Convergence of this sequence .

[bs vasanth]

Homework Statement

find the limit n\rightarrow∞ of 10ⁿ/ n!

Homework Equations

L hospital rule

The Attempt at a Solution

took log and separated the num and denom as:
n ln10-ln(n!)
n ln10-n ln(n)+n
1/n ( ln10 - ln(n)+1)
now i applied l hospital rule
then i got lim n\rightarrow∞ as 0.So the actual answer is 1. (e⁰)
I just want to know if the approach is correct.

 

[Fightfish]

n ln10-n ln(n)+n
1/n ( ln10 - ln(n)+1)

I'm afraid I don't see how you transition between these two steps.
Also, intuitively the factorial function should dominate and the limit should be zero.

 

[bs vasanth]

I'm afraid I don't see how you transition between these two steps.
Also, intuitively the factorial function should dominate and the limit should be zero.

my bad ,in my desperate attempt to get a ratio, I did that stupid thing. How do we solve it beyond intuition?

 

[Fightfish]

From n ln10-n ln(n)+n, you can combine all the terms into a single expression. Taking the limit, you will find that the expression tends to minus infinity. The limit of the original expression is then 0.

 

[bs vasanth]

n*ln(10e/n) this is what i am getting , and it is not -infinity .

 

[Dick]

n*ln(10e/n) this is what i am getting , and it is not -infinity .

Yes, the limit of that is -infinity. n goes to +infinity. What does ln(10e/n) do?

 

[bs vasanth]

Yes, the limit of that is -infinity. n goes to +infinity. What does ln(10e/n) do?

I took n out , then
n( ln10-lnn +1 )
n(ln10-lnn+lne)
n ln(10e/n)
i don't know what to do after this.

 

[bs vasanth]

Are you saying that ln0 is taken as -inf, but there is also n before that right which becomes +inf.

 

[HallsofIvy]

The ratio test gives: a_{n+1}/a_n= (10^{n+1}/(n+ 1)!)(n!/10^n)= (10^{n+1})/10^n)(n!/(n+1)!)= 10/(n+1) goes to 0 as n goes to 0. Strictly speaking, the ratio test shows that \sum a_n converges but if that is so, then, of course, the sequence \{a_n\} converges to 0.

(The converse is not necessarily true. For example, \{1/n\} clearly converges to 0 but the sum \sum 1/n does NOT converge so the ratio test does not work.)

 

[Office_Shredder]

Are you saying that ln0 is taken as -inf, but there is also n before that right which becomes +inf.

n*ln(10e/n) if n is really big is a really big positive number multiplied by a really big negative number. What is that going to be?

 

[bs vasanth]

n*ln(10e/n) if n is really big is a really big positive number multiplied by a really big negative number. What is that going to be?

Now i get it , it is going to really big negative number, so -inf.
therefor the answer for the original question is 0.
thankyou everyone.