# Using geometric series for computing Integrals

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.

• Why the downvote? – Paras Khosla Mar 5 at 15:38

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 $$|x| >1$$.

• 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? – Paras Khosla Feb 27 at 7:02
• 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. – Qiaochu Yuan Feb 27 at 7:03
• Thanks @QiaochuYuan for clarifying. Thanks Prof. Murthy. – Paras Khosla Feb 27 at 7:05
• 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. – Qiaochu Yuan Feb 27 at 7:06

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 $$|x| and divergent if $$|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 $$|x| (with new series having same radius of convergence). However there can be cases (like in current question) where the given series is convergent only when $$|x| 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 $$|x|\leq 1$$.

• Thanks for the insightful answer. Cheers :)) – Paras Khosla Mar 26 at 9:27
• @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. – Paramanand Singh Mar 26 at 9:30