Convergence of Infinite Series

americanforest
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Question: Test for convergence:

\sum\frac{n!}{10^n}

(the sum is from 1 to infinity)

I tried using

\frac{n^n}{10^n}\geq\frac{n!}{10^n}\geq\frac{n}{10^n}

and showing that either the first one was convergent or the last one was divergent using various tests but didn't get anywhere.

Any hints?
 
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Try using the Ratio Test. That's what I first try to do anytime I see n!
 
Ratio Test:

\frac{(n+1)!}{10^{n+1}}\frac{10^n}{n!}=\frac{n+1}{10}

Since that's more than 1 as n goes to infinity it diverges. Am I right?
 
I did a bunch of these. I need to get a good grade so I will just post them up here with my answers and if I got one wrong please just let me know to look over it again.

1. \sum\frac{1}{ln(n)} (sum from 2 to inf.) diverges

b/c 1/ln(n)>1/n which diverges

2. \sum\frac{1}{2n(2n+1)} converges

b/c 1/n^2 converges

3. \sum\frac{1}{(n(n+1))^{.5}} diverges

b/c 1/n diverges and then limit comparison test to show this diverges too.

4. \sum\frac{1}{2n+1} diverges

b/c 1/2n diverges and then limit comparison to show this diverges too.

I'd really appreciate it if somebody could check me on these. Thanks.
 
Those all look right to me
 
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...

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