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To what does the serie $\sum_{j=0}^{n}a^j b^{n-j}$ converge? Does this serie have a name?

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It converges, because it is finite. About a name, I'd suggest "Charlie". – yohBS Aug 19 '12 at 11:38
Well it is $\frac {a^{n+1}-b^{n+1}}{a-b}$ of course (if $a\not =b$ : multiply by $a-b$) – Raymond Manzoni Aug 19 '12 at 11:41
@yohBS The question was not whether the sum converges, but to what it converges. – MJD Aug 19 '12 at 11:55
up vote 5 down vote accepted

I'm assuming you are asking for the convergence for $n\to\infty$.

We have (assuming $a\ne b$) $$\sum_{j=0}^na^jb^{n-j} = b^n\sum_{j=0}^n\left(\frac{a}{b}\right)^j = b^n\frac{(a/b)^{n+1}-1}{a/b-1} = \frac{a^{n+1}-b^{n+1}}{a-b}$$ Therefore the sequence converges if both $|a|<1$ and $|b|<1$, with limit $0$. Moreover if one of $a$ or $b$ is $1$ and the absolute value of the other is $<1$, the sequence converges to $1/(1-b)$ (for $a=1$) or $1/(1-a)$ (for $b=1$). Otherwise, the sequence diverges.

If $a=b$, then the sum is $na^n$, which converges to $0$ iff $|a|<1$ and diverges otherwise.

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This sum is related to the Cauchy Product and has the structure of a convolution.

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