# How find this sum $\sum\limits_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$

Find this sum

$$I=\sum_{n=0}^{\infty}\dfrac{1}{(3n+1)(3n+2)(3n+3)}$$

My try: let

$$f(x)=\sum_{n=0}^{\infty}\dfrac{x^{3n+3}}{(3n+1)(3n+2)(3n+3)},|x|\le 1$$

then we have
$$f^{(3)}(x)=\sum_{n=0}^{\infty}x^{3n}=\dfrac{1}{1-x^3}$$
then we find the $f(x)$,Following is very ugly(can you someone can post your follow solution,) have other simple methods? Thank you very much.

#### Solutions Collecting From Web of "How find this sum $\sum\limits_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$"

Your try is excellent… Note that $f(0)=f'(0)=f”(0)=0$ hence
$$f(1)=\int_0^1f'(x)\,\mathrm dx=\int_0^1\int_0^xf”(y)\,\mathrm dy\,\mathrm dx=\int_0^1\int_0^x\int_0^yf”'(z)\,\mathrm dz\,\mathrm dy\,\mathrm dx,$$
that is,
$$2f(1)=\int_0^1(1-z)^2f”'(z)\,\mathrm dz=\int_0^1\frac{1-z}{1+z+z^2}\,\mathrm dz.$$
The rest is routine. The change of variable $2z+1=\sqrt3t$ yields
$$2f(1)=\int_{1/\sqrt3}^\sqrt3\frac{\sqrt3-t}{1+t^2}\,\mathrm dt=\left[\sqrt3\arctan t-\frac12\log(1+t^2)\right]_{1/\sqrt3}^\sqrt3.$$
Note that $\arctan\sqrt3=\pi/3$ and $\arctan1/\sqrt3=\pi/6$, hence
$$2f(1)=\sqrt3\cdot\left(\frac\pi3-\frac\pi6\right)-\frac12\log4+\frac12\log\frac43,$$
that is,
$$f(1)=\frac14\left[\frac\pi{\sqrt3}-\log3\right]\approx0.1788.$$
Second method: The rational fraction is such that
$$\frac2{(3n+1)(3n+2)(3n+3)}=\frac1{3n+1}-\frac2{3n+2}+\frac1{3n+3},$$
hence
$$2f(x)=x^2g_1(x)-2xg_2(x)+g_3(x),\qquad g_k(x)=\sum_{n\geqslant0}\frac{x^{3n+k}}{3n+k}.$$
Thus, for each $k$,
$$g’_k(x)=\sum_{n\geqslant0}x^{3n+k-1}=\frac{x^{k-1}}{1-x^3}.$$
Since $g’_k(0)=0$ for every $k\geqslant1$, this yields
$$2f(x)=x^2\int_0^x\frac{1}{1-t^3}\mathrm dt-2x\int_0^x\frac{t}{1-t^3}\mathrm dt+\int_0^x\frac{t^2}{1-t^3}\mathrm dt.$$
The change of variable $t=xu$ yields
$$2f(x)=x^3\int_0^1\frac{1}{1-x^3u^3}(1-2u+u^2)\mathrm du,$$
that is,
$$2f(x)=x^3\int_0^1\frac{1-u}{1-xu}\frac{1-u}{1+xu+x^2u^2}\mathrm du.$$
When $x\to1$, one obtains once again
$$2f(1)=\int_0^x\frac{1-u}{1+u+u^2}\mathrm du.$$
More generally, for every integer $k\geqslant2$,
$$\sum_{n\geqslant0}\frac{(k-1)!}{(kn+1)(kn+2)\cdots(kn+k)}=\int_0^1\frac{(1-u)^{k-2}}{1+u+\cdots+u^{k-1}}\,\mathrm du.$$

We all know that sums of the form $\displaystyle{\sum\frac1{(n+a)(n+b)}}$ or $\displaystyle{\sum\frac1{(n+a)(n+b)(n+c)}}$ etc., are telescopic in nature, and solving them is trivial: for integer values of a, b, and c, that is ! But what if a, b, and c are not integers? What then? Suddenly, things aren’t so simple and trivial anymore, and the formerly banal and deceitfully tame-looking problem takes on whole new twist, gaining entirely unexpected dimensions of depth, meaning, and insight. The key, when dealing with such surprising turns of events, is to just take a step back, and try to parse or rephrase the older, well-worn solution in terms which might prove useful or relevant to circumventing the obstacles raised by the newly-encountered situation.

