Series for $\log 3$ I have the following series:
$$\sum_{k=0}^{\infty} \left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)$$
Wolfram says this is just $\log 3$. I have been trying to figure out how this works purely through series manipulation (without integrals etc.).
I've tried splitting it up into several series but nothing seems to fit nicely because the pattern is 3-period. The series I know for $\log$ which tried first was:
$$\log(1+x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}x^k$$
perhaps with $x=-\frac{2}{3}$, but this introduces powers which don't seem natural to derive from the original expression.
Any help would be great.
 A: Let $H_n=\sum_{k=1}^n{\frac{1}{k}}$, we have
$$ \sum_{k=0}^{n} \left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)=(H_{3n+3}-1)-\sum_{k=1}^n{\frac{3}{3k+3}}=(H_{3n+3}-1)-(H_{n+1}-1) $$
Since $H_n=\ln n +\gamma+o(1)$ you have
$$ \sum_{k=0}^{n} \left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)=\ln(3)+o(1) $$
and thus $$ \sum_{k=0}^{+\infty} \left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)=\ln(3) $$
A: Let us go with brute force:
$$ \frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3} = \int_{0}^{1}x^{3k}(1+x-2x^2)\,dx $$
leads to:
$$ \sum_{k\geq 0}\left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)=\int_{0}^{1}\frac{1+x-2x^2}{1-x^3}\,dx=\int_{0}^{1}\frac{2x+1}{x^2+x+1}\,dx $$
where the RHS equals
$$ \left[\log(x^2+x+1)\right]_{0}^{1} = \color{red}{\log 3} $$
as wanted.
A: We have that
$$\sum_{k=0}^{N} \left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)=\\=\sum_{k=0}^{N} \left(\frac{1}{3k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}\right)-\sum_{k=0}^{N} \left(\frac{1}{3k+3}+\frac{2}{3k+3}\right)=$$
$$=\sum_{k=0}^{3N} \frac{1}{k+1}-\sum_{k=0}^{N} \frac{1}{k+1}=\log (3N)-\log N+O\left(\frac1N\right) \to\log 3$$
A: In general, we have
$$\sum_{k=0}^\infty(\frac{1}{nk+1}+\frac{1}{nk+2}+\frac{1}{nk+3}+\dots+\frac{1}{nk+n-1}-\frac{n-1}{nk+n})=\ln(n)$$ for any integer $n>1$.
Check for $n=3$, you will get the result.
Check my related post.
