1. [SOLVED] Convergence Series

The set of real numbers x for which the series $\sum_{n=1}^{\infty}\frac{n!x^{2n}}{n^n(1+x^{2n})}$ converges is?

How should this be approached?

2. well do know how to solve these types of problems or are they new for you?

3. Take note: $\left( {\sqrt[n]{{\frac{{n!}}
{{n^n }}}}} \right) \to \frac{1}
{e}$

4. Originally Posted by Plato
Take note: $\left( {\sqrt[n]{{\frac{{n!}}
{{n^n }}}}} \right) \to \frac{1}
{e}$
So use the root test and then test values of $x\in\mathbb{R}$?

5. Originally Posted by allsmiles
well do know how to solve these types of problems or are they new for you?
I know how to solve series problems. The issue here is method identifying the interval for $x\in\mathbb{R}$

6. $\lim_{n\to\infty}\sqrt[n]{\frac{n!x^{2n}}{n^n(1+x^{2n})}}=\sqrt[n]{\frac{n!}{n^n}}\frac{(x^{2n})^{\frac{1}{n}}}{(1+x ^{2n})^{\frac{1}{n}}}=\frac{x^2}{e(1+x^{2n})^{\fra c{1}{n}}}$

$(1+x^{2n})^{\frac{1}{n}}$ this term is posing some issues now.

To use L'Hopitals Rule, I need either $\frac{\infty}{\infty},\ \frac{0}{0}, \ 0^0,\ \infty^{\infty}$, but I have $\infty^0$.

7. Correction: You need either $\frac{\infty}{\infty}, \frac{0}{0}, 0^0, 1^\infty$ or $\infty^0$

$\infty^\infty$ is not an indeterminate.

8. $\lim_{n\to\infty}\frac{x^2}{e}\left[(1+x^{2n})^{\frac{-1}{n}}\right]=\frac{x^2}{e}\lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}}$

Then just exponentiate to convert to $\frac{\infty}{\infty}$? Or can I just ln $(1+x^{2n})^{\frac{-1}{n}}$ WLOG by leaving $\frac{x^2}{e}$ on the outside?

9. Yea I'd do natural log because $\lim_{n\to\infty}\frac{x^2}{e}\left[(1+x^{2n})^{\frac{-1}{n}}\right] = \lim_{n\to\infty} \frac{x^2}{e} * \lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}}$

10. Originally Posted by lilaziz1
Yea I'd do natural log because $\lim_{n\to\infty}\frac{x^2}{e}\left[(1+x^{2n})^{\frac{-1}{n}}\right] = \lim_{n\to\infty} \frac{x^2}{e} * \lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}}$
But can I ln only the right portion and let the left portion be?

11. Yea. Remember to e the answer you get from $\lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}}$ and multiply your final answer with $\frac{x^2}{e}$

12. Originally Posted by dwsmith
But can I ln only the right portion and let the left portion be?
Yes...

$\frac{x^2}{e} * \lim_{n\to\infty} (1+x^{2n})^{\frac{-1}{n}}$

Let $y = (1+x^{2n})^{\frac{-1}{n}}$

$lny = \frac{-1}{n} ln ( 1+x^{2n} )$

$\lim_{ n \to \infty } lny = \lim_{ n \to \infty } \frac{-1}{n} ln ( 1+x^{2n} ) = \lim_{ n \to \infty } \frac{-1}{n} \lim_{ n \to \infty } ln ( 1+x^{2n} ) = 0 * \infty$

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...

$\frac{x^2}{e} * \lim_{n\to\infty} (1+x^{2n})^{\frac{-1}{n}} =\frac{x^2}{e} * \lim_{n\to\infty} y = \frac{x^2}{e} * \lim_{n\to\infty} e^{lny} = x^2$

I think the above is valid...but i'm about 7 beers deep by now...soo dont trust me to much

13. Originally Posted by AllanCuz
Yes...

$\frac{x^2}{e} * \lim_{n\to\infty} (1+x^{2n})^{\frac{-1}{n}}$

Let $y = (1+x^{2n})^{\frac{-1}{n}}$

$lny = \frac{-1}{n} ln ( 1+x^{2n} )$

$\lim_{ n \to \infty } lny = \lim_{ n \to \infty } \frac{-1}{n} ln ( 1+x^{2n} ) = \lim_{ n \to \infty } \frac{-1}{n} \lim_{ n \to \infty } ln ( 1+x^{2n} ) = 0 * \infty$

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...

$\frac{x^2}{e} * \lim_{n\to\infty} (1+x^{2n})^{\frac{-1}{n}} =\frac{x^2}{e} * \lim_{n\to\infty} y = \frac{x^2}{e} * \lim_{n\to\infty} e^{lny} = x^2$

I think the above is valid...but i'm about 7 beers deep by now...soo dont trust me to much
That must be why you answered the question then because I don't need someone to solve it. My questions were on rules of certain operations.

14. Originally Posted by dwsmith
That must be why you answered the question then because I don't need someone to solve it. My questions were on rules of certain operations.
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....

$\lim_{x \to 0 } cf(x) = c \lim_{x \to 0 } f(x)$

Which you most definately know, so I'm confused at why you are confused?

15. Originally Posted by dwsmith
Can I just ln $(1+x^{2n})^{\frac{-1}{n}}$ WLOG by leaving $\frac{x^2}{e}$ on the outside?
This was the question.

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