Question

Explain why $\sum _{k=0}^{\infty }{\left(x-k\right)}^k$ is not a power series.

Short answer

The base of the factors ${\left(x-k\right)}^k$varies with the index.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$f\left(x\right)=\sum _{k=0}^{\infty }{\left(x-k\right)}^k$

Step 2. Explain the series.

If $\sum _{k=0}^{\infty }{a}_k$ is a sequence of real numbers and x be a variable and $x_0$be a real number. Then a power series in $x-x_0$is a series of the form localid="1649342585929" $\underset{k=0}{\overset{\infty }{\sum a_k}}{\left(x-x_0\right)}^k$

Rewriting the series,

$f\underset{k=0}{\overset{\infty }{\sum a_k}}{\left(x-x_0\right)}^k=a_0+a_1\left(x-x_0\right)+a_2{\left(x-x_0\right)}^2+a_3{\left(x-x_0\right)}^3+...$

The series of $f\left(x\right)$is not a power series, this is because $x_0$in the series $f\left(x\right)=\sum _{k=0}^{\infty }{\left(x-k\right)}^k$, that is k, is not a constant real number.

We can say that the base of the factors ${\left(x-k\right)}^k$varies with the index.