Factorials within alternating series

ahmed markhoos
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Homework Statement



∑ [ (-1)^n * n!/(10^n) ]

2. The attempt at a solution

the problem is that I cannot use derivative to make sure that a(n) is decreasing neither L hopital rule to find the limit.
 
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ahmed markhoos said:

Homework Statement



∑ [ (-1)^n * n!/(10^n) ]

2. The attempt at a solution

the problem is that I cannot use derivative to make sure that a(n) is decreasing neither L hopital rule to find the limit.

Have you thought about whether the nth term of the series goes to zero?
 
LCKurtz said:
Have you thought about whether the nth term of the series goes to zero?
I don't know if this is correct or not! because I've used L' hopital rule for one side and left the side of n! without derivation

##\lim_{n\rightarrow \infty} {\frac{n!}{10^n}}##

## ln(f(n)) = ln{\frac {n!}{10^n}} ##
##ln(f(n)) = ln(n!) - ln(10^n)##
##ln(f(n)) = ln(n!) - n*ln(10)##

##\lim_{n\rightarrow \infty} {ln(n!) - n*ln(10)}##

Using L' hopital rule ##\lim_{n\rightarrow \infty} {ln(n!) - ln(10)}\ = ∞ ##

since: ##ln(f(n)) = ∞## $$f(n) = e^∞ = ∞$$

which make the series diverges, is this correct ?
 
That isn't the form for L'Hospital's rule and you certainly don't need L'Hospital's rule for this problem. Why don't you just note the terms are positive and check that they are increasing for large ##n##?
 
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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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