The sequence (-1)^(n)*e^(1/n) is divergent or convergent?

Question

tkhunny · Answer

Do the terms approach zero (0)?

anonymous · Answer

Well, (1/n) approaches zero when n tends to infinite.

paounn · Answer

The term (-1)^n makes it just float sign every n, so the interesting factor is e^1/n.
Remember how both 1/n and e^something behaves. If n diverges, 1/n tends to zero. Now, if "something" approaches zero, how does e^"something" behaves?

anonymous · Answer

Well it behaves like e^0 which is 1 I think.

paounn · Answer

True. Now what does the "jumping sign" term do to that 1?

tkhunny · Answer

No one should care about the oscillating sign.  There terms do not disappear (tend toward zero).  They approach 1 (not zero).  It will not ever converge without zero.

anonymous · Answer

Changes it to -1 if n is odd or 1 is n is even.

paounn · Answer

Then it doesn't neither converge nor diverge.

anonymous · Answer

Well thanks for the help, Paounn.