Hi,
I am trying to prone that the sum of the following series is log3.
1+1/2-2/3+1/4+1/5-2/6+1/7+1/8-2/9+1/10+1/11-2/12...
Thank's in advance.
Let $\displaystyle H_n$ be the $\displaystyle n$-th Harmonic number (the sum of the first $\displaystyle n$ terms of the Harmonic series). Consider partial sums of three terms of your series at a time. Show that $\displaystyle S_n = H_{3n}-H_n$ gives the correct partial sums for your series. So, your series is given by $\displaystyle \lim_{n \to \infty}S_n$.
Next, use common inequalities for the Harmonic numbers. For example, Young (1991) showed that $\displaystyle \dfrac{1}{2n+2} < H_n - \ln n - \gamma < \dfrac{1}{2n}$. So,
$\displaystyle \dfrac{1}{6n+2} < H_{3n} - \ln (3n) - \gamma < \dfrac{1}{6n}$ and $\displaystyle -\dfrac{1}{2n} < \gamma + \ln n - H_n < -\dfrac{1}{2n+2}$
Adding these inequalities together gives:
$\displaystyle \dfrac{1}{6n+2} - \dfrac{1}{2n} < H_{3n} + \gamma + \ln n - \ln (3n) - \gamma - H_n < \dfrac{1}{6n} - \dfrac{1}{2n+2}$
Simplifying gives:
$\displaystyle -\dfrac{2n+1}{2n(3n+1)} < S_n - \ln 3 < \dfrac{1-2n}{6n(n+1)}$
Finally, use the Squeeze Theorem to show that $\displaystyle \lim_{n \to \infty} (S_n - \ln 3) = 0$, so $\displaystyle \lim_{n \to \infty} S_n = \ln 3$
Btw, to show that $\displaystyle S_n = H_{3n} - H_n$, you can write your series as follows:
$\displaystyle \begin{align*}1+\frac{1}{2} -\frac{2}{3} +\cdots + \frac{1}{3n-2} + \frac{1}{3n-1} - \frac{2}{3n} & = \sum_{k= 1}^n \left(\dfrac{1}{3k-2} + \dfrac{1}{3k-1} -\dfrac{2}{3k}\right) \\ & = \sum_{k= 1}^n \left(\dfrac{1}{3k-2} + \dfrac{1}{3k-1} + \dfrac{1}{3k} -\dfrac{3}{3k}\right) \\ & = \sum_{k= 1}^n \left(\dfrac{1}{3k-2} + \dfrac{1}{3k-1} + \dfrac{1}{3k}\right) -\sum_{k=1}^n\dfrac{1}{k} \\ & = \sum_{k=1}^{3n}\dfrac{1}{k} - \sum_{k=1}^n \dfrac{1}{k} \\ & = H_{3n} - H_n\end{align*}$
I'm curious what course (if any) this would be learned in. I've had a fair bit of math and never ran into a detailed analysis of the Harmonic series like this.
Is this something you just naturally come across post-grad in the course of solving other problems?
Are you asking hedi or me? If you are asking me, then this comes about from examining Euler's constant, which is $\displaystyle \gamma = \lim_{n \to \infty} H_n - \ln n$. Young was showing how quickly the sequence $\displaystyle H_n - \ln n$ converges to $\displaystyle \gamma$.
Actually, I just thought of a more direct way to show that the series converges to $\displaystyle \ln 3$. By definition, $\displaystyle \gamma := \lim_{n \to \infty} (H_n - \ln n)$, so
$\displaystyle \begin{align*}0 & = \gamma - \gamma \\ & = \lim_{n \to \infty} \left(H_{3n} - \ln (3n) \right) - \lim_{n \to \infty} \left( H_n - \ln n \right) \\ & = \lim_{n \to \infty} \left( H_{3n} - H_n - \ln 3 \right)\end{align*}$