Closed form of the series $\sum_{n=1}^{\infty}\frac1{(3n+1)(5n+2)}$.

Question

I want to compute the closed form of the series $$S=\sum_{n=1}^{\infty}\frac1{(3n+1)(5n+2)}.$$ (1) the seres is convergent; (2) I can prove that $S\leq\frac1{13}$; (3) wolframalpha gives the sum is $$S=\frac{\pi }{2 \sqrt{3}}\left(1-\sqrt{3 \left(1-\frac{2}{\sqrt{5}}\right)}\right)+\frac{1}{4}\left[\sqrt5\log\frac{3+\sqrt5}{2}-2-\log\left(\frac{3125}{729}\right)\right]=0.0706492\cdots.$$ See Wolframalpha

Two methods to compute $S$ (What I did): (i) $$\sum_{n=1}^{\infty}\frac1{(3n+1)(5n+2)} =\sum_{n=1}^{\infty}\left(\frac3{3n+1}-\frac5{5n+2}\right) =\sum_{n=1}^{\infty}\left(3\int_0^1x^{3n}\ \text dx-5\int_0^1x^{5n+1}\ \text dx\right)$$ $${\color{red}=?}\ \int_0^1\sum_{n=1}^{\infty}(3x^{3n}-5x^{5n+1})\ \text dx =\int_0^1\left(\frac{3x^3}{1-x^3}-\frac{5x^6}{1-x^5}\right)\ \text dx =0.581474\cdots.$$ I think the mistake is the red part ${\color{red}=}$, i.e. the change of order of Infinite summation and integral.

(ii) $$S=\sum_{n=1}^{\infty}\frac1{(3n+1)(5n+2)} =15\sum_{n=1}^{\infty}\left(\frac1{15n+5}-\frac1{15n+6}\right) =15\sum_{n=1}^{\infty}\left(\int_0^1x^{15n+4}-x^{15n+5}\text dx\right)$$ $${\color{blue}=}\ 15\int_0^1 \sum_{n=1}^{\infty}x^{15n+4}(1-x)\text dx =15\int_0^1\frac{x^{19}(1-x)}{1-x^{15}}\text dx =0.0706492\cdots.$$ This gives the right result, but it is hard to get the closed form of the integral.

My question is the following:

1. Why in the method (i) the ${\color{red}=}$ is not right and why in the method (ii) the ${\color{blue}=}$ is right?

2. How to improve method (ii) to get the closed form of $$S=\frac{\pi }{2 \sqrt{3}}\left(1-\sqrt{3 \left(1-\frac{2}{\sqrt{5}}\right)}\right)+\frac{1}{4}\left[\sqrt5\log\frac{3+\sqrt5}{2}-2-\log\left(\frac{3125}{729}\right)\right].$$

Any help and hints will welcome, thank you!

Answer 1

The closed form of the integral is not bad for the general case since, using the Gaussian hypergeometric function, $$\int \frac{x^a}{1-x^b}\,dx=\frac{x^{a+1}}{a+1} \, \, \, _2F_1\left(1,\frac{a+1}{b};\frac{a+1}{b}+1;x^b\right)$$ Combining $$\int \frac{x^a(1-x)}{1-x^b}\,dx=\frac{x^{a+1} \, _2F_1\left(1,\frac{a+1}{b};\frac{a+b+1}{b};x^b\right)}{a+1}-\frac{x^{a+2} \, _2F_1\left(1,\frac{a+2}{b};\frac{a+b+2}{b};x^b\right)}{a+2}$$ If $a$ and $b$ are positive $$\int_0^1 \frac{x^a(1-x)}{1-x^b}\,dx=\frac 1b\left(\psi \left(\frac{a+2}{b}\right)-\psi\left(\frac{a+1}{b}\right)\right)$$ is the simplest form. The problem is that the explicit formulae of the digamma function are quite messy as you already showed.

For your specific problem, the integral is just $$\psi \left(\frac{7}{5}\right)-\psi\left(\frac{4}{3}\right)$$ Expanding $$\psi \left(\frac{7}{5}\right)=\frac{5}{2}-\gamma -\frac{1}{2} \sqrt{1-\frac{2}{\sqrt{5}}} \pi -\frac{5 \log(5)}{4}+\frac{1}{2} \sqrt{5} \coth ^{-1}\left(\sqrt{5}\right)$$ $$\psi\left(\frac{4}{3}\right)=3-\gamma -\frac{\pi }{2 \sqrt{3}}-\frac{3 \log(3)}{2}$$

Then, not more much simpler $$I=-\frac 12+\frac{1}{2} \sqrt{5} \coth ^{-1}\left(\sqrt{5}\right)-\frac{1}{4} \log\left(\frac{3125}{729}\right)+\frac{\pi }{2 \sqrt{3}}\left(1-\sqrt{3 \left(1-\frac{2}{\sqrt{5}}\right)}\right)$$

