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

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