# Power series limit help

• Apr 30th 2008, 02:18 PM
cowboys111
Power series limit help
Hi im having trouble understanding part of the power series. Im trying to find the radius of convergence using the ratio test, the initial problem is the sum of k=1 as k goes to infinity of x^k/(1+k^2). When i use the ratio test i get it down to xk^2/(k+1)^2, now my book goes from this to just the absolute value of x. If you expand (k+1)^2 the k^2's cancel but what happens to the 2k+1 are you allowed to just drop it or am i missing something?
Thanks for the help.
• Apr 30th 2008, 02:25 PM
Plato
Quote:

Originally Posted by cowboys111
Hi im having trouble understanding part of the power series. Im trying to find the radius of convergence using the ratio test, the initial problem is the sum of k=1 as k goes to infinity of x^k/(1+k^2). When i use the ratio test i get it down to xk^2/(k+1)^2, now my book goes from this to just the absolute value of x. If you expand (k+1)^2 the k^2's cancel but what happens to the 2k+1 are you allowed to just drop it or am i missing something?

I fear that you may be missing a lot.
$\frac{{k^2 }}{{\left( {1 + k} \right)^2 }} = \left( {\frac{1}{{\frac{1}{k} + 1}}} \right)^2 \to 1$
• Apr 30th 2008, 02:33 PM
TheEmptySet
Quote:

Originally Posted by cowboys111
Hi im having trouble understanding part of the power series. Im trying to find the radius of convergence using the ratio test, the initial problem is the sum of k=1 as k goes to infinity of x^k/(1+k^2). When i use the ratio test i get it down to xk^2/(k+1)^2, now my book goes from this to just the absolute value of x. If you expand (k+1)^2 the k^2's cancel but what happens to the 2k+1 are you allowed to just drop it or am i missing something?
Thanks for the help.

$\sum_{k=1}^{\infty}\frac{x^k}{1+k^2}$

$\lim_{k \to \infty} \left| \frac{\frac{x^{k+1}}{1+(k+1)^2}}{\frac{x^k}{1+k^2} } \right|= \lim_{k \to \infty} \left| \frac{x^{k+1}}{k^2+2k+2} \cdot \frac{k^2+1}{x^k} \right|=\lim_{k \to \infty} \left| \frac{x(1+\frac{1}{k^2})}{1+\frac{2}{k}+\frac{2}{k ^2}} \right|=|x|$

the 2nd to last step comes from multiplying the numerator and denominator by $\frac{1}{k^2}$