Determine whether the following series

∞∑n=1enn\sum_{n=1}^\infty {\frac{e^n}{n}}

diverges or converges.

The solution given goes as follow:

Note that limn→∞enn=limx→∞exx=limx→∞ex1=∞\lim_{n \rightarrow \infty}{\frac{e^n}{n}} = \lim_{x \rightarrow \infty}{\frac{e^x}{x}} = \lim_{x \rightarrow \infty}{\frac{e^x}{1}} = \infty

By the Divergence test, the series diverges.

Question: I thought the L’hopital rule cannot be used to show limit does not exist. But in the solution provided, it does so. May I know whether the solution is correct or not?

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It’s correct. The limit must exists, no matter if it is finite or not.

– Antioquia3943

2 days ago

@Antioquia3943: you refer to correct solution or correct application of L’Hopital rule?

– Idonknow

2 days ago

The solution is correct.

– Antioquia3943

2 days ago

This has nothing whatsoever to do with l’Hôpital’s rule.

– copper.hat

2 days ago

You can use the L’hopital to show that ana_n diverges which implies ∑an\sum a_n diverges.

– MathematicsStudent1122

2 days ago

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1 Answer

1

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Using the ratio test an=enn⟹an+1an=enn+1∼e−ena_n=\frac{e^n}{n}\implies \frac{a_{n+1}}{a_n}=\frac{e n}{n+1}\sim e-\frac e n So, divergence and not much to do with L’Hأ´pital’s rule which, I agree, shows that an→∞a_n\to \infty from which you could also conclude.