As we know the expansion of $\ln(1+x)$ is as follows:
$\ln (1+x) = x - x^2/2 + x^3/3 - x^4/4+\cdots$
Let $S_1 = 1 + 1/3 + 1/5 + 1/7+\cdots$
Let $S_2 = 1/2 + 1/4 + 1/6 + 1/8+\cdots$
$S_1 - S_2 = \ln(2) -----------------(1)$
$S_1 + S_2 = 1 + 1/2 + 1/3 + 1/4 + 1/5+\cdots$
$S_1 + S_2 = 2(1/2 + 1/4 + 1/6 + 1/8+\cdots)$
$S_1 + S_2 = 2S_2$
$ S_1 = S_2 -------------------- (2)$
From (1) and (2):
$\ln(2) = 0$
What did I do wrong?
On
There is a theorem commonly known as Riemann rearrangement theorem (I believe) that states the following; If a series $\sum a_n$ converges but does not converge absolutely, it can be rearranged to yield any sum. [Note: Absolute convergence means the series $\sum|a_n|$ converges] The series you chose converges but does not converge absolutely. The divergence of second series $S_2$ is well-known. Hence, $S_1$ can be rearranged to yield any result. If you want to understand exactly how that construction is possible, I have attached the proof here.
Say $p_n=\frac{a_n+|a_n|}{2}$ and $q_n=\frac{a_n-|a_n|}{2}$. $\therefore, p_n-q_n=|a_n|, p_n+q_n=|a_n|$. $\sum p_n, \sum q_n$ cannot both be convergent otherwise $\sum(p_n+q_n)=\sum|a_n|$ would be convergent. Also, since $\sum a_n=\sum p_n-\sum q_n$ is convergent, and both series $\sum p_n, \sum q_n$ are not convergent, this implies they are both divergent. Now let $P_1, P_2,....$ be the first non-negative terms of $\sum a_n$, and $N_1, N_2, ....$ denote the absolute values of the first negative terms of $\sum a_n$, then the series $\sum P_n, \sum N_n$ have the same elements as the series $\sum p_n, \sum q_n$ in the same order. $\implies \sum P_n, \sum N_n$ are also divergent. Say you want to rearrange the series $\sum a_n$ to yield a result in the neighborhood $[x,y]$ such that $-\infty\le x\le y\le \infty$. Then choose $P_1, P_2,...P_{n_1}$ such that $P_1+P_2...P_{n_1}\to y$. Choose $Q_1, Q_2,...Q_{k_1}$ such that $P1+P_2...P_{m_1}-Q_1-Q_2-....Q_{k_1}\to x$. The existence of such $P, Q$ are guaranteed since $\sum P_n, \sum Q_n$ are divergent.
The taylor-series expansion of $\ln (1+x)$ is not absolutely convergent in nature. Hence you cannot perform manipulations with its expanded infinite series. You may rearrange $S_1$ and $S_2$ to get infinitely many different answers.