# Infinite Series $1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$

Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of $\ln(1+x)$ and $\arctan(x)$: $$1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$$

The question isn't homework or anything, just a thought tease. I tried for a long while but couldn't find anything remotely close. Thanks in advance for the help.

Rewrite $-\frac23$ as $\frac13-1$, $-\frac26$ as $\frac16-\frac12$, and in general, $-\frac{2}{3n}$ as $\frac1{3n}-\frac1n$. Then the sum to $3n$ terms is:

$$1 + \frac12 + \frac13 + \frac14 + \cdots + \frac1{3n} - \left(1 + \frac12 + \frac13 + \frac14 + \cdots + \frac1n\right) = \sum_{r=n+1}^{3n}\frac1r$$

This tends to $$\int_n^{3n}\frac{dx}x = \log 3$$

as $n \to\infty$.

Since $$\lim_{n\to\infty}\left(1+\frac1 2+\frac1 3+\frac1 4+\frac1 5+\frac1 6+...+\frac1 n-\log(n)\right)=\gamma$$ ($\gamma$ is Euler constant, for more information see here.)

we see that $$\gamma=\lim_{n\to\infty}\left(\color{\red}{1}+\color{\red}{\frac1 2}+\color{#0000ff}{\frac1 3}+\color{\red}{\frac1 4}+\color{\red}{\frac1 5}+\color{#0000ff}{\frac1 6}+...+\color{\red}{\frac1 {3n-2}}+\color{\red}{\frac1 {3n-1}}+\color{#0000ff}{\frac1 {3n}}-\log(3n)\right)$$ and $$\gamma=\lim_{n\to\infty}\color{#0000ff}{3\left(\frac1 3+\frac1 6+...+\frac1 {3n}\right)}-\log(n)$$ Subtracting, we obtain $$0=\lim_{n\to\infty}\left(1+\frac1 2-\frac2 3+\frac1 4+\frac1 5-\frac2 6++-...+\frac1{3n-2}+\frac1{3n-1}-\frac2 {3n}\right)-\log(3)$$ which implies that $$1 + \frac12 - \frac23 + \frac14 + \frac15 - \frac26 + ...=\log(3)$$

This is a direct proof using $$\int_0^1 x^n dx= \frac{1}{n+1},$$ the sum of a geometric progression and integral evaluation.

\begin{align} S&=\sum_{k=0}^\infty \left(\frac{1}{3k+1}+\frac{1}{3k+2}-\frac{2}{3k+3}\right)\\ &=\sum_{k=0}^\infty \int_0^1 \left(x^{3k}+x^{3k+1}-2x^{3k+2}\right)dx\\ &=\int_0^1 \sum_{k=0}^\infty \left(x^{3k}+x^{3k+1}-2x^{3k+2}\right)dx\\ &=\int_0^1 \frac{1+x-2x^2}{1-x^3}dx\\ &=\int_0^1 \frac{1+2x}{1+x+x^2}dx\\ &=\log(1+x+x^2)|_{0}^1\\ &=\log(3) \end{align}

Thus, similarly to

$$\log(1+x)=1-\frac{1}{2}x+\frac{1}{3}x^2-\frac{1}{4}x^3...$$

we have

$$\log(1+x+x^2)=1+\frac{1}{2}x-\frac{2}{3}x^2+\frac{1}{4}x^3+\frac{1}{5}x^4-\frac{2}{6}x^5+...$$

Moreover, the pattern generalizes.

Alternatively, note that $$S:=1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\ldots=\sum_{k=1}^\infty\,\frac{-\omega^{k}-\bar{\omega}^{k}}{k}\,,$$ where $$\omega:=\exp\left(+\frac{2\pi\text{i}}{3}\right)=\frac{-1+\sqrt{3}\text{i}}{2}\,,\text{ so }\bar{\omega}=\exp\left(-\frac{2\pi\text{i}}{3}\right)=\frac{-1-\sqrt{3}\text{i}}{2}\,.$$ Since the logarithmic power series $$\ln(1-z)=\displaystyle- \sum_{k=1}^\infty\,\frac{z^k}{k}$$ converges for all $$z\in\mathbb{C}\setminus\{1\}$$ such that $$|z|\leq 1$$, we get $$S=\ln(1-\omega)+\ln(1-\bar{\omega})=\left(\frac{1}{2}\,\ln(3)-\text{i}\,\frac{\pi}{6}\right)+\left(\frac{1}{2}\,\ln(3)+\text{i}\,\frac{\pi}{6}\right)\,,$$ making $$S=\ln(3)$$.

As every third term is "annoying", let us restore a know pattern,

$$S:=1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots=\left(1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots\right)-\left(\frac33+\frac36\cdots\right).$$

It is tempting to simplify and deduce

$$S=\left(1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots\right)-\left(1+\frac12\cdots\right)=\color{red}{0},$$

but this would be wrong as the sum is not absolutely summable and changing the order of the terms is not allowed.

The right way (given by others) is by evaluating the partial sums with

$$S_{3n}=H_{3n}-H_n$$whicht tends to$$(\ln(3n)-\gamma)-(\ln(n)-\gamma)=\color{green}{\ln(3)}.$$

The reason why the simplification doesn't work is that the terms of the second series are more "sparse". In the partial sum approach, the sparsity is respected.