# [SOLVED] Convergence Series

#### dwsmith

MHF Hall of Honor
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?

#### allsmiles

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

#### Plato

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

dwsmith

#### dwsmith

MHF Hall of Honor
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

MHF Hall of Honor
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

MHF Hall of Honor
$$\displaystyle \lim_{n\to\infty}\sqrt[n]{\frac{n!x^{2n}}{n^n(1+x^{2n})}}=\sqrt[n]{\frac{n!}{n^n}}\frac{(x^{2n})^{\frac{1}{n}}}{(1+x^{2n})^{\frac{1}{n}}}=\frac{x^2}{e(1+x^{2n})^{\frac{1}{n}}}$$

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

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#### lilaziz1

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.

dwsmith

#### dwsmith

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

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?

#### lilaziz1

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

MHF Hall of Honor
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?