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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?

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    $\begingroup$ Riemann series theorem loosely paraphrased says that if you rearrange terms of a conditionally convergent series, you can change the resulting sum to whatever you want. You rearranged the terms and so got a different sum. $\endgroup$ – JMoravitz Jan 15 '20 at 15:03
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    $\begingroup$ $S1$ and $S2$ diverge. It makes no sense to subtract them. $\endgroup$ – Andrés E. Caicedo Jan 15 '20 at 15:05
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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.

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  • $\begingroup$ doesn't $\ln(1+x)$ converge to $\ln 2$ when $x=1$? $\endgroup$ – J. W. Tanner Jan 15 '20 at 15:14
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    $\begingroup$ Sam probably left out the crucial word "...is not absolutely convergent in nature..." $\endgroup$ – JMoravitz Jan 15 '20 at 15:15
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    $\begingroup$ This is a tad misleading as phrased. The Taylor series expansion of $\ln(1+x)$ is absolutely convergent within its interval of convergence, and that is the standard way of understanding the sentence as written. It would be best to add at the end of the first sentence "when $x=1$". $\endgroup$ – Andrés E. Caicedo Jan 15 '20 at 15:29
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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.

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