Infinite Series $\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$ How do I find the sum of the following infinite series:
$$\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$$
The series definitely seems to be convergent.
 A: Here's $\let\leq\leqslant\let\geq\geqslant$a complex analysis approach. (It may look a bit tedious, but that's because of some issues with convergence. After all, the concept is beautiful.)
Let $\zeta=e^{2\pi i/3}$, we have $\zeta^k+\zeta^{2k}=-1$ if $3\nmid k$ and $\zeta^k+\zeta^{2k}=2$ if $3\mid k$.
Hence
$$\sum_{n=1}^\infty\left(\frac1{3n-1}+\frac1{3n+1}-\frac2{3n}\right)=-\sum_{k=2}^\infty\frac{\zeta^k+\zeta^{2k}}k.$$
Let $f(z)=\sum_{k=2}^\infty\frac{z^k+z^{2k}}k$. Note that $f(\zeta)$ exists because from the original formula we have
$$f(\zeta)=\sum_{n=1}^\infty\frac2{3n(3n+1)(3n-1)}=O\left(\sum_{n=1}^\infty\frac1{n^3}\right)=O(1).$$
For $|z|<1$,
$$\begin{align*}f(z)
&=\sum_{k=2}^\infty\int_0^z\left(t^{k-1}+2t^{2k-1}\right)dt\end{align*}.$$
Because the series $\sum_{n\geq1}t^n$ converges uniformly for $|t|\leq z$, we can write
$$\begin{align*}f(z)
&=\int_0^z\sum_{k=2}^\infty\left(t^{k-1}+2t^{2k-1}\right)dt\\
&=\int_0^z\left(\frac t{1-t}+2\frac{t^3}{1-t^2}\right)dt.\end{align*}$$
Finally, by Abel's theorem,
$$\begin{align*}f(\zeta)
&=\int_0^\zeta\left(\frac t{1-t}+2\frac{t^3}{1-t^2}\right)dt\\
&=\left[t^2+t+\log(t^3-t^2-t+1)\right]_0^\zeta\\
&=-1+\log3.\end{align*}$$
Because we had our initial sum with a minus sign, the answer is $1-\log3$.
A: Your sum is equal to:
\begin{align}
\sum_{i=1}^{\infty} \left ( \frac{1}{3i-1}+\frac{1}{3i+1}-\frac{2}{3i}\right )  &=\sum_{i=1}^{\infty} \frac{(3i+1) \cdot 3i+3i \cdot (3i-1)-2(3i-1) \cdot (3i+1)}{(3i-1) \cdot 3i \cdot (3i+1)} \\
&=\sum_{i=1}^{\infty}\frac{9i^2+3i+9i^2-3i-2(9i^2-1)}{(3i-1) \cdot 3i \cdot (3i+1)} \\
&=\sum_{i=1}^{\infty}\frac{18i^2-18i^2+2}{(3i-1) \cdot 3i \cdot (3i+1)} \\
&=\sum_{i=1}^{\infty}\frac{2}{(3i-1) \cdot 3i \cdot (3i+1)} \\
&= 2 \sum_{i=1}^{\infty} \frac{1}{(3i-1) \cdot 3i \cdot (3i+1)}
\end{align}
A: Let
$$S_m=\sum_{k=1}^m\left(\frac{1}{3k-1}+\frac{1}{3k+1}-\frac{2}{3k}\right)$$
this can be written as follows
$$S_m=\sum_{k=1}^m\left(\frac{1}{3k-1}+\frac{1}{3k}+\frac{1}{3k+1}-\frac{1}{k}\right)
=H_{3m+1}-1-H_m=\frac{1}{3m+1}-1+H_{3m}-H_m$$
where $H_n=\sum_{k=1}^n1/k=\ln n+\gamma+o(1)$, is the $n$th harmonic number. Using this asymptotic expansion we see that
$$S_m=\ln 3-1+o(1)$$
Hence $\lim\limits_{m\to\infty}S_m=-1+\ln3$.
A: Hint:
$$S=\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots\infty$$
$$S
=\sum_{k=0}^{\infty}\left(\frac1{2+3k}+\frac1{4+3k}-\frac2{3+3k}\right)\\
=\sum_{k=0}^{\infty}\left(\frac1{2+3k}+\frac1{4+3k}-\frac2{3+3k}\right)\\
$$
A: \begin{array}{l}
 s_0 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {\frac{{x^{3k + 3} }}{{3k + 3}}}  \Rightarrow s'_0 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {x^{3k + 2} }  = \frac{{x^2 }}{{1 - x^3 }} \\ 
 s_1 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {\frac{{x^{3k + 4} }}{{3k + 4}}}  \Rightarrow s'_1 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {x^{3k + 3} }  = \frac{{x^3 }}{{1 - x^3 }} \\ 
 s_2 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {\frac{{x^{3k + 2} }}{{3k + 2}}}  \Rightarrow s_2 ^\prime  \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {x^{3k + 1} }  = \frac{x}{{1 - x^3 }} \\ 
 s_0 \left( x \right) + s_1 \left( x \right) + s_2 \left( x \right) = \sum\limits_{x = 0}^{ + \infty } {\left( {\frac{{x^{3k + 3} }}{{3k + 3}} + \frac{{x^{3k + 4} }}{{3k + 4}} + \frac{{x^{3k + 2} }}{{3k + 2}}} \right)}  \\ 
 \frac{x}{{1 - x^3 }} + \frac{{x^3 }}{{1 - x^3 }} - \frac{{2x^2 }}{{1 - x^3 }} = \frac{{x^3  + x - 2x^2 }}{{1 - x^3 }} = \frac{{x\left( {x - 1} \right)}}{{1 + x + x^2 }} \\ 
 \end{array}
