# 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? Mar 5, 2019 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? Feb 27, 2019 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. Feb 27, 2019 at 7:03
• Thanks @QiaochuYuan for clarifying. Thanks Prof. Murthy. Feb 27, 2019 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. Feb 27, 2019 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 :)) Mar 26, 2019 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. Mar 26, 2019 at 9:30