Why is the value of the series representation for $\arctan$ at $ x = 1$ necessarily $\arctan(1)$

Question

Exercise 6.6.1. The derivation in Example 6.6.1 shows the Taylor series for $\arctan(x)$ is valid for all $x \in (−1,1)$. Notice, however, that the series also converges when $x = 1$. Assuming that $\arctan(x)$ is continuous, explain why the value of the series at $x = 1$ must necessarily be $\arctan(1)$. What interesting identity do we get in this case?

This is a question from Abbott's Understanding Analysis. My question is what exactly constitutes a proof that the series equals the function at a point.

Answer 1

In what follows all letters in various equations represent real numbers.

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The series formula $$\frac{1}{1+x^2}=1-x^2+x^4-x^6+\dots \tag{1}$$ holds only when $x\in(-1,1)$ but the above result when integrated with respect to $x$ leads to the formula $$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\dots\tag{2}$$ and unlike the series in $(1)$ the above result is true not only when $x\in(-1,1)$ but also for $x=\pm 1$. You want to know why?

This is a good question and +1 for the same. The explanation is based on the proper derivation of $(2)$ from $(1)$. We start with the algebraic identity $$1-t^2+t^4-\dots+(-1)^{n-1}t^{2n-2}=\frac{1-(-1)^nt^{2n}}{1+t^2}$$ (just use formula for sum of a finite geometric progression) and recast it in the form $$\frac{1}{1+t^2}=1-t^2+t^4-\cdots+(-1)^{n-1}t^{2n-2}+(-1)^n\frac{t^{2n}}{1+t^2}\\=1-t^2+\dots+(-1)^{n-1}t^{2n-2}+(-1)^nR_n(t)\tag{3}$$ where $R_n(t) =t^{2n}/(1+t^2)$. Note that $\lim_{n\to\infty} R_n(t) =0$ if and only if $t\in(-1,1)$ which explains the fact that the formula $(1)$ holds only when $x\in(-1,1)$.

Integrating the identity $(3)$ with respect to $t$ on interval $[0,x]$ we get $$\arctan x=x-\frac{x^3}{3}+\dots+(-1)^{n-1}\frac{x^{2n-1}}{2n-1}+(-1)^{n}\int_{0}^{x}R_n(t)\,dt$$ and by almost a stroke of luck the expression $\int_{0}^{x}R_n(t)\,dt$ tends to $0$ as $n\to\infty$ when $x$ lies not only in $(-1,1)$ but also when $x=\pm 1$. Since $R_n(t) $ is even function of $t$ its integral is an odd function and changing sign of $x$ changes the sign of the whole integral. Thus it is sufficient to deal with positive values of $x$ only. If $0\leq x\leq 1$ and $0\leq t\leq x$ then $$0\leq R_{n} (t) \leq t^{2n}$$ and therefore on integrating we get $$0\leq \int_{0}^{x}R_{n}(t)\,dt\leq \frac{x^{2n+1}}{2n+1}\leq \frac{1}{2n+1}$$ and hence the integral tends to $0$ as $n\to\infty$ for all $x\in[-1,1]$ and the formula $(2)$ is therefore true for $x=\pm 1$ also.

The same conclusion can be reached using Taylor's theorem also by analyzing the remainder for $x=\pm 1$ and you may try that (it should be the same as given in Example 6.6.1 in your book and not too difficult). The proof based on the geometric series and its integration (as given above) is much simpler and my preferred approach to the arctan series.

This is a special case of a more general theorem about power series given by the famous mathematician Abel:

Abel's Theorem: Let $\sum_{n=0}^{\infty}a_nx^n$ be a power series with radius of convergence $R$. If the series also converges for $x=R$ then we have $$\lim_{x\to R^{-}} \sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty}a_nR^n$$ A similar conclusion holds if the series converges for $x=-R$.

Answer 2

Proof. Since $arctan(x)$ is continuous, for all $\epsilon > 0 $ there exists some $\delta > 0 $ such that $|1-x| < \delta $ implies that $|arctan(1) - arctan(x)| < \epsilon $. We also know that $$arctan(x) = x - \frac 1 3 x^3 + \frac 1 5 x^5 - \frac 1 7 x^7 + \cdots $$ when $ x \in (-1, 1)$. Hence $|1-x| < \delta $ implies that $$|arctan(1) - ( x - \frac 1 3 x^3 + \frac 1 5 x^5 - \frac 1 7 x^7 + \cdots )| < \epsilon $$ for all $\epsilon>0$. Hence $$arctan(1) = 1 - \frac 1 3 1^3 + \frac 1 5 1^5 - \frac 1 7 1^7 + \cdots$$ as desired.