Exponent shift method for power series summation

Description of the method

This method is used for summing up various power series that look similar to existing power series but have some terms missing, extra, or shifted. It is closely related to another method, the substitution method for power series summation.

Missing terms in summation

Suppose we know how to do a certain summation $\sum _{k=k_0}^{\mathrm{\infty }}{a}_k{x}^k$, and we are asked to do a summation $\sum _{k=k_1}^{\mathrm{\infty }}{a}_k{x}^k$, where $k_1\ge k_0$ is a positive integer. This is given by:

$\sum _{k=k_1}^{\mathrm{\infty }}{a}_k{x}^k=\sum _{k=k_0}^{\mathrm{\infty }}{a}_k{x}^k-\sum _{k=k_0}^{k_1-1}{a}_k{x}^k$

Shifted index of summation

Suppose we know how to do a summation of the form:

$\sum _{k=k_0}^{\mathrm{\infty }}{a}_k{x}^k$

Let's now consider another summation, with $k_1,r$ integers such that $k_1+r\ge k_0$:

$\sum _{k=k_1}^{\mathrm{\infty }}{a}_{k+r}{x}^k$

We simplify this as follows

$\sum _{k=k_1}^{\mathrm{\infty }}{a}_{k+r}{x}^k=\frac{1}{x^r}\sum _{k=k_1}^{\mathrm{\infty }}{a}_{k+r}{x}^{k+r}=\frac{1}{x^r}\sum _{l=k_1+r}^{\mathrm{\infty }}{a}_l{x}^l$

We now use the missing terms in summation idea to simplify this final summation in terms of the original.

Examples

Series we need to sum	Type	Rewrite in terms of known series	Final answer
$\sum _{k=1}^{\mathrm{\infty }}\frac{x^k}{k!}$	Missing terms only	$(\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{k!})-\frac{x^0}{0!}$	$e^x-1$
$\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{(k+2)!}$	Shifted index of summation + missing terms	$\frac{1}{x^2}\sum _{l=2}^{\mathrm{\infty }}\frac{x^l}{l!}=\frac{1}{x^2}[(\sum _{l=0}^{\mathrm{\infty }}\frac{x^l}{l!})-(\frac{x^0}{0!}+\frac{x^1}{1!})]$	$\frac{e^x-1-x}{x^2}$, expression valid for $x\ne 0$. At $x=0$, we get a value of $1/2$.
$\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{k+1}$	Shifted index of summation	$\frac{1}{x}\sum _{l=1}^{\mathrm{\infty }}\frac{x^l}{l}\left|\right|<math>\frac{-\ln (1-x)}{x}$ for $x\in [-1,0)\cup (0,1)$, 1 for $x=0$.