# Thread: Prove Convergence to (3/2)ln2

1. ## Prove Convergence to (3/2)ln2

Hello,
I have to prove that
1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + 1/11 - 1/6 + ...
converges to (3/2)ln2 (which is also equal to ln8/2 if it is easier to work with).
I have no idea what to do with this, as the alternating harmonic series has conditional and not absolute convergence. How would I go about actually proving this convergence?

2. Try to write the sum with the form $\sum_k a_k$. I think we will have to use the result $\displaystyle \sum_{k=1}^{+\infty}\frac{(-1)^k}k=\ln 2$.

3. Let's indicate $s_{j k}$ the sum...

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} +... + \frac{1}{j k}$ (1)

... $t_{2 j k}$ the sum...

$\displaystyle 1 - \frac{1}{2} + \frac{1}{3} +... + \frac{1}{2 j k-1} - \frac{1}{2 j k}$ (2)

... and $u_{3k}$ the sum...

$\displaystyle 1 + \frac{1}{3} - \frac{1}{2} +\frac{1}{5}+ \frac{1}{7} - \frac{1}{4}... + \frac{1}{4k-3} + \frac{1}{4k-1} - \frac{1}{2k}$ (3)

With a little of patience You can verify that is...

$\displaystyle u_{3n} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{4n} - \frac{1}{2}\ (1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n}) -$

$\displaystyle - \frac{1}{2}\ (1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}) = s_{4n}- s_{2n} + \frac{1}{2}\ (s_{2n} - s_{n})$ (4)

But is also...

$\displaystyle t_{2n} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n} - 2\ (1 + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{2n}) =$

$\displaystyle = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n} - (1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}) = s_{2n} - s_{n}$ (5)

Comparing (4) and (5) You obtain that is...

$\displaystyle u_{3n}= t_{4n}+ \frac{1}{2}\ t_{2n}$ (6)

... and now You have to find the $\displaystyle \lim_{n \rightarrow \infty}$ of (6)...

Kind regards

$\chi$ $\sigma$