Sequence e^2/2^n, converge or diverge (easy)

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Homework Statement


does the series [itex]e^n/2^n[/itex] converge or diverge?

does the series [itex]2^n/e^n[/itex] converge or diverge?

The Attempt at a Solution



I took [itex]lim→∞ e^x/2^x[/itex] and am getting ∞, so it diverges, right?

I also used L'Hopital's rule and got the same result.

My prof hinted that I was wrong... what am I missing?

Thanks!
 
Last edited:
8point1 said:

Homework Statement


does the series [itex]e^n/2^n[/itex] converge or diverge?

The Attempt at a Solution



I took [itex]lim→∞ e^x/2^x[/itex] and am getting ∞, so it diverges, right? My prof hinted that I was wrong.

Thanks!

Yes, you are right. e/2 is greater than 1. The series diverges.
 
8point1 said:
what about 2/e?

You tell me. What do you know about the geometric series r^n?
 
8point1 said:

Homework Statement


does the [itex]> >[/itex]series [itex]< <[/itex][itex]e^n/2^n...[/itex]

does the [itex]> >[/itex]series [itex]< <[/itex][itex]2^n/e^n...[/itex]

8point1,

there is a distinction. Those expressions are the general terms
of the sequences and the series.

The summations of those terms are the respective series.
 
The sequence diverges, so the series must diverge also, correct?
 
8point1 said:
The sequence diverges, so the series must diverge also, correct?

That's correct for (e/2)^n. (2/e)^n is different.