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The question is :

Is $\sum_{k=1}^\infty \frac{(-3)^k(k!)}{k^k}$ convergent?

Note : I can't find the limit of its main term. I know the answer must be related to some test about convergence of series ... I don't know which one and i can't find the limit.

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    Ratio test.. ${}$ – 2017-01-05
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    @OpenBall that's the problem ! which one is bigger? the denominator or the other one ? – 2017-01-05
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    Same question, replacing `3` with `e`: what's the answer? – 2017-01-05

2 Answers 2

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Using the ratio test:

$$\lim_{\text{k}\to\infty}\left|\frac{\left(\frac{\left(-\text{n}\right)^{\text{k}+1}\cdot\left(\left(\text{k}+1\right)!\right)}{\left(\text{k}+1\right)^{\text{k}+1}}\right)}{\left(\frac{\left(-\text{n}\right)^\text{k}\cdot\left(\text{k}!\right)}{\text{k}^\text{k}}\right)}\right|=\lim_{\text{k}\to\infty}\left|-\text{n}\left(\frac{\text{k}}{1+\text{k}}\right)^\text{k}\right|=\left|\text{n}\right|\lim_{\text{k}\to\infty}\left|\left(\frac{\text{k}}{1+\text{k}}\right)^\text{k}\right|=\frac{\left|\text{n}\right|}{e}>1$$

Which leads to:

$$\left|\text{n}\right|>e$$

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    Where is $n$ coming from? I can't see an answer from 5xum. – 2017-01-05
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    @LeBtz He was giving the hint: Ratio test, I did that for a general $\text{n}$. – 2017-01-05
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    I find this answer confusing. The answer by @5xum has now been deleted. And all that was mentioned was the ratio test. Including this $n$ is a bit weird. – 2017-01-05
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Apply Stirling's approximation $$k!\sim \sqrt{2k\pi}\left(\frac{k}{e}\right)^k$$ and use Root Test.

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    But the series is alternating... – 2017-01-05
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    $$L=\text{lim}\,\text{sup}_{n\to\infty}\sqrt[n]{ |a_n|}$$ – 2017-01-05