Infinite geometric series derivation problem

Question

I don't understand how's the pointed portion of the second line derived. can anyone explain please.

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They used the result that they show in blue. You do know that $\sum\limits_{k=1}^\infty a^k = \frac{a}{1-a}$ I hope? How about the related formula for $\sum\limits_{k=1}^\infty ka^k$? If your question is how to apply those results, it is as simple as renaming $x$ to $k$ and renaming $\frac{e^t}{2}$ to $a$. If your question is where the results they applied came from in the first place, that can be found elsewhere on this site easily. — JMoravitz, Oct 28 '19 at 15:11
is there any formula for ka^k series from 1 to infinity ... @JMoravitz — Himel Mazumder, Oct 28 '19 at 15:21
@HimelMazumder See How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? and the questions linked there. — Martin Sleziak, Oct 28 '19 at 15:27

score 1 · Accepted Answer · answered Oct 28 '19 at 15:23

You know $S_0=\sum_1^\infty a^k = \dfrac{a}{1-a}$ from the blue

Now consider $S_1=\sum_1^\infty k \,a^k$

You have $aS_1 = \sum_1^\infty a k \,a^k = \sum_1^\infty k \,a^{k+1} = \sum_2^\infty (k-1) \,a^{k} = \sum_1^\infty (k-1) \,a^{k} = S_1-S_0$

so $aS_1 = S_1-S_0$ and thus $(1-a)S_1=S_0=\dfrac{a}{1-a}$

giving $S_1=\dfrac{a}{(1-a)^2} = \dfrac{1}{(1-a)^2} - \dfrac{1}{1-a}$

Here $a=\dfrac{e^t}{2}$

1 Answers1