[SOLVED] Convergence Series

**dwsmith** · May 20th 2010, 03:11 PM

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?

**allsmiles** · May 20th 2010, 03:22 PM

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

**Plato** · May 20th 2010, 03:24 PM

Take note: $\left( {\sqrt[n]{{\frac{{n!}}<br /> {{n^n }}}}} \right) \to \frac{1}<br /> {e}$

**dwsmith** · May 20th 2010, 03:27 PM

Originally Posted by Plato

Take note: $\left( {\sqrt[n]{{\frac{{n!}}<br /> {{n^n }}}}} \right) \to \frac{1}<br /> {e}$

So use the root test and then test values of $x\in\mathbb{R}$ ?

**dwsmith** · May 20th 2010, 03:29 PM

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 $x\in\mathbb{R}$

**dwsmith** · May 20th 2010, 03:54 PM

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

$(1+x^{2n})^{\frac{1}{n}}$ 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 $\infty^0$ .

**lilaziz1** · May 20th 2010, 06:06 PM

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

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

**dwsmith** · May 20th 2010, 06:19 PM

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

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?

**lilaziz1** · May 20th 2010, 06:31 PM

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

**dwsmith** · May 20th 2010, 06:32 PM

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?

**lilaziz1** · May 20th 2010, 06:40 PM

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 $\frac{x^2}{e}$

**AllanCuz** · May 20th 2010, 06:42 PM

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

**dwsmith** · May 20th 2010, 06:44 PM

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.

**AllanCuz** · May 20th 2010, 06:47 PM

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?

**dwsmith** · May 20th 2010, 06:49 PM

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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