The set of real numbers x for which the series $\displaystyle \sum_{n=1}^{\infty}\frac{n!x^{2n}}{n^n(1+x^{2n})}$ converges is?
How should this be approached?
$\displaystyle \lim_{n\to\infty}\sqrt[n]{\frac{n!x^{2n}}{n^n(1+x^{2n})}}=\sqrt[n]{\frac{n!}{n^n}}\frac{(x^{2n})^{\frac{1}{n}}}{(1+x ^{2n})^{\frac{1}{n}}}=\frac{x^2}{e(1+x^{2n})^{\fra c{1}{n}}}$
$\displaystyle (1+x^{2n})^{\frac{1}{n}}$ this term is posing some issues now.
To use L'Hopitals Rule, I need either $\displaystyle \frac{\infty}{\infty},\ \frac{0}{0}, \ 0^0,\ \infty^{\infty}$, but I have $\displaystyle \infty^0$.
$\displaystyle \lim_{n\to\infty}\frac{x^2}{e}\left[(1+x^{2n})^{\frac{-1}{n}}\right]=\frac{x^2}{e}\lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}}$
Then just exponentiate to convert to $\displaystyle \frac{\infty}{\infty}$? Or can I just ln $\displaystyle (1+x^{2n})^{\frac{-1}{n}}$ WLOG by leaving $\displaystyle \frac{x^2}{e}$ on the outside?
Yes...
$\displaystyle \frac{x^2}{e} * \lim_{n\to\infty} (1+x^{2n})^{\frac{-1}{n}} $
Let $\displaystyle y = (1+x^{2n})^{\frac{-1}{n}} $
$\displaystyle lny = \frac{-1}{n} ln ( 1+x^{2n} ) $
$\displaystyle \lim_{ n \to \infty } lny = \lim_{ n \to \infty } \frac{-1}{n} ln ( 1+x^{2n} ) = \lim_{ n \to \infty } \frac{-1}{n} \lim_{ n \to \infty } ln ( 1+x^{2n} ) = 0 * \infty$
The above is not defined but we note that the first function goes towards 0 faster then the second function goes toward infinity. So I would say this would approach 0.
With this result we can use...
$\displaystyle \frac{x^2}{e} * \lim_{n\to\infty} (1+x^{2n})^{\frac{-1}{n}} =\frac{x^2}{e} * \lim_{n\to\infty} y = \frac{x^2}{e} * \lim_{n\to\infty} e^{lny} = x^2 $
I think the above is valid...but i'm about 7 beers deep by now...soo dont trust me to much
oh why would you need to know that? Clearly if our limit is of some variable to some number, then all multiplication that has nothing to do with that variable can be taken outside of the limit.
It's no different then....
$\displaystyle \lim_{x \to 0 } cf(x) = c \lim_{x \to 0 } f(x) $
Which you most definately know, so I'm confused at why you are confused?