0
$\begingroup$

$$a_n=(-1)^{n+1}$$ $$S_n={1-1+1-1+...(-1)^{n+1}}$$ 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?

$\endgroup$
  • 4
    $\begingroup$ $2S_n$ 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
5
$\begingroup$

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 $2S_n = 1+(-1)^{n+1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.