# Calculating the infinite series $1-\frac13+\frac15-\frac17+\frac19-\frac1{11}\cdots$

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 :)

• en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80 – lemon Sep 2 '14 at 14:35
• To answer one of them, the series of the title of your question is convergent due to this test. – fuglede Sep 2 '14 at 14:35
• 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. – Patrick Da Silva Sep 2 '14 at 14:37
• 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) – lemon Sep 2 '14 at 14:39
• 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. – 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 $$1-\frac13+\frac15-\frac17+\frac19-\cdots=\frac\pi4$$

• 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.... – 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 $$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}=\arctan(1)=\frac{\pi}{4}.$$

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

• Your answer is not constructive. You have to explain why we have this equality ! – 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.

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