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

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?

• I don't follow a few things. One of them is where the sum of three limits that are each $0$ is equal to $\frac{5}{6}$. Commented Oct 23, 2018 at 4:38
• Actually, $$1+1/3-1/2+1/5+1/7-1/4+1/9+1/11-1/8+1/13+1/15-1/16+\cdots=\infty$$
– bof
Commented Oct 23, 2018 at 5:02
• Um. What's the pattern. How do you know when to add and when to subtract? And how do you know what to add or subtract? Is the 1/4 a typo? Do you always add two and subtract one? Commented Oct 23, 2018 at 6:26

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 $$\frac32\ln2$$.

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}

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.

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