For instance, it is by no means difficult to show that $\displaystyle{\sum_{n=1}^\infty\frac1{(n+a)(n+b)}}=\frac{H_a-H_b}{a-b}$ , or that $$\sum_{n=1}^\infty\frac1{(n+a)(n+b)(n+c)}=\frac{(a-b)\cdot H_c\ +\ (b-c)\cdot H_a\ +\ (c-a)\cdot H_b}{(a-b)(b-c)(c-a)}$$ for natural values of a, b, c. Indeed, the user Random Variable has already proved it on this thread, though I now realize that my initial formula is slightly mistaken, in the sense that either the sum should begin at $0$, with the $\gamma$ present, or at $1$, with the $\gamma$ absent. Now, the whole question becomes how to redefine $H_m$ , so as to be able to extend its meaning to non-natural arguments. For natural arguments, we have the well-known formula $H_m=\displaystyle\sum_{k=1}^m\frac1k$ . Now, let us take the simple function $f_k(x)=\displaystyle\frac{x^k}k$ , and notice that $f_k'(x)=x^{k-1}$ . And since we know that $\displaystyle{\sum_{k=1}^mx^{k-1}=\frac{1-x^m}{1-x}}$ , this finally allows us to conclude that $\displaystyle{H_m=\int_0^1\frac{1-x^m}{1-x}dx}$ , which, unlike the previous formula, can easily be extended to non-natural arguments as well. This has already been done by Euler two and a half centuries ago, so it’s hardly new territory. Now, in our case, $\{a,b,c\}=\left\{\frac13,\frac23,1\right\}$ , and the value of $H_1$ is $1$, so all that’s left to do is to compute $H_\frac13$ and $H_\frac23$ using the above formula, since $$\sum_{n=0}^\infty\frac1{(3n+1)(3n+2)(3n+3)}=\frac16+\frac1{27}\cdot\sum_{n=1}^\infty\frac1{\left(n+\frac13\right)\left(n+\frac23\right)(n+1)}$$

$$H_\frac13=\int_0^1\frac{1-\sqrt[3]x}{1-x}dx=\int_0^1\frac{1-t}{1-t^3}d\left(t^3\right)=\int_0^1\frac1{1+t+t^2}d\left(t^3\right)=\int_{\sqrt[3]0}^{\sqrt[3]1}\frac{3\,t^2}{1+t+t^2}dt=$$

$$=3\int_0^1\left(1-\frac{1+t}{1+t+t^2}\right)dt=3\,\bigg[\int_0^11\cdot dt-\tfrac12\int_0^1\frac{2+2t}{1+t+t^2}dt\bigg]=$$

$$=3\,\bigg[t|_0^1-\tfrac12\bigg(\int_0^1\frac1{1+t+t^2}dt+\int_0^1\frac{1+2t}{1+t+t^2}dt\bigg)\bigg]=$$

$$=3\,\bigg[1-\tfrac12\int_0^1\frac1{\left(t+\frac12\right)^2+\frac34}dt-\tfrac12\cdot\ln\left(1+t+t^2\right)_0^1\bigg]=$$

$$=3\,\bigg[1-\tfrac12\cdot\frac1{\sqrt3/2}\cdot\arctan\bigg(\frac{t+\frac12}{\sqrt3\big/2}\bigg)_0^1-\frac{\ln3}2\bigg]=3-\frac\pi{2\sqrt3}-\tfrac32\ln3.$$

Similarly for $\displaystyle{H_{\frac23}=\tfrac32+\frac\pi{2\sqrt3}-\tfrac32\ln3}$ . Then, by substituting these values back into the original formula above, we finally arrive at the desired result $\displaystyle{I=\frac{\pi\sqrt3-3\ln3}{12}}$ .

$\ds{ I_{N} \equiv \sum_{n = 0}^{N}{1 \over \pars{3n + 1}\pars{3n + 2}\pars{3n + 3}}\,, \qquad I = I_{\infty} =\ {\large ?}}$

\begin{align}
I_{N}&\equiv
{1 \over 6}\sum_{n = 0}^{N}{1 \over n + 1/3}

{1 \over 3}\sum_{n = 0}^{N}{1 \over n + 2/3}
+
{1 \over 6}\sum_{n = 0}^{N}{1 \over n + 1}
\\[3mm]&=
{1 \over 6}\sum_{n = 0}^{N}\pars{{1 \over n + 1/3} – {1 \over n + 2/3}}
+
{1 \over 6}\sum_{n = 0}^{N}\pars{{1 \over n + 1} – {1 \over n + 2/3}}
\\[3mm]&=
{1 \over 18}\sum_{n = 0}^{N}{1 \over \pars{n + 1/3}\pars{n + 2/3}}

{1 \over 18}\sum_{n = 0}^{N}{1 \over \pars{n + 1}\pars{n + 2/3}}
\end{align}
We’ll use the well known identity:
$\ds{\sum_{n = 0}^{\infty}{1 \over \pars{n + z_{0}}\pars{n + z_{1}}} ={\Psi\pars{z_{0}} – \Psi\pars{z_{1}} \over z_{0} – z_{1}}}$ where $\Psi\pars{z}$ is the $\it\mbox{digamma function}$

\begin{align}
I & = I_{\infty}
=
{1 \over 18}\,{\Psi\pars{1/3} – \Psi\pars{2/3} \over 1/3 – 2/3}

