$$a_n=(-1)^{n+1}$$ $$S_n={1-1+1-1+...(-1)^{n+1}}$$ If $S_n$ sums $S_n$ in the following order:$a_1+(a_2+a_1)+(a_3+a_2)+(a_4+a_3)+...(a_{n+1}+{a_n})$

Then $$2S_n=1+(-1+1)+(1-1)+(-1+1)+...0$$ $$2S_n=1$$ So $S_n$ converges to $1/2$.

But we know that $a_n$ doesn't converge to $0$, and so the infinite sum of $a_n$ should not converge by the nth term test for divergence. So what's incorrect in the above reasoning?

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    $\begingroup$ $2S_n$ is either $0$ or $2$ depending on parity of $n$ where do you get $a_{n+1}$ in the sum? I think you are assuming $S_n=S_{n+1}$ but that's not true $\endgroup$ – user160738 Jan 26 '16 at 18:46
  • $\begingroup$ The sum can be summed to $\frac 12$ by standard summation method (Cesaro, Abel, etc.) so you are not that wrong in the end ;) $\endgroup$ – Renato Faraone Jan 26 '16 at 19:17

you have shown that $S_n+S_{n-1}=1$, not that $2S_n=1$. Otherwise, what happened to the second $a_{n+1}$? If you do it correctly, you get $2S_n = 1+(-1)^{n+1}$.

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