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 :)
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$\begingroup$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 $$ 1-\frac13+\frac15-\frac17+\frac19-\cdots=\frac\pi4 $$
$\endgroup$ 1 $\begingroup$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 $$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}=\arctan(1)=\frac{\pi}{4}.$$
$\endgroup$ $\begingroup$$$\sum\limits_{n = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^n }}{{2n + 1}}} = \frac{\pi }{4}$$
$\endgroup$ 1 $\begingroup$$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+ \dots =\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$$
Have you got taught the Dirichlet's criterion?
Use this,and you prove that your infinite sum converges.
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