# where am I going wrong

#### kolahalb

Joined Sep 25, 2007
6
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 ∞

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

#### kolahalb

Joined Sep 25, 2007
6
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...

#### DMaiorca

Joined Oct 15, 2007
1
I think I spotted your error:

Changing variables as 1/n=t,the limit becomes (1/R)=Lt(t->0) [(t!)^t]

In actuality, the sub should have been (1/R)=Lt(1/t->0) [(t!)^t]

Its the simple stuff that always messes with me, too!

Joined May 1, 2007
19
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 ∞

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
hey kolahalb

your series does indeed converge everywhere in the complex numbers. it is just exp(z). obviously (1/n!) tends to zero and so a-fortiori its n-th root tends to zero. hence 1/R = 0 and R is infinite.

note incidentally that (1/n!) is 1/(n!) and not (1/n)! the latter is undefined for n<>1 [what is (1/2)! for example?]

peace
stm