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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$\begingroup$ It converges, because it is finite. About a name, I'd suggest "Charlie". $\endgroup$– yohBSAug 19, 2012 at 11:38
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1$\begingroup$ Well it is $\frac {a^{n+1}-b^{n+1}}{a-b}$ of course (if $a\not =b$ : multiply by $a-b$) $\endgroup$– Raymond ManzoniAug 19, 2012 at 11:41
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1$\begingroup$ @yohBS The question was not whether the sum converges, but to what it converges. $\endgroup$– MJDAug 19, 2012 at 11:55
2 Answers
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.
This sum is related to the Cauchy Product and has the structure of a convolution.