Yes, that is completely correct. 1/n goes to 0 as n goes to infinity so goes to 1. For large n, then, goes to 0 and the alternating series converges.
My series is:
(-1)^(n-1) * e^(1/n)
Now I say that it is convergent. My reasoning for this is that I believe that the following is true:
That series is equivalent to:
(-1)^(n-1) * [e^(1/n) / n]
For an alternating series if the part of the series that does not alternate is decreasing, and the limit of it is equal to 0, then the original series is convergent.
And since [e^(1/n) / n] > [e^(1/(n+1))/(n+1)] the non-alternating part is decreasing.
and since the limit as n goes to infinity of it is simply just 1/infiniti which equals 0, then the original alternating series is convergent.