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Calculate $a\displaystyle \sum_{n=1}^{\infty}n(1-a)^{n-1}$

where $a \in (0,1)$.

Lord_Farin
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lele
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  • When you move a factor $a(1-a)$ out of the sum, do you recognize the series you then get? – Daniel Fischer Jun 29 '13 at 22:51
  • I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – Lord_Farin Jun 29 '13 at 23:01
  • I post my above comment because I saw that you didn't accept any answer from your question. Please, consider my previous comment. Thanks. – Cortizol Jun 29 '13 at 23:03
  • This question, again. It gives me an excuse to link this once more. – David Mitra Jun 29 '13 at 23:21
  • http://en.wikipedia.org/wiki/Arithmetico-geometric_sequence#Series.2C_sum_to_n_terms and http://www.artofproblemsolving.com/Wiki/index.php/Arithmetico-geometric_series – lab bhattacharjee Jun 30 '13 at 04:39

3 Answers3

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Hint: Consider the series:

$$\sum_{n=0}^{\infty} b^n = \frac{1}{1-b}$$

And take the derivative of both sides with respect to $b$. Now apply this to your series.

George V. Williams
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Hints:

$$\forall\,x\in\Bbb C\;,\;|x|<1\;,\;\;f(x):=\frac1{1-x}=\sum_{n=0}^\infty x^n\implies$$

$$f'(x)=\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}\ldots\ldots$$

DonAntonio
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For fun (and this is probably not the solution you are expected to give) we give a mean proof. A certain coin has probability $a$ of landing heads, and probability $1-a$ of landing tails. The coin is tossed repeatedly. Let random variable $X$ be the number of tosses until we get the first head.

The probability that $X=1$ is $a$, the probability that $X=2$ is $(1-a)a$, the probability that $X=3$ is $(1-a)^2a$, and so on. It follows that the infinite sum we are asked to evaluate is $E(X)$, the mean of $X$.

Let us calculate this mean another way. Let the mean be $\mu$. Toss the coin. With probability $a$, we get a head on the first toss, and $X=1$. With probability $1-a$ we get a tail on the first toss, and we have wasted a toss. The mean number of additional tosses needed in that case is $\mu$. Thus $$\mu=(a)(1)+(1-a)(1+\mu).$$ Solve this linear equation for $\mu$. It turns out that $\mu=\dfrac{1}{a}$.

André Nicolas
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