Answer 2

Note that $$\sum_{n=1}^{\infty} \frac1{(n+a)(n+b)} = \frac{\psi(a+1)-\psi(b+1)}{a-b}$$

Proof: $$\sum_{n=1}^{\infty} \frac1{(n+a)(n+b)} = \frac1{b-a}\sum_{n=1}^{\infty} \left(-\frac1{n}+\frac{1}{n+a} - \frac1{n+b}+\frac1{n}\right) = \frac{\psi(b+1)-\psi(a+1)}{b-a}$$ which follows from the identity here.

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Then the sum we need is just $$S=\frac{\psi(4/3)-\psi(7/5)}{2}$$ By the known recurrence for $\psi$ you can write $$S=\frac{\psi(1/3) -\psi(2/5)}{2}+\frac14$$

Then you may use Gauss's Digamma theorem to evaluate both the digammas.

Answer 3

Actually, it is more convenient to solve harmonic-type series by finding the explicit value of digamma function, other than doing the integral of its generating function, we may begin with $$ S=\sum_{n=1}^{\infty}\left(\frac3{3n+1}-\frac5{5n+2}\right)=\sum_{n=1}^{\infty}\left(\frac3{3n+1}-\frac1{n}\right) - \sum_{n=1}^{\infty}\left(\frac5{5n+2}-\frac1{n}\right) $$ key to this sum is to find the explicit value of digamma function of $1/3$ and $1/5$, yet actually, we do not need a complete knowledge over digamma function.

A generalized device to do this by hand, we may recall this post. Here is the exact example how to solve those simple cases like $1/3$ and $1/5$ by the same manner.

One should be noted that $$ \sum_{n=1}^{\infty}\left(\frac1{3n+1}-\frac1{3n}\right) = \sum_{n=1}^{\infty}\left(\frac0{3n+2} + \frac1{3n+1}+\frac{-1}{3n}\right) $$ we consider the unit vector on the unit circle of complex plane, periodically rotating around the origin with angular step of $\frac{2\pi}{3}$, start with $-\frac{5\pi}{6}$, this shows $$ \begin{aligned} \cos\left(\frac{\pi}{6}\right)\left(\sum_{n=1}^{\infty}\left(\frac1{3n+1}-\frac1{3n}\right) + s_0 \right) &= \Re\left\{\sum_{n=1}^{\infty}\frac1{n}\exp\left[\left(\frac{2\pi n}{3}-\frac{5\pi}{6}\right)i\right]\right\} \\ &= \Re\left\{-e^{-\frac{5\pi i}{6}}\log\left(1-e^{\frac{2\pi i}{3}}\right)\right\} \end{aligned} $$ where $s_0=1$ is the very first item as $n=0$ that we do not count in the left sum. This identity easily derives $$ \sum_{n=1}^{\infty}\left(\frac1{3n+1}-\frac1{3n}\right) = -1 + \frac2{\sqrt{3}}\,\Re\left\{-e^{-\frac{5\pi i}{6}}\log\left(1-e^{\frac{2\pi i}{3}}\right)\right\} $$ complex simplify is tedious, but such result already promise a closed form of this digamma-related sum, which finally simplified as $$ \sum_{n=1}^{\infty}\left(\frac3{3n+1}-\frac1{n}\right) = -3 + \frac{\pi}{2\sqrt{3}} + \frac{3\ln3}{2} $$

By the same arguments, denote that $$ a=\sin\frac{2\pi}{5},\quad b=\sin\frac{\pi}{5}, \quad a= \frac{1+\sqrt{5}}{2} b = \phi b $$ consider the periodical unit vector with angular step of $-\frac{2\pi}{5}$ and $-\frac{4\pi}{5}$ (note the minus direction), this time staring at $\frac{9\pi}{10}$ and $-\frac{7\pi}{10}$, we have $$ \begin{aligned} \sum_{n=1}^{\infty}\left(\frac{a}{5n+2}-\frac{a}{5n} + \frac{b}{5n+3} - \frac{b}{5n+4} \right) + s_0 &= \Re\left\{\sum_{n=1}^{\infty}\frac1{n}\exp\left[\left(-\frac{2\pi n}{5}+\frac{9\pi}{10}\right)i\right]\right\} \\ &= \Re\left\{-e^{\frac{9\pi i}{10}}\log\left(1-e^{-\frac{2\pi i}{5}}\right)\right\} \end{aligned} $$ $$ \begin{aligned} \sum_{n=1}^{\infty}\left(\frac{b}{5n+2}-\frac{b}{5n} - \frac{a}{5n+3} + \frac{a}{5n+4} \right) + s'_0 &= \Re\left\{\sum_{n=1}^{\infty}\frac1{n}\exp\left[\left(-\frac{4\pi n}{5}-\frac{7\pi}{10}\right)i\right]\right\} \\ &= \Re\left\{-e^{\frac{-7\pi i}{10}}\log\left(1-e^{-\frac{4\pi i}{5}}\right)\right\} \end{aligned} $$ where $s_0=\frac{a}{2}+\frac{b}{3}-\frac{b}4$, $s'_0=\frac{b}{2}-\frac{a}{3}+\frac{a}4$, also note that $\frac{a}{b} + \frac{b}{a} = \phi + \frac1{\phi} = \sqrt{5}$, hence cancelling the 3rd and 4th item not needed $$ \sum_{n=1}^{\infty}\left(\frac{1}{5n+2}-\frac{1}{5n} \right) = -\frac1{2} + \frac{1}{\sqrt{5}b}\Re\left\{-e^{\frac{9\pi i}{10}}\log\left(1-e^{-\frac{2\pi i}{5}}\right)\right\} + \frac{1}{\sqrt{5}a}\Re\left\{-e^{\frac{-7\pi i}{10}}\log\left(1-e^{-\frac{4\pi i}{5}}\right)\right\} $$ with some very tedious simplify (may suggest doing this step with Mathematica), finally we have $$ \sum_{n=1}^{\infty}\left(\frac{5}{5n+2}-\frac{1}{n} \right) = -\frac{5}{2} + \frac{\pi}{2}\sqrt{1-\frac{2}{\sqrt{5}}} - \frac{\sqrt{5}}{2}\operatorname{arcoth}\sqrt{5} + \frac{5\ln5}{4} $$ those grant you the final answer done by hand.

Answer 4

A simple more direct way: $$S=\sum_{k=1}^{\infty} \left( \frac{1}{k+1/3}-\frac{1}{k+2/5}\right).$$ Using, the Polygamma function $\psi(z)$ (Digamma Function) https://en.wikipedia.org/wiki/Digamma_function $$\psi(z+1)=-\gamma+\sum_{K=1}^{\infty} \left( \frac{1}{k+1}- \frac{1}{k+z} \right).$$ We can write $S=\psi(7/5)-\psi(4/3)=\psi(2/5)-\psi(1/3)-1/2$ after using $\psi(z+1)=\psi(z)+1/z$. Further, $\psi(r/m)$ when $r<m \in \mathbb{N}$ can be expressed in terms of elementry functions of $r,m$.

Answer 5

Here is another way to avoid manipulating the difference of two divergent serie, and to show why the method (ii) seemingly works. We effectively seek $f(1)$, where
$$f(x) = 15\sum_{n=1}^\infty \frac{x^{15n+6}}{(15n+5)(15n+6)}$$ $$\implies f''(x) = 15\sum_{n=1}^\infty x^{15n+4} = \frac{15x^{19}}{1-x^{15}}$$

As $f(0)=f'(0)=0$, we have $$f(1) = \int_0^1f'(x)\,dx = \int_0^1\int_0^xf''(t)\, dt \,dx= \int_0^1(1-t)\cdot f''(t)\, dt = 15\int_0^1\frac{t^{19}(1-t)}{1-t^{15}}\,dt$$

You will recognise the above integral as the same you got using your method (ii).

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Evaluating the integral in $f(1)$ in closed form is challenging, though you already have a good answer using the digamma function, which is the simplest way. If an alternate simple way occurs, I will add here.

Answer 6

For the summation, what you could do is consider in the genarl case $$S_p=\sum_{n=1}^{p}\frac1{(a\,n+b)(c\,n+d)} =\frac{1}{b c-a d}\sum_{n=1}^{p}\left(\frac{c}{c\, n+d}-\frac{a}{a\, n+b}\right)$$ and use the polygamma function $$S_p=\frac{1}{b c-a d}\Bigg(\psi \left(p+\frac{d}{c}+1\right)-\psi \left(\frac{d}{c}+1\right)-\psi \left(p+\frac{b}{a}+1\right)-\psi \left(\frac{b}{a}+1\right) \Bigg)$$ and use the asymptotic $$\psi \left(p+k\right)=\log (p)+\frac{2 k-1}{2 p}+\frac{-6 k^2+6 k-1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ and then $$S_p=\frac{1}{b c-a d}\left(\psi \left(\frac{b}{a}+1\right)-\psi \left(\frac{d}{c}+1\right)\right)-\frac{1}{a c\, p}+\frac{a c+a d+b c}{2 a^2 c^2 \,p^2}+O\left(\frac{1}{p^3}\right)$$ which gives the limit and a shortcut solution for a finite sum.