I want to calculate the series: $$ \sum_{n=0}^\infty\frac{(-1)^n}{2n+1}~=\frac{\pi}{4} $$ which converges by Dirichlet. Since $\arctan x=\int_0^x\frac1{1+t^2}dt$, and for $x\in[0,1)$, we have $$ \frac{1}{1+x^2}=\sum_{n=0}^\infty(-x^2)^n=\sum_{n=0}^\infty(-1)^nx^{2n} $$ So $$ \arctan x=\int_0^x\sum_{n=0}^\infty(-1)^nx^{2n} $$ holds for every $x\in[0,1)$. By M-test, the power series $$ \sum_{n=0}^\infty(-1)^nx^{2n} $$ converges uniformly to $\frac1{1+x^2}$ on the closed interval $[0,1-\epsilon]$, where $\epsilon\in(0,1)$. So we can interchange the order of taking integral and taking limit, which means $$ \arctan x=\sum_{n=0}^\infty(-1)^n\int_0^xt^{2n}dt=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1} $$ Let $x\to1^-$ to get $$ \frac{\pi}{4}=\lim_{x\to1^-}\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1} $$ If we can prove $$ \lim_{x\to1^-}\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1} $$ then we are done. Actually, it is not the first time that I have met such a problem: take the limit at exactly the convergence radius. Please help me with this last step. Thanks a lot!
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Are you familiar with the conditions under which you can interchange the limits and summation? EG See this. – Calvin Lin Mar 08 '23 at 05:54
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But I have no idea how it can be applied here – SlInevitable2003 Mar 08 '23 at 06:03
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I edited your question to somewhat improve formatting (though fixing the title was my main goal). I hope the changes are reasonable. If not, remember that you can change things back or roll back the edit. – Christian E. Ramirez Mar 08 '23 at 08:05
1 Answers
You want Abel's Theorem, which states if that $$ A(x) \; = \; \sum_{n=0}^\infty a_n x^n $$ has radius of convergence $1$, and if $$ \alpha \; = \; \sum_{n=0}^\infty a_n $$ exists, then $$ \lim_{x \to 1-}A(x) = \alpha $$
An outline of the proof is as follows:
We can write $$ (1-x)^{-1}[A(x) - \alpha] \; = \; \sum_{n=1}^\infty (s_n -\alpha)x^n \hspace{1cm} |x| < 1 $$ where $s_n = \sum_{m=1}^n a_m$. But then $$ (1-x)^{-1}|A(x)-\alpha| \; \le \; \sup_{n \ge 1}|s_n-\alpha| \frac{1-x^N}{1-x} + \sup_{n \ge N}|s_n - \alpha|\frac{x^N}{1-x} $$ for $N \ge 1$ and $0<x<1$, so that $$ |A(x)-\alpha| \;\le \; \sup_{n \ge 1}|s_n-\alpha|(1-x^N) + \sup_{n \ge N}|s_n - \alpha| $$ for $N \ge 1$ and $0<x<1$. Given $\varepsilon > 0$ find $N$ to make the second term smaller than $\frac12\varepsilon$, and then use continuity of $x^N$ to find $\delta > 0$ so that the first term is smaller than $\frac12\varepsilon$ for $1-\delta < x < 1$.
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