Summation

Shifting the Summation Index

Let's say we have

$$\sum _{n=2}^{\mathrm{\infty }}n(n-1)a_n{x}^{n-2}$$

and want to write it as a series where the generic term is $x^k$ instead of $x^{n-2}$.

We can let $k=n-2,n=k+2$ and make the substitutions:

$$\sum _{k+2=2}^{\mathrm{\infty }}(k+2)(k+2-1)a_{k+2}{x}^{k+2-2}=\sum _{k=0}^{\mathrm{\infty }}(k+2)(k+1)a_{k+2}{x}^k.$$

This is useful when we want to combine series with different generic terms.

For example, consider:

$$\begin{array}{cc}\sum _{n=0}^{\mathrm{\infty }}n{a}_n{x}^{n-1}+\sum _{n=0}^{\mathrm{\infty }}2a_n{x}^{n+1}.& \text{(a)}\end{array}$$

If we want to combine these into a single series, we need to rearrange them so their generic term is the same. We'll make it $x^k$ in both cases.

First, note that since the first term in the first series is $0$, nothing is lost be shifting its starting point to $n=1$, and now we have:

$$\begin{array}{cc}\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}+\sum _{n=0}^{\mathrm{\infty }}2a_n{x}^{n+1}.& \text{(b)}\end{array}$$

Now, in the first, let $k=n-1,n=k+1$, and in the second, let $k=n+1,n=k-1$, and make the substitutions:

$$\begin{array}{cc}\sum _{k+1=1}^{\mathrm{\infty }}(k+1)a_{k+1}{x}^{k+1-1}+\sum _{k-1=0}^{\mathrm{\infty }}2a_{k-1}{x}^{k-1+1},& \text{(c)}\end{array}$$

which simplifies to

$$\begin{array}{cc}\sum _{k=0}^{\mathrm{\infty }}(k+1)a_{k+1}{x}^k+\sum _{k=1}^{\mathrm{\infty }}2a_{k-1}{x}^k.& \text{(d)}\end{array}$$

Since the first series starts at $k=0$ and the second at $k=1$, we have some work left to do. We can split the first series into a constant term plus a series starting at $k=1$ by noting that when $k=0$, the first term in the first series is $a_1$:

$$\begin{array}{cc}{a}_1+\sum _{k=1}^{\mathrm{\infty }}(k+1)a_{k+1}{x}^k+\sum _{k=1}^{\mathrm{\infty }}2a_{k-1}{x}^k.& \text{(e)}\end{array}$$

Now we can combine the two series to get:

$$a_1+\sum _{k=1}[(k+1)a_{k+1}+2a_{k-1}]x^k.$$