Convergence of this series

Hi guys, I was trying to do this problem with the ratio test to see it's convergence. I get divergent as the ratio test gives me an answer > 1 but wolframalpha tells me it's convergent by ratio test. Can anyone test this?

The series is:

∑(e^n-(-pi)^n)/4^n from n =1 to infinity


Thanks!

This does converge.

Using the geometric series,
∑(n = 1 to ∞) (e^n - (-π)^n)/4^n
= ∑(n = 1 to ∞) (e/4)^n - ∑(n = 1 to ∞) (-π/4)^n
= (e/4) / (1 - e/4) - (-π/4) / (1 - (-π/4)), since |e/4| < 1 and |-π/4| < 1
= e/(4 - e) + π/(4 + π).

I hope this helps!
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P.S.: Using the ratio test,

r = lim(n→∞) |[(e^(n+1) - (-π)^(n+1))/4^(n+1)] / [(e^n - (-π)^n)/4^n]|
..= (1/4) * lim(n→∞) |e^(n+1) - (-π)^(n+1)| / |e^n - (-π)^n|
..= (1/4) * lim(n→∞) |(e/(-π))^(n+1) - 1| / |e^n/(-π)^(n+1) - 1/(-π)|, dividing each term by (-π)^(n+1)
..= (1/4) * lim(n→∞) |(-e/π)^(n+1) - 1| / |e (-e/π)^(n+1) + 1/π|
..= (1/4) * lim(n→∞) |0 - 1| / |0 + 1/π|, since |-e/π| < 1
..= π/4.

Since r = π/4 < 1, the series converges.