TITLE: The closed form of $\sum_{n=1}^{\infty} \left(\frac{1}{\lfloor\sqrt{3n}\rfloor^2}-\frac{1}{3n}\right)$
QUESTION [9 upvotes]: I need some ideas to exploit for finding the closed form of 
$$\sum_{n=1}^{\infty} \left(\frac{1}{\lfloor\sqrt{3n}\rfloor^2}-\frac{1}{3n}\right)$$

REPLY [15 votes]: 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.