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

Question

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.

Answer 1

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 $n \to\infty$.

Answer 2

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 $$1 + \frac12 - \frac23 + \frac14 + \frac15 - \frac26 + ...=\log(3)$$

Answer 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.