically. $\endgroup$ – achille hui May 24 '14 at 9:25
  • $\begingroup$ @achillehui Interesting. Do you have a proof for it? $\endgroup$ – user 1357113 May 24 '14 at 9:30
  • $\begingroup$ Sort of, it is proving it the boring way by grouping the sum according to $p = \lfloor\sqrt{3n}\rfloor$. I feel uneasy because I can't numerical verify it reproduces the right number. $\endgroup$ – achille hui May 24 '14 at 9:39
  • $\begingroup$ @achillehui OK. Thank you for your comments. $\endgroup$ – user 1357113 May 24 '14 at 9:41
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We will evaluate the sum by grouping the terms according to $p = \lfloor 3n \rfloor$.

$$\sum_{n=1}^\infty \left(\frac{1}{\lfloor\sqrt{3n}\rfloor^2} - \frac{1}{3n}\right) = \sum_{p=1}^\infty \sum_{\frac{p^2}{3} \le n < \frac{(p+1)^2}{3}} \left(\frac{1}{p^2} - \frac{1}{3n}\right) $$ and then analyze the inner sums in group of three. We have

  • $p = 3k-2$
    $$\frac{(3k-2)^2}{3} \le n < \frac{(3k-1)^2}{3} \iff 3k^2 - 4k + 2 \le n \le 3k^2 - 2k$$
  • $p = 3k-1$ $$\frac{(3k-1)^2}{3} \le n < \frac{(3k)^2}{3} \iff 3k^2-2k+1 \le n \le 3k^2-1$$
  • $p = 3k$ $$\frac{(3k)^2}{3} \le n < \frac{(3k+1)^2}{3} \iff 3k^2 \le n \le 3k^2+2k$$

This implies the number of $n$ with a given $p$ is equal to

$$\Big|\{\; n : \lfloor \sqrt{3n} \rfloor = p\;\}\Big| = \begin{cases} 2k - 1,& p = 3k - 2\\ 2k - 1,& p = 3k - 1\\ 2k + 1,& p = 3k \end{cases}$$

As a result, we have $$ \sum_{p=1}^{3K} \sum_{\frac{p^2}{3} \le n < \frac{(p+1)^2}{3}} \left(\frac{1}{p^2} - \frac{1}{3n}\right) = \sum_{k=1}^K \left[\frac{2k-1}{(3k-2)^2} + \frac{2k-1}{(3k-1)^2} + \frac{2k+1}{(3k)^2}\right] - \frac13 H_{3K^2 + 2K} $$ where $H_n$ is the $n^{th}$ Harmonic number. A little bit of algebra allow us to rewrite what's in the square bracket as

$$\frac23\left[ \frac{1}{3k-2} + \frac{1}{3k-1} + \frac{1}{3k} \right] + \frac13 \left[\frac{1}{(3k-2)^2} - \frac{1}{(3k-1)^2}\right] + \frac{1}{(3k)^2}$$

From this, we find the partial sum above is equal to

$$\frac23 H_{3K} - \frac13 H_{3K^2+2_K} + \sum_{k=1}^K\left\{ \frac19 \frac{1}{k^2} + \frac{1}{27} \left[\frac{1}{(k-\frac23)^2} - \frac{1}{(k-\frac13)^2}\right] \right\}$$ It is trivial to show $$\lim_{K\to\infty} \frac23 H_{3K} - \frac13 H_{3K^2+2K} = \frac13 (\log 3 + \gamma).$$ Together with the facts $$ \sum_{k=1}^\infty \frac{1}{k^2} = \zeta(2) = \frac{\pi^2}{6} \quad\text{ and }\quad \sum_{k=1}^\infty \frac{1}{(k-\alpha)^2} = \psi'(1-\alpha) $$ where $\psi(x)$ is the Digamma function, the sum we desired is

$$\frac13 (\log 3 + \gamma) + \frac{\pi^2}{54} + \frac{1}{27} \left(\psi'\left(\frac13\right) - \psi'\left(\frac23\right)\right)$$

Numerically, WA evaluate this to an number $\approx 1.0018139073607347758976943818370...$
Unfortunately, the original sum converges very slowly. I does not have an independent validation of the correctness of above closed form.

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  • $\begingroup$ Numerically does not seem to work, but I have not yet checked the details to tell where the problem is. By the way the answer seems to be $1$ or at least something very close to this value. $\endgroup$ – Start wearing purple May 24 '14 at 10:27
  • $\begingroup$ @O.L. It cannot be $1$. When I created this question some weeks ago I checked it with Mathematica (and after a long time, for some partial sum, it exceeded $1$) $\endgroup$ – user 1357113 May 24 '14 at 10:29
  • $\begingroup$ @O.L. I have typos in the first version I write down. the coefficient in front of the digamma functions is $1/27$ instead of $1/3$. $\endgroup$ – achille hui May 24 '14 at 10:31
  • $\begingroup$ @O.L. N[Sum[1/Floor[Sqrt[3 n]]^2 - 1/(3 n), {n, 1, 50000}]] $\sim 1.00009$ $\endgroup$ – user 1357113 May 24 '14 at 10:34
  • $\begingroup$ The partial sum for the original series is about $1.001428940904307$ for the first $10^6$ terms and $1.00169218180237$ for the first $10^7$ terms. It converges very slowly :-( $\endgroup$ – achille hui May 24 '14 at 10:36

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