I have found an exercise in an exam; however, I don't know where to start. Find the sum of the series:
$$\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n+1)} $$
I just learned how to do it with polynomials in the numerator.
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Do you know about Taylor series? – saulspatz May 02 '19 at 14:21
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for arctan, in particular? – J. W. Tanner May 02 '19 at 14:22
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1Have a look at https://math.stackexchange.com/questions/1238780/is-my-proof-that-frac-pi4-sum-limits-n-geq-0-1n-frac12n1-corr. – Martin R May 02 '19 at 14:29
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Hint:
Consider the function
$$f(x)=\sum_{n=0}^\infty\frac{x^n}{2n+1}$$ which converges in $[-1,1)$. It reminds of a geometric series, but there are those annoying denominators.
From there, consider
$$xf(-x^2)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1},$$ where the exponents are the same as the denominators. You can differentiate term-wise to get a true geometric series, that you can sum:
$$(xf(-x^2))'=\sum_{n=0}^\infty(-1)^nx^{2n}=\frac 1{1+x^2}.$$
Now integrating from $0$ to $1$,
$$\sum_{n=0}^\infty\frac{(-1)^n1^{2n+1}}{2n+1}=\arctan 1.$$