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I have a summation that I calculated, but I'm having a hard time to turn it into a geometric series.

I have the summation:

$y(n) = \sum_{k=-\infty}^{\infty} k*a^ku(k)*u(n-k)$ where $u(k)$ is the step function.

I reduced this to:

$y(n) = \sum_{k=0}^{n} k*a^k$ where $n >= 0$.

Since this is a finite series, I was taught that I could find a ratio of the second to first term, and use the following equation:

$y(n) = (firstterm) * {1-R^{n+1}\over1-R}$ Where "R" is the ratio of first term to second term.

However, I can't figure out what "R" should be since the first term is zero. How can I go about calculating "R" and $y(n)$?

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Not every finite series is a geometric series. There is no $R$ for your series, and the formula you are trying to apply, doesn't apply. – Gerry Myerson Mar 1 '13 at 1:33
up vote 0 down vote accepted

Let $$y(n)=a+2a^2+3a^3+\cdots+na^n$$ Then $$ay=a^2+2a^3+\cdots+na^{n+1}$$ Subtracting, $$(1-a)y=a+a^2+\cdots+a^n-na^{n+1}$$ Now the first $n$ terms do form a geometric series, so $$(1-a)y={a(1-a^n)\over1-a}-na^{n+1}$$ Now you can divide both sides by $1-a$ (provided $a\ne0$) to get a formula for $y$.

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This is really helpful! Now I understand your comment. – Chris Harris Mar 1 '13 at 1:41

And, by induction, once you have a formula for $\sum_{j=0}^n a^j j^k$, you can get a formula for $\sum_{j=0}^n a^j j^{k+1}$.

Gerry Myerson showed how to go from $k=0$ to $k=1$.

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You want to use the formula for a finite geometric series $$ \sum_{k=0}^{m-1}a^k=\frac{1-a^m}{1-a} \qquad\text{for $a\neq1$} $$ (note that I stop the summation before including $a^m$), to find the sum of the series $$ \sum_{k=0}^{m-1}ka^k \qquad\text{for $a\neq1$.} $$ Since the latter is not a geometric series, you cannot apply the formula directly. However you may note that the terms of the latter series can obtained from those of the former by formally differentiating with respect to $a$ and the multiplying by$~a$. Since the formula for the geometric summation is valid for (almost) all$~a$, its validity survives formal differentiation and multiplication. Differentiation gives $$\def\d{\mathrm d} \frac\d{\d a}\left(\frac{1-a^m}{1-a}\right)=\frac{ma^{m+1}-(m+1)a^m+1}{(1-a)^2} $$ so that you find $$ \sum_{k=0}^{m-1}ka^k = a\,\frac{ma^{m+1}-(m+1)a^m+1}{(1-a)^2} \qquad\text{for $a\neq1$.} $$

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