Find sum of the series $ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$

Question

I have to find the sum of the given series $$ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$$

My attempt $$ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$$ or, $$ S=1 + \left( \frac{1}{3}-\frac{1}{2}+\frac{1}{5} \right)+\left(\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}\right)+\left(\frac{1}{11}-\frac{1}{6}+\frac{1}{13}\right)+\left( \frac{1}{15}-\frac{1}{8}+\frac{1}{17}\right)+ \cdots$$ or, $$S= 1+ \sum_{n=1}^\infty \left( \frac{1}{4n-1}-\frac{1}{2n}+\frac{1}{4n+1}\right)$$ or, $$S= 1+ \sum_{n=1}^\infty \left( \frac{1}{4n-1}-\frac{1}{4n}-\frac{1}{4n}+\frac{1}{4n+1}\right)$$ or, $$S= 1+ \sum_{n=1}^\infty \left( \frac{1}{(4n-1)(4n)}-\frac{1}{(4n)(4n+1)} \right)$$ Since rearrrangement is allowed now, we have $$S= 1+\sum_{n=1}^\infty \frac{1}{(4n-1)(4n)}-\sum_{n=1}^\infty\frac{1}{(4n)(4n+1)}$$ or, $$ S = 1+ I_1 - I_2$$ But even after many trials, I was unable to find the value of either $I_1$ or $I_2$, any help regarding this will be much appreciated.

Hint: Compare $S$ to the (termwise) sum of the following two series $$1-\frac12+\frac13-\frac14+\frac15-\cdots$$ and $$0+\frac12+0-\frac14+0+\frac16+0-\frac18+\cdots.$$ The former is "known" (depending on where are you are in the course) to be $\ln 2$, and the latter is one half of the former. — Jyrki Lahtonen, May 26 '20 at 07:45
Another way of finding the sums $I_1$ and $I_2$ is to interpret them as integrals of suitable power series. See Did's comment under an old answer of mine. Like $$I_1=\int_0^1(x^2-x^3+x^6-x^7+x^{10}-x^{11}+\cdots).$$ You need to turn that series into a rational function for this to help. — Jyrki Lahtonen, May 26 '20 at 07:50
I have corrected your typo, it should be a $+$ instead of $-$. — Culver Kwan, May 26 '20 at 08:31
@Jyrki your hint regarding the use of power series was quite helpful. But I am still unable to see how to work with your first hint. — L--, May 26 '20 at 08:35
Does this answer your question? Evaluating the sum after reordering an infinite series $1-\frac {1} {2}+\frac {1} {3}-\frac {1} {4}+\ldots $ — Culver Kwan, May 26 '20 at 08:42

J.G. · Answer 1 · 2020-05-26T10:54:49.977

2

Since $S=1+\sum_{n\ge1}\int_0^1x^{4n-2}(1-2x+x^2)dx$, by monotone convergence$$S=1+\int_0^1\frac{x^2(1-x)^2dx}{1-x^4}=1+\int_0^1\frac{x^2(1-x)dx}{(1+x)(1+x^2)}.$$You can do the rest with partial fractions:$$S=1+\left[-x+\ln(x+1)+\frac12\ln(x^2+1)\right]_0^1=\frac32\ln2.$$

edited May 26 '20 at 10:54

answered May 26 '20 at 08:54

J.G.

115,835

score 1 · Answer 2 · answered May 26 '20 at 08:50

Let $s_n$ be the sum of the first $n$ terms of the series. It is clear that if $s_{3n}$ converges, then the series converges to the same sum. $$\begin{align} s_{3n}&=\sum_{k=1}^{2n}\frac1{2k-1}-\sum_{k=1}^n\frac1{2k}\\ &=\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^{2n}\frac1{2k}-\sum_{k=1}^{n}\frac1{2k}\\ &=\sum_{k=1}^{4n}\frac1k-\frac12\sum_{k=1}^{2n}\frac1{k}-\frac12\sum_{k=1}^{n}\frac1{k} \end{align}$$

Can you continue from here? Use the standard approximation for $\sum\frac1k$.

Find sum of the series $ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$

2 Answers2