Formula for nth derivative of partial sum of geometric series.

Question

I am trying to find a formula for either

(1) the $n$th derivative for the following $m$th partial sum:

$$\frac{d^n}{dx^n} \sum_{i=0}^m x^i$$

or (2) the $n$th derivative of the infinite series given by

$$\frac{d^n}{dx^n} \sum_{i=m}^\infty x^i$$

I know that $\sum_{i=0}^m x^i = \frac{1-x^{m+1}}{1-x}$ and that $\sum_{i=m}^\infty x^i = \frac{x^m}{1-x}$. However, sequentially taking the derivatives and searching for a pattern is proving difficult. Can someone show me the formula? Thanks

Answer 1

Here is a derivation of (1). We obtain \begin{align*} \frac{d^n}{dx^n} \sum_{k=0}^m x^k&=\sum_{k=0}^m k^{\underline{n}}x^{k-n}\tag{1}\\ &=\sum_{k=n}^mk^{\underline{n}}x^{k-n}\tag{2}\\ &=\color{blue}{\sum_{k=0}^{m-n}(k+n)^{\underline{n}}x^k}\tag{3} \end{align*}

Comment:

• In (1) we use the falling factorial notation $k^{\underline{n}}=k(k-1)\cdots(k-n+1)$.

• In (2) we note that $k^{\underline{n}}=0$ if $n>k$ and therefore we can start with index $k=n$. We also note the representation is valid if $n>m$, since then we use the common convention that a series with upper limit less than lower limit is equal to zero.

• In (3) we shift the index $k$ to start with $k=0$.

A derivation of (2) can be done similarly.