Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$

Question

Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$

My try

It can be verified that $\lim_{k \to \infty} S_{3k} < + \infty$ and $\lim_{k \to \infty} S_{3k} = \lim_{k \to \infty} S_{3k+1} = \lim_{k \to \infty} S_{3k+2}$.

So letting $a_n := S_{3n}$, $a_{n+1} - a_n = \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2} = \frac{8k+5}{(4k+1)(4k+3)(2k+2)}$.

Since $\lim (a_{n+1} - a_n) = \lim S_n - a_1$, suffice to compute $\lim_{n\to\infty}(a_{n+1} - a_n)$.

$$ \begin{aligned} \lim_{n \to \infty} (\frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2}) &=\lim_{n \to \infty} (\frac{1}{4n+1}) + \lim_{n \to \infty} (\frac{1}{4n+3}) - \lim_{n \to \infty} (\frac{1}{2n+2}) \\ &= \frac{5}{6} \end{aligned} $$

And $a_1 = S_3 = 5/6$, thus $\lim S_n = 5/3$. Am I right?

Answer 1

Write, as usual, $$H_n=\sum_{k=1}^n\frac1 k.$$ Then $$\frac11+\frac13+\frac15+\cdots+\frac1{2n-1}=H_{2n}-\frac{H_n}2.$$ In your notation, $$S_{3n}=H_{4n}-\frac{H_{2n}}{2}-\frac{H_n}{2}.$$ But $$H_n=\ln n+\gamma+O(1/n)$$ where $\gamma$ is Euler's constant, so $$S_{3n}=\ln 4n-\frac{\ln 2n}{2}-\frac{\ln n}{2}+O(1/n)=\frac{3\ln2}2 +O(1/n)$$ so the limit is $\frac32\ln2$.

Answer 2

The sum in the question can be re-written as $$ \begin{align} \sum_{k=1}^\infty\left(\frac1{4k-3}+\frac1{4k-1}-\frac1{2k}\right) &=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1{4k-3}+\frac1{4k-1}-\frac1{2k}\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^{2n}\frac1{2k}-\sum_{k=1}^n\frac1{2k}\right)\\ &=\frac12\lim_{n\to\infty}\left(\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^{2n}\frac1k\right)+\frac12\lim_{n\to\infty}\left(\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^n\frac1k\right)\\ &=\frac12\lim_{n\to\infty}\sum_{k=2n+1}^{4n}\frac nk\frac1n+\frac12\lim_{n\to\infty}\sum_{k=n+1}^{4n}\frac nk\frac1n\\ &=\frac12\int_2^4\frac1x\,\mathrm{d}x+\frac12\int_1^4\frac1x\,\mathrm{d}x\\[6pt] &=\frac12\log(2)+\frac12\log(4)\\[9pt] &=\frac32\log(2) \end{align} $$

Answer 3

It seems to me the pattern is you are adding the following triplets: $\frac 1 {4k+1} +\frac 1 {4k+3}-\frac 1 {2^{k+1}} $.

If this converged we could rearrange the terms. The infinite sum of $\sum\frac {-1} {2^{k+1}}$ is $-1$ which is finite so that would mean the sum $\sum (\frac 1 {4k+1}+\frac 1 {4k+3})= \sum \frac 1 {2k+1} $ is finite.

But it's not because it is harmonic.

So the sum does not converge.

Unless the pattern is something else.

Answer 4

\begin{eqnarray*} % \nonumber to remove numbering (before each equation) && 1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9} + \cdots\\ &=&1 + \frac{1}{3}-\frac{2}{4}+\frac{1}{5}+\frac{1}{7}-\frac{2}{8}+\frac{1}{9} + \cdots\\ &=& \int_0^1\frac{1+x^2-2x^3}{1-x^4}dx=\int_0^1\frac{(1-x)(1+x+2x^2)}{(1-x)(1+x)(1+x^2)}dx\\ &=& \int_0^1\left(\frac{1}{1+x}+\frac{x}{1+x^2}\right)dx\\ &=&\left(\ln(1+x)+\frac{1}{2}\ln(1+x^2)\right)\Big|_0^1=\frac{3}{2}\ln 2. \end{eqnarray*}