what is the the summation $S_n = \sum\limits_{n = 1}^\infty{b_n} $ with $b_n = n{\alpha}^n$?
Here $0 \le \alpha < 1$. any closed form solution?
I know maybe we should build a geometric sequence or arithmetic sequence from $b_n$, but how to do that?
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1See http://en.wikipedia.org/wiki/Arithmetico-geometric_sequence – lab bhattacharjee Nov 01 '14 at 05:17
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See this for ideas. – David Mitra Nov 01 '14 at 06:14
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If you dont know differentiation then use the fact that $$\begin{aligned}S_n-\alpha S_n & =\sum_{n=1}^{\infty} n\alpha^n-\sum_{n=1}^{\infty} n\alpha^{n+1}\\ & =\alpha+\sum_{n=2}^{\infty} n\alpha^n-\sum_{n=2}^{\infty} (n-1)\alpha^{n} \\ & =\alpha+\sum_{n=2}^{\infty} (n-n+1)\alpha^n \\ & = \sum_{n=1}^{\infty} \alpha^{n}\end{aligned}$$ Finish it! If you want to be more rigorous first show that $S_n$ converges
Sayan
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You dont need to use convergence if you use limits, i.e., just using undefined finite sums and after see what happen when $\text{upper limit}\to\infty$ – Masacroso Nov 01 '14 at 12:59
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The usual geometric series formula is $\displaystyle\sum_{n = 0}^{\infty}\alpha^n = \dfrac{1}{1-\alpha}$.
Differentiate both sides with respect to $\alpha$ to get $\displaystyle\sum_{n = 1}^{\infty}n\alpha^{n-1} = \dfrac{1}{(1-\alpha)^2}$.
Do you see how to get the formula for the series you asked about?
JimmyK4542
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We know that $$\sum_{n=1}^{\infty}x^{n}=\frac{1}{1-x}$$ if $|x|<1$. Differentiate both sides with respect to $x$ to get $$\sum_{n=1}^{\infty}(n-1)x^{n}=\frac{1}{(1-x)^2}$$ Hence $$\sum_{n=1}^{\infty}nx^n=\sum_{n=1}^{\infty}(n-1)x^{n}+\sum_{n=1}^{\infty}x^{n}=\frac{1}{(1-x)^2}+\frac{1}{1-x}$$ Put $x=\alpha$ to get the required result.
QED
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