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Olin [163]
3 years ago
9

Consider the series ∑n=1[infinity]2nn!nn. Evaluate the the following limit. If it is infinite, type "infinity" or "inf". If it d

oes not exist, type "DNE". limn→[infinity]∣∣∣an+1an∣∣∣=L Answer: L= What can you say about the series using the Ratio Test? Answer "Convergent", "Divergent", or "Inconclusive". Answer: Determine whether the series is absolutely convergent, conditionally convergent, or divergent. Answer "Absolutely Convergent", "Conditionally Convergent", or "Divergent".
Mathematics
1 answer:
Vikki [24]3 years ago
6 0

I guess the series is

\displaystyle\sum_{n=1}^\infty\frac{2^nn!}{n^n}

We have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac{2^nn!}{n^n}}\right|=2\lim_{n\to\infty}\left(\frac n{n+1}\right)^n

Recall that

e=\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n

In our limit, we have

\dfrac n{n+1}=\dfrac{n+1-1}{n+1}=1-\dfrac1{n+1}

\left(\dfrac n{n+1}\right)^n=\dfrac{\left(1-\frac1{n+1}\right)^{n+1}}{1-\frac1{n+1}}

\implies\displaystyle2\lim_{n\to\infty}\left(\frac n{n+1}\right)^n=2\frac{\lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)^{n+1}}{\lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)}=\frac{2e}1=2e

which is greater than 1, which means the series is divergent by the ratio test.

On the chance that you meant to write

\displaystyle\sum_{n=1}^\infty\frac{2^n}{n!n^n}

we have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}}{(n+1)!(n+1)^{n+1}}}{\frac{2^n}{n!n^n}}\right|=2\lim_{n\to\infty}\frac1{(n+1)^2}\left(\frac n{n+1}\right)^2

=\displaystyle2\left(\lim_{n\to\infty}\frac1{(n+1)^2}\right)\left(\lim_{n\to\infty}\left(\frac n{n+1}\right)^n\right)=2\cdot0\cdot e=0

which is less than 1, so this series is absolutely convergent.

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