Posted by Subhotosh Khan on August 05, 2002 at 08:14:54:
In Reply to: ratio test posted by Don on August 02, 2002 at 18:23:09:
: : To show that the summation of k^k / k! diverges my book gives this as an example:
: : a sub(k+1)/a sub(k) =(k+1)^k+1 / (k+1)! * k!/ k^k
: : =(k+1/ k)^k
: : = (1 + 1/k)^k = e as k goes to infinity.
: : What I don't see is how they factored (k+1)!, cancel out terms and got from
: : (k+1)^k+1 / (k+1)! * k!/ k^k
: : to
: : (1 + 1/k)^k .
: : When I tried I got a different answer.
: : Can someone help me?
: : Thanks!
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[(k+1)^k+1] / {(k+1)!} * k!/ k^k
=[(k)^(k+1)*(1 + 1/k)^(k+1)]/{(k+1)*k!} * k!/k^k
=[k*(1 + 1/k)^(k+1)]/{(k+1)}
=[k*(1 + 1/k)]/{(k+1)}*(1 + 1/k)^(k)
Now do you see it....