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Given $$\sum_{n=0}^\infty \frac{x^2}{(1+x^2)^n}$$

I'm trying to find the nth partial sum, in order to test for convergence of the series. How would I go about doing this?

I know there are other ways of going about it but I'm trying to learn this particular one.

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  • $\begingroup$ It is a geometric series, "$a$" equal to $x^2$, "$r$" equal to $1/(1+x^2)$. $\endgroup$
    – André Nicolas
    May 12, 2014 at 19:35
  • $\begingroup$ The numerator doesn't depend on $n$, so we can pull that out from the sum. What is left may look familiar. $\endgroup$
    – Daniel Fischer
    May 12, 2014 at 19:36

1 Answer 1

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Hint 1: If $r\neq 1$ we have the formula $$ \sum_{n=0}^{k-1}a\cdot r^n = a\frac{1-r^k}{1-r} $$ and if $r=1$ we have the formula $$ \sum_{n=0}^{k-1}a\cdot r^n=a\cdot k $$ Can you prove these two formulas? Subhint: Google "geometric series".

Hint 2: The partial sums of your series can be written as $$ \sum_{n=0}^{k-1}\frac{x^2}{(1+x^2)^n}=\sum_{n=0}^{k-1}x^2\cdot\left(\frac{1}{1+x^2}\right)^n $$

Can you combine these two hints to obtain a formula?

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  • $\begingroup$ A nicely-formatted hint. $\endgroup$
    – Emily
    May 12, 2014 at 19:43

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