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Need a hint to compute $\displaystyle \sum_{x=0}^\infty xa^x$ and $\displaystyle \sum_{x=0}^\infty x^2a^x$, where $a \in (0,1)$.

Adayah
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zesy
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  • This will help: http://math.stackexchange.com/questions/647587/sum-of-a-power-series-n-xn – Brenton May 31 '15 at 18:39
  • Why that strange way to write the limits in the sum? – AdLibitum May 31 '15 at 18:40
  • I've posted answers to this question several times. I wonder if it's our most frequently re-posted question. – Michael Hardy May 31 '15 at 18:56
  • @MichaelHardy: He seems to be the first to also ask about $\sum \limits _{n=0} ^\infty n^2 a^n$. – Alex M. May 31 '15 at 19:00
  • @Brenton: Apparently, the original is http://math.stackexchange.com/questions/67364/sequence-sum-question – Alex M. May 31 '15 at 19:01
  • @AlexM. : I'm pretty sure I've posted answers to that one too. If you throw a die repeatedly until you get the first $1$, what is the variance of the random variable that is the number of times you throw it? The answer to that question involves the series you say no one has asked about. ${}\qquad{}$ – Michael Hardy May 31 '15 at 19:14
  • $\displaystyle\sum_{n=0}^\infty n^2 a^n = \left(a^2\sum_{n=0}^\infty n(n-1)a^{n-2}\right) + \left(a\sum_{n=0}^\infty n a^{n-1}\right)={}$ $a^2$ times the second derivative of a function defined by a geometric series, plus $a$ times the first derivative. ${}\qquad{}$ – Michael Hardy May 31 '15 at 19:17

2 Answers2

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Hint: Differentiate $\sum_{x=0}^{\infty}a^x$ with respect to $a$.

Start wearing purple
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Consider the genral term and rewrite as $$ia^i=a(ia^{i-1})=a\frac{d}{da}(a^i)$$ So we can write the series as $$a\frac{d}{da}\sum_{i=0}^{\infty}a^i=a\frac{d}{da}\frac{1}{1-a}=\frac{a^2}{(1-a)^2}$$ I leave the second one to you. Hope this helps.

Crunch
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