Difficult limit of a sequence

[gcc00]

Hi everyone, I was trying to do this limit

As n tends to infinity, n belongs to Naturals,

My idea was to do this:



But I'm not able to calculate the limit circled in red

Thanks in advance to everyone who is going to help

 

[BeansNRice]

Have you tried using Stirling's approximation for the factorial?

 

[gcc00]

It's an exercise, I have to solve it with convergence tests, in theory it should be doable with only root test and the ratio test

 

[logistic_guy]

Hi everyone, I was trying to do this limit
View attachment 39284
As n tends to infinity, n belongs to Naturals,

\(\displaystyle \frac{\sqrt{999!} + 2^{\sqrt{999}}}{3^{999} + 999^3} \approx 4.55 \times 10^{805}\)


\(\displaystyle \frac{\sqrt{9999!} + 2^{\sqrt{9999}}}{3^{9999} + 9999^3} \approx 9.81 \times 10^{13056}\)


This suggests that:

\(\displaystyle \lim_{n\rightarrow \infty}\frac{\sqrt{n!} + 2^{\sqrt{n}}}{3^{n} + n^3} = \infty\)

 

[fresh_42]

It's an exercise, I have to solve it with convergence tests, in theory it should be doable with only root test and the ratio test

This looks divergent. Try to find a lower bound that increases with n. You only need to consider [imath] \dfrac{\sqrt{n!}}{3^n}. [/imath]

 

[Dr.Peterson]

It's an exercise, I have to solve it with convergence tests, in theory it should be doable with only root test and the ratio test

I'm familiar with the