{1 \over 18}\,{\Psi\pars{1} – \Psi\pars{2/3} \over 1 – 2/3}
\\[3mm]&=
{1 \over 6}\,\bracks{\Psi\pars{2 \over 3} – \Psi\pars{1 \over 3}}

{1 \over 6}\,\bracks{\Psi\pars{1} – \Psi\pars{2 \over 3}}
=
-\,{1 \over 6}\,\Psi\pars{1} – {1 \over 6}\,\Psi\pars{1 \over 3}
+
{1 \over 3}\,\Psi\pars{2 \over 3}
\end{align}
However,
\begin{align}
\Psi\pars{1} &= -\gamma\quad\mbox{where}\ \gamma\ {\it\mbox{is the Euler-Mascheroni constant}}\ \pars{~\gamma = 0.5772\ldots~}
\\[3mm]
\Psi\pars{1 \over 3}&=-\gamma – {\root{3} \over 6}\,\pi – {3 \over 2}\,\ln\pars{3}
\\[3mm]
\Psi\pars{2 \over 3}&=-\gamma + {\root{3} \over 6}\,\pi – {3 \over 2}\,\ln\pars{3}
\\
I & = I_{\infty} =
-\,{1 \over 6}\pars{-\gamma}

{1 \over 6}\bracks{-\gamma – {\root{3} \over 6}\,\pi – {3 \over 2}\,\ln\pars{3}}
+
{1 \over 3}\bracks{-\gamma + {\root{3} \over 6}\,\pi – {3 \over 2}\,\ln\pars{3}}
\\[3mm]&=
{\root{3} \over 12}\,\pi – {1 \over 4}\,\ln\pars{3}
\end{align}

$$\color{#0000ff}{\large \sum_{n = 0}^{N}{1 \over \pars{3n + 1}\pars{3n + 2}\pars{3n + 3}} = {\root{3} \over 12}\,\pi – {1 \over 4}\,\ln\pars{3}} \approx 0.1788$$

$$\frac{d^3}{dx^3} f(x) = \frac{1}{1-x^3}$$

where $f(0)=f'(0)=f”(0)=0$. Note that we may use partial fractions and a little rearranging to find that

$$\frac{1}{1-x^3} = \frac13 \left [\frac{1}{1-x} + \frac{x+\frac12}{\left (x+\frac12\right)^2+\frac{3}{4}}+\frac{3}{2} \frac{1}{\left (x+\frac12\right)^2+\frac{3}{4}}\right ]$$

Integrating and using $f”(0)=0$ we get that

$$f”(x) = -\frac13 \log{(1-x)} + \frac16 \log{(1+x+x^2)} + \frac{1}{\sqrt{3}} \arctan{\frac{2 x+1}{\sqrt{3}}} – \frac{\pi}{6 \sqrt{3}}$$

From here the challenge will be managing the various terms rather than individual integrations, which are easily done through parts. Note that, for example,

$$\int dx \, \log{(1+x+x^2)} = x \log{(1+x+x^2)} – \int dx \frac{2 x^2+x}{x^2+x+1}$$

You may use partial fractions on the latter integral and determine that

$$\int dx \frac{2 x^2+x}{x^2+x+1} = 2 x – \frac12 \log{(1+x+x^2)} – \sqrt{3} \arctan{\frac{2 x+1}{\sqrt{3}}} +C$$

Using a substitution ($u=(2 x+1)/\sqrt{3}$) and integration by parts, we similarly find that

$$\int dx \, \arctan{\frac{2 x+1}{\sqrt{3}}} = \frac{\sqrt{3}}{2}\left [\frac{2 x+1}{\sqrt{3}} \arctan{\frac{2 x+1}{\sqrt{3}}} – \frac12 \log{(1+x+x^2)}\right ] + C$$

Omitting a lot of algebra, I find that, using $f'(0)=0$:

$$f'(x) = \frac13 (1-x) \log{(1-x)} – \frac16 (1-x) \log{(1+x+x^2)} + \frac{1+x}{\sqrt{3}} \arctan{\frac{2 x+1}{\sqrt{3}}} – \frac{\pi}{6 \sqrt{3}} (1+x)$$

Now we integrate once more. Again, the challenge is bookkeeping, as all the integrals may be done by parts. Using $f(0)=0$, I get that

$$f(x) = -\frac16(1-x)^2 \log{(1-x)} + \frac{1}{12} (x^2-2 x-2) \log{(1+x+x^2)} + \frac{(x^2+2 x)}{2 \sqrt{3}} \arctan{\frac{2 x+1}{\sqrt{3}}} – \frac{\pi}{12 \sqrt{3}} (x^2+2 x)$$

The sum is then

$$f(1) = \frac{\sqrt{3}}{12} \pi – \frac14 \log{3} \approx 0.178$$

which checks out with, e.g., WA.