How to evaluate the sum of $\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}$ I hava an infinite sum
$$\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}$$
I factored the denominator
$$\sum_{n=0}^{\infty}\frac{1}{\left(3n+1\right)\left(n+1\right)}$$
Then I separated the fraction
$$\frac{1}{2}\sum_{n=0}^{\infty}\frac{3}{\left(3n+1\right)}-\frac{1}{\left(n+1\right)}$$
Then I set 1 (the numerator) to be equal to x to some power which I don't know if I can do
$$\frac{3}{2}\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1}-\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$$
Then I set the integral which would satisfy the previous terms
$$\frac{3}{2}\sum_{n=0}^{\infty}\int_{0}^{1}x^{3n}dx$$
and
$$-\frac{1}{2}\sum_{n=0}^{\infty}\int_{0}^{1}x^{n}dx$$
Then I changed the order of summation and integration and I got
$$\frac{3}{2}\int_{0}^{1}\frac{1}{1-x^{3}}dx$$ and $$-\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}dx$$
The first integral can be factored to
$$\frac{3}{2}\int_{0}^{1}\frac{1}{\left(1-x\right)\left(1+x+x^{2}\right)}dx$$
Then separated
$$\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}+\frac{x+2}{1+x+x^{2}}dx$$
The first one will cancel out with
$$-\frac{1}{2}\int_{0}^{1}\frac{1}{1-x}dx$$
And I'm left with
$$\frac{1}{2}\int_{0}^{1}\frac{x+2}{1+x+x^{2}}dx$$
which is
$$\frac{\sqrt{3}\pi}{12}+\frac{\ln\left(3\right)}{4}$$
But the correct answer is
$$\frac{\sqrt{3}\pi}{12}+\frac{3\ln\left(3\right)}{4}$$
So I would like to ask if this approach is invalid or if I'm just missing something.
 A: $$ \begin{align}\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1} &= \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{3}{3n+1}-\frac{1}{n+1} \right) \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \left(3\int_{0}^{1}x^{3n} \, \mathrm dx- \color{red}{3\int_{0}^{1} x^{3n+2} \, \mathrm dx}\right) \\ &= \frac{3}{2} \sum_{n=0}^{\infty} \int_{0}^{1} \left(x^{3n}-x^{3n+2} \right) \, \mathrm dx \\ &= \frac{3}{2} \sum_{n=0}^{\infty} \int_{0}^{1} x^{3n}\left(1-x^{2}\right) \, \mathrm dx \end{align}$$
Since $x^{3n}(1-x^{2})$ is nonnegative for all $n$ and all $x$, Tonelli's theorem allows us to interchange the order of integration and summation.
Therefore, we have $$ \begin{align} \sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1} &= \frac{3}{2} \int_{0}^{1} (1-x^{2}) \sum_{n=0}^{\infty} x^{3n}\, \mathrm dx \\ &= \frac{3}{2} \int_{0}^{1} \frac{1-x^{2}}{1-x^{3}} \, \mathrm dx  \\ &= \frac{3}{2} \int_{0}^{1} \frac{x+1}{x^{2}+x+1} \, \mathrm dx \\ &= \frac{3}{4} \int_{0}^{1} \left( \frac{2x+1}{x^{2}+x+1} +\frac{1}{x^{2}+x+1} \right)\, \mathrm dx \\ &=  \frac{3}{4} \, \ln(3) + \frac{3}{4} \int_{0}^{1} \frac{\mathrm dx}{\left(x+\frac{1}{2} \right)^{2} + \frac{3}{4}} \, \mathrm dx \\ &= \frac{3}{4} \, \ln(3) + \frac{3}{2\sqrt{3}} \, \arctan \left(\frac{2}{\sqrt{3}} \left( x+\frac{1}{2}\right) \right) \Bigg|_{0}^{1} \\ & = \frac{3}{4} \, \ln(3) + \frac{3}{2 \sqrt{3}} \left(\frac{\pi}{3}- \frac{\pi}{6} \right) \\ &=\frac{3}{4} \, \ln(3) + \frac{\sqrt{3} \pi}{12}. \end{align}$$
A: If you're looking to rewrite the series as an integral, here's a way of doing it:
Consider
$$\sum_{n=0}^{\infty}{1\over{(3n+1)(n+1)}}x^{3n+1}$$
So that its derivative will cancel out one of the factors in the denominator:
$$\sum_{n=0}^{\infty}{1\over{(n+1)}}x^{3n}$$
Similar to how you showed, we can now write the original series as a definite integral:
$$\sum_{n=0}^{\infty}{1\over{(3n+1)(n+1)}}=\sum_{n=0}^{\infty}\int_{0}^{1}{1\over{(n+1)}}x^{3n}dx
$$
$$
=\int_{0}^{1}\sum_{n=1}^{\infty}{(x^3)^{n-1}\over n} dx=\int_{0}^{1}{1\over x^3}\sum_{n=1}^{\infty}{(x^3)^n\over n} dx$$
The last sum is recognizable as the power series expansion of $-\ln(1-x^3)$, so that:
$$
\sum_{n=1}^{\infty}{1\over {(3n+1)(n+1)}} =-\int_{0}^{1}{1\over x^3}\ln(1-x^3) dx$$
The integral is a little beyond me at the moment, however.
A: Hi welcome to MSE:As a hint for telescopic series$$\sum_{n=0}^{\infty}\frac{1}{3n^{2}+4n+1}
\\=\sum_{n=0}^{\infty}\frac{1}{(3n+1)(n+1)}
\\=\sum_{n=0}^{\infty}\frac{3}{(3n+1)(3n+3)}
\\=3\sum_{n=0}^{\infty}\frac{(3n+2)}{(3n+1)(3n+2)(3n+3)}
\\=3\sum_{n=0}^{\infty}\frac{(3n+3)-1}{(3n+1)(3n+2)(3n+3)}
\\=3\sum_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)}-3\sum_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$$
