[SOLVED] Convergence Series

dwsmith

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Mar 2010
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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?
 
May 2010
53
1
Altlanta GA
well do know how to solve these types of problems or are they new for you?
 

Plato

MHF Helper
Aug 2006
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Take note: \(\displaystyle \left( {\sqrt[n]{{\frac{{n!}}
{{n^n }}}}} \right) \to \frac{1}
{e}\)
 
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dwsmith

MHF Hall of Honor
Mar 2010
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Take note: \(\displaystyle \left( {\sqrt[n]{{\frac{{n!}}
{{n^n }}}}} \right) \to \frac{1}
{e}\)
So use the root test and then test values of \(\displaystyle x\in\mathbb{R}\)?
 

dwsmith

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Mar 2010
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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 \(\displaystyle x\in\mathbb{R}\)
 

dwsmith

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Mar 2010
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Florida
\(\displaystyle \lim_{n\to\infty}\sqrt[n]{\frac{n!x^{2n}}{n^n(1+x^{2n})}}=\sqrt[n]{\frac{n!}{n^n}}\frac{(x^{2n})^{\frac{1}{n}}}{(1+x^{2n})^{\frac{1}{n}}}=\frac{x^2}{e(1+x^{2n})^{\frac{1}{n}}}\)

\(\displaystyle (1+x^{2n})^{\frac{1}{n}}\) 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 \(\displaystyle \infty^0\).
 
Last edited:
Mar 2010
107
14
Correction: You need either \(\displaystyle \frac{\infty}{\infty}, \frac{0}{0}, 0^0, 1^\infty \) or \(\displaystyle \infty^0\)

\(\displaystyle \infty^\infty \) is not an indeterminate.
 
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dwsmith

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Mar 2010
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\(\displaystyle \lim_{n\to\infty}\frac{x^2}{e}\left[(1+x^{2n})^{\frac{-1}{n}}\right]=\frac{x^2}{e}\lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}}\)

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?
 
Mar 2010
107
14
Yea I'd do natural log because \(\displaystyle \lim_{n\to\infty}\frac{x^2}{e}\left[(1+x^{2n})^{\frac{-1}{n}}\right] = \lim_{n\to\infty} \frac{x^2}{e} * \lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}} \)
 

dwsmith

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Mar 2010
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Florida
Yea I'd do natural log because \(\displaystyle \lim_{n\to\infty}\frac{x^2}{e}\left[(1+x^{2n})^{\frac{-1}{n}}\right] = \lim_{n\to\infty} \frac{x^2}{e} * \lim_{n\to\infty}(1+x^{2n})^{\frac{-1}{n}} \)
But can I ln only the right portion and let the left portion be?