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I was exploring the Wikipedia page on Geometric series wherein they've used the formula for sum to compute the power series for $\tan^{-1}x$. Geometric Power Series -Wikipedia

$$\int \dfrac{1}{1+x^2}\mathrm dx=\int\sum_{k=0}^{\infty}(-1)^kx^{2k}\mathrm dx=\sum_{k=0}^{\infty}\dfrac{(-1)^k}{2k+1}x^{2k+1}$$

The computation of the integral of the sum makes perfect sense to me but isn't the formula for sum namely $S_n=\displaystyle\sum_{i=0}^{\infty}ar^i=\left(\dfrac{a}{1-r}\right)$ valid when $\mid r\mid \lt1$.


Why do we simply neglect the fact that the sum converges iff $\mid r\mid \lt1$? Or does it mean that the Power Series for $\tan^{-1} x$ would converge for $\mid x\mid \lt1$? Thanks in anticipation.

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  • $\begingroup$ Why the downvote? $\endgroup$ – Paras Khosla Mar 5 at 15:38
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The argument is valid only for $|x| <1$. Also the series has radius of convergence $1$ and $\tan ^{-1} x$ does not have a power series expansion around $0$ for $|x| >1$.

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  • $\begingroup$ Based on this fact, sir can we say that this power series representation is valid only for $\mid x \mid \lt 1$. If that is so why can we replace the general integral $\tan^{-1}x$(valid $\forall \ x$) with something that only converges for a particular range of $x$ values? $\endgroup$ – Paras Khosla Feb 27 at 7:02
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    $\begingroup$ We can't. It's a common abuse of notation to write that a function is equal to its Taylor series; strictly speaking this only holds inside the radius of convergence. $\endgroup$ – Qiaochu Yuan Feb 27 at 7:03
  • $\begingroup$ Thanks @QiaochuYuan for clarifying. Thanks Prof. Murthy. $\endgroup$ – Paras Khosla Feb 27 at 7:05
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    $\begingroup$ More generally, an implicit norm people are often using when they write down identities involving series or other infinite constructions is that the identity only holds when the series converge. $\endgroup$ – Qiaochu Yuan Feb 27 at 7:06
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The series which has been used here is of a special kind called power series of the form $$\sum_{n=0}^{\infty}a_nx^n$$ and such series have associated with them a number $R\geq 0$ called radius of convergence. The series is convergent if $|x|<R$ and divergent if $|x|>R$. The case for $|x|=R$ has to be handled for each series individually. Also there are cases when $R=\infty$ so that the series converges for any given value of $x$.

A key property of such series is that they can be integrated or differentiated term by term inside the region of convergence $|x|<R$ (with new series having same radius of convergence). However there can be cases (like in current question) where the given series is convergent only when $|x|<R$ but the new series after integration also converges for $|x|=R$.

It is much simpler to handle your specific series directly by noting that $$\arctan x=\int_{0}^{x}\frac{dt}{1+t^2}=\int_{0}^{x}\left(1-t^2+t^4-\dots+(-1)^{n-1}t^{2n-2}+(-1)^{n}\frac{t^{2n}}{1+t^2}\right)\,dt$$ which leads to $$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots +(-1)^{n-1}\frac{x^{2n-1}}{2n-1}+(-1)^nR_n\tag{1}$$ where $$R_n=\int_{0}^{x}\frac{t^{2n}}{1+t^2}\,dt$$ The identity $(1)$ holds for all $x\in\mathbb {R} $ but the expression $R_{n} \to 0$ as $n\to\infty $ only when $|x|\leq 1$ (prove this!) and hence the identity $$\arctan x=\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}$$ is valid only when $|x|\leq 1$.

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  • $\begingroup$ Thanks for the insightful answer. Cheers :)) $\endgroup$ – Paras Khosla Mar 26 at 9:27
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    $\begingroup$ @ParasKhosla: a similar analysis can be done for series of $\log(1+x)$ as integral of $1/(1+t)$ on interval $[0,x]$ and in that case the result holds only when $-1<x\leq 1$. You should grab a copy of Hardy's A Course of Pure Mathematics for dealing with such fine points of calculus. $\endgroup$ – Paramanand Singh Mar 26 at 9:30

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