I stumbled at the answer of a problem solved in Reily-Hobson-Bence.Please check if I am wrong:
Find the parts of the z-plane for which the following series is convergent:
∑[(1/n!)(z^n)] where n runs from 0 to ∞
Cauchy's radius: (1/R)=Lt (n->∞) [(1/n!)^(1/n)]
Changing variables as 1/n=t,the limit becomes (1/R)=Lt(t->0) [(t!)^t]
I am having the limit as 1 and hence,R=1...but the book says Since
[(n!)^(1/n)] behaves like n as n → ∞ we find lim[(1/n!)^(1/n)] = 0. Hence R = ∞ and the series is convergent for all z
Thank you for your help.
I resolved the problem in this manner: (1/n!) falls at a much faster rate than that of (1/n).As a result the function is decreasing all the way starting from the highest value one and always being positive.If you draw a graph,the curve should tend to zero as n tends to infinity...