Sum of Specific Convergent Series Let L be the sum of the following alternating convergent series $$L = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$$
Now consider the rearrangement
$$1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+... = \sum_{n=1}^{\infty}\left(\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}\right)$$
Show that it sums to $\frac{3L}{2}$.
We are given 2 extra hints:


*

*the difference between $\frac{1}{2n+1}+\frac{1}{2n+3}+...+\frac{1}{4n-1}$ and $\frac{1}{2n+2}+\frac{1}{2n+4}+...+\frac{1}{4n}$ is small;

*the sum $A(n) = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ can be rewritten when considering the expression $A(n)-A(n-1)$.


I need help starting... what can I conclude from evaluating the hinted upon difference? And rewriting $A(n)$ leads to a recursion relation, with solution in the form of a harmonic partial sum, but again I can't see how it helps in any way. Thanks.

Multiply the given relation by a factor of $\frac{1}{2}$ and add it to the original sequence. Now we get:
$$L = 1 - \frac{1}{2} + \frac 13 - \frac 14 + \frac 15 - \frac 16 + \frac 17 - \frac 18 + ...$$
$$\frac 12 L = 0 + \frac 12 + 0 - \frac 14 + 0 + \frac 16 + 0 - \frac 18+ ...$$

$$\frac 32 L = 1 + 0 + \frac 13 - \frac 12 + \frac 15 + 0 + \frac 17 - \frac 14 +...$$
$$\frac 32 L = 1 + \frac 13 - \frac 12 + \frac 15 + \frac 17 - \frac 14 +... = \sum_{n=1}^{\infty}\left(\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}\right)$$
Additionally it's a well-known fact that $L = \ln(2)$. Also this kind of an rearragements are possible due to the fact that the series is conditionally convergent. Actually due to the Riemann Series Theorem we can rearrange any conditionally sequence in such a manner that it will converge to any giver number in $\mathbb{R}$
