My main question is the one in the title, however I was also wondering, in general when it comes to infinite series, how can you find out whether the series converges to a value or not? And can you tell if there will be something strange about it? What I mean is that you would expect the sum of all natural numbers to be infinity, but it is -1/12. Is there a way of knowing if something like this will happen?

Sorry for the ton of questions! And thank you for any answers :)

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    $\begingroup$ en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80 $\endgroup$ – lemon Sep 2 '14 at 14:35
  • $\begingroup$ To answer one of them, the series of the title of your question is convergent due to this test. $\endgroup$ – fuglede Sep 2 '14 at 14:35
  • $\begingroup$ There is usually a lecture in the math curriculum of any university which covers convergence of series. If you are not studying math at university you could try learning it on Khan Academy. $\endgroup$ – Patrick Da Silva Sep 2 '14 at 14:37
  • $\begingroup$ Regarding the sum of all natural numbers - it does not converge to -1/12, it diverges to infinite (so there's nothing "strange" about it). (What you're referring to is the Ramanujan summation) $\endgroup$ – lemon Sep 2 '14 at 14:39
  • $\begingroup$ What do you exactly mean by you would expect the sum of all natural numbers to be infinity, but it is -1/12? There are methods to sum a divergent series, while the partial sum do not converge, but it's a different matter than converging series like the one in title. $\endgroup$ – Jean-Claude Arbaut Sep 2 '14 at 14:39

Recall that $$ \frac{1}{1+x^2}=1-x^2+x^4+\cdots $$ If you integrate from $0$ to $1$ $$ \begin{align}\int_0^1\frac{1}{1+x^2} dx& =\int_0^11-x^2+x^4-x^6+\cdots\,dx\\ & =1-\frac13+\frac15-\frac17+\cdots \end{align} $$ but $$\int_0^1\frac{1}{1+x^2} dx=\arctan 1-\arctan 0 = \frac\pi4, $$ then $$ 1-\frac13+\frac15-\frac17+\frac19-\cdots=\frac\pi4 $$

  • $\begingroup$ Hahaha- I was initially trying to calculate what the [pi/2 x integral of 1/(1+x^2) from -pi/4 to pi/4] was and this led me to formulate that series! :) I'm trying to find the integarl without knowing that it is arctan1.... $\endgroup$ – Meep Sep 2 '14 at 14:56

An other way: $$\arctan x=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^n.$$ if $|x|<1$. Morevoer $$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$$ is an Alternating series, then it converge. By Abel's theorem you can conclude that $$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}=\arctan(1)=\frac{\pi}{4}.$$


$$\sum\limits_{n = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{2n + 1}}} = \frac{\pi }{4}$$

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    $\begingroup$ Your answer is not constructive. You have to explain why we have this equality ! $\endgroup$ – idm Sep 2 '14 at 14:55

$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+ \dots =\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$$

Have you got taught the Dirichlet's criterion?

Use this,and you prove that your infinite sum converges.

  • $\begingroup$ It should be put as a comment ! $\endgroup$ – idm Sep 2 '14 at 14:45
  • $\begingroup$ Why should I put it as a comment??? $\endgroup$ – user159870 Sep 2 '14 at 14:45
  • $\begingroup$ because it's a remark and not an answer ! $\endgroup$ – idm Sep 2 '14 at 14:46
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    $\begingroup$ I'm not familiar with this, but I'll look up the proof- thanks! :) $\endgroup$ – Meep Sep 2 '14 at 14:59

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