Talk:Derivative of Geometric Sequence/Corollary

This result is correct, but an alternate formulation seems more intuitive.

Instead of:

$\ds \sum_{n \mathop \ge 1} n \paren {n + 1} x^{n - 1} = \frac 2 {\paren {1 - x}^3}$

Could also be shown as:

$\ds \sum_{n \mathop \ge 2} n \paren {n - 1} x^{n - 2} = \frac 2 {\paren {1 - x}^3}$

And taking further derivatives yields:

$\ds \sum_{n \mathop \ge 3} n \paren {n - 1} \paren {n - 2} x^{n - 3} = \frac 6 {\paren {1 - x}^4}$

$\ds \sum_{n \mathop \ge 4} n \paren {n - 1} \paren {n - 2} \paren {n - 3} x^{n - 4} = \frac {24} {\paren {1 - x}^5}$

$\cdots$

$\ds \sum_{n \mathop \ge k} \dfrac {n!} {\paren {n - k}!} x^{n - k} = \frac {k!} {\paren {1 - x}^{k+1}}$

Is there an advantage to the formulation shown? --Robkahn131 (talk) 14:46, 21 April 2021 (UTC)

The most obvious advantage is that the summation starts at $1$ rather than $k$. Why is the format you suggest "more intuitive" than this? --prime mover (talk) 14:50, 21 April 2021 (UTC)

We still get:

$\ds \sum_{n \mathop \ge 1} n \paren {n + 1} \paren {n + 2} x^{n - 1} = \frac 6 {\paren {1 - x}^4}$

$\ds \sum_{n \mathop \ge 1} n \paren {n + 1} \paren {n + 2} \paren {n + 3} x^{n - 1} = \frac {24} {\paren {1 - x}^5}$

$\cdots$

$\ds \sum_{n \mathop \ge 1} n^{\overline {k + 1} } x^{n - 1} = \frac {k!} {\paren {1 - x}^{k + 1} }$

or whatever.

From Power Rule for Derivatives:

\(\ds \frac \d {\d x} \sum_{n \mathop \ge 1} x^n\)

\(=\)

\(\ds \sum_{n \mathop \ge 1} n x^{n - 1}\)

\(\ds \frac \d {\d x} \sum_{n \mathop \ge 1} n x^{n - 1}\)

\(=\)

\(\ds \sum_{n \mathop \ge 1} n \paren {n - 1} x^{n - 2}\)

\(\ds \frac \d {\d x} \sum_{n \mathop \ge 1} n \paren {n - 1} x^{n - 2}\)

\(=\)

\(\ds \sum_{n \mathop \ge 1} n \paren {n - 1} \paren {n - 2} x^{n - 3}\)

Come to think of it, the summation can still start at 1 and there will merely be a few zero terms at the beginning.

It's probably just me - I saw the $\paren {n + 1}$ term in the corollary and thought "That doesn't belong there - that should be $\paren {n - 1}$. --Robkahn131 (talk) 15:44, 21 April 2021 (UTC)

Tom-ay-to, tom-ah-to. Feel free to document this, but please do the general result, seeing case after case for $1, 2, 3, 4, \ldots$ makes me feel ill.

We already have a result that calculates the general $n$th derivative, so you're most of the way there.

Please leave this corollary alone, because a) it's already being used, and b) because that's how we roll. --prime mover (talk) 16:28, 21 April 2021 (UTC)