Integration and differentiation method for power series summation

Description of the method

This is a general method used for summing up power series. The general idea is to consider a power series of the form:

$\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k$

Instead of trying to sum it up directly, we try either of these:

• Differentiate the power series, find the sum of that, and then integrate the function obtained, choosing the antiderivative whose value at 0 equals the constant term of the power series; OR
• Integrate the power series, find the sum of that, and then differentiate the function obtained,

Examples

For these examples, we do not worry for the moment about the interval of validity, but simply try to compute the sums formally.

Summing up the derivative power series

Power series	Derivative power series	Function that the derivative power series converges to	Antiderivative of that function whose value at 0 is the constant term of the power series = Our answer
$\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{x}^{2k+1}}{k!(2k+1)}$	$\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{x}^{2k}}{k!}$	$e^{-x^2}$	$\underset{0}{\overset{x}{\int }}{e}^{-t^2}dt$, also called the ERF of $x$ (it is not among the elementary functions usually seen in calculus.
$\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k-1}{x}^k}{k}=x-\frac{x^2}{2}+\frac{x^3}{3}-\dots$	$\sum _{l=0}^{\mathrm{\infty }}(-1{)}^l{x}^l=1-x+x^2-x^3+\dots$	$\frac{1}{1+x}$ (geometric series)	$\ln (1+x)$
$\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{x}^{2k+1}}{2k+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\dots$	$\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{x}^{2k}$	$\frac{1}{1+x^2}$	$\arctan x$
$\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{x}^{4k+1}}{4k+1}=x-\frac{x^5}{5}+\frac{x^9}{9}-\dots$	$\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{x}^{2k}$	$\frac{1}{1+x^4}$	We would need to integrate $1/(1+x^4)$.

Summing up the antiderivative power series

Power series	Antiderivative power series (we fix the constant term based on our convenience, it could be anything)	Function that the antiderivative power series converges to	Derivative of that function = Our answer
$\sum _{k=0}^{\mathrm{\infty }}(k+1)x^k$	$\sum _{l=0}^{\mathrm{\infty }}{x}^l$	$\frac{1}{1-x}$	$\frac{1}{(1-x{)}^2}$

Interval of validity

We will assume that the summation formula that we have is valid on the entire interval of convergence of the power series.

The radius of convergence is not affected by differentiation or integration, i.e., the radius of convergence of a power series is the same as that of its derivative power series and antiderivative power series.

However, the endpoints can change as follows:

• Going from a function to its derivative, the interval of convergence cannot get bigger, but it may get smaller, In other words, we may lose endpoints, but we cannot gain endpoints.
• Going from a function to its antiderivative, the interval of convergence may get bigger, but it cannot get smaller. In other words, we may gain endpoints, but we cannot lose endpoints.