Infinite Series $\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$

Question

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.

Answer 1

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 $\lim\limits_{m\to\infty}S_m=-1+\ln3$.

Answer 2

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 $1-\log3$.

Answer 3

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}

Answer 4

Hint: $$S=\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots\infty$$ $$S =\sum_{k=0}^{\infty}\left(\frac1{2+3k}+\frac1{4+3k}-\frac2{3+3k}\right)\\ =\sum_{k=0}^{\infty}\left(\frac1{2+3k}+\frac1{4+3k}-\frac2{3+3k}\right)\\ $$

Answer 5

\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}