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I can convert $1 + 2 + 3 + 4 + 5 + ... = -\frac {1}{12}$ to summation notation: $$\sum_{n = 1}^\infty n = -\frac {1}{12}$$ But, how can I convert the following series to summation notation: $$1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + ... = \frac 12$$ I tried using the WolframAlpha Summation calculator (I input $1 - 1 + 1 - 1$ until some point), but it kept returning 0. One idea I have is to use summation notation without any bounds, but I doubted myself on that.

Is there a way to convert the above series to summation notation?

Any answers are appreciated.

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    $\begingroup$ What about $\sum_{n=1}^\infty (-1)^n$? $\endgroup$ – Henricus V. Feb 13 '16 at 1:52
  • $\begingroup$ You have a sum of alternating positive and negative ones.... note that $(-1)^n $ bounces from positive to negative as we go from even to odd integers $\endgroup$ – Eleven-Eleven Feb 13 '16 at 1:52
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$1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + ...=\sum \limits_{n=0}^{\infty} (-1)^n=\sum \limits_{n=1}^{\infty} (-1)^{n+1}=\sum \limits_{n=0}^{\infty} \cos(\pi n)$

Also, $\sum \limits_{n = 1}^\infty n = -\frac {1}{12}$ in the context of $\zeta(-1)$. See here for more information.

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    $\begingroup$ also $\sum_{n=1}^{\infty}sin(\pi/2+\pi n)$ if for some reason we're coming up with fun ways. $\endgroup$ – Noah Harris Feb 13 '16 at 2:13
  • $\begingroup$ @noah very interesting! $\endgroup$ – zz20s Feb 13 '16 at 2:29

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