Infinite limit of factorial series

flyerpower
Messages
46
Reaction score
0

Homework Statement


MSP5419g73d00047g3g1200002bha5g9gah09ab35.gif


The Attempt at a Solution



I have no idea where to start from.
I know that e = 1/0! + 1/1! + 1/2! + 1/3! + ... + 1/n! + ...
and tried to separate the factorial off the geometric series but had no success.
 
Physics news on Phys.org
We start from \frac{2^{k-1}(k-1)}{(k+1)!} =\frac{2^{k-1}(k+1-2)}{(k+1)!} =<br /> \frac{2^{k-1}}{k!}-\frac{2^k}{(k+1)!} and we recognize a telescopic series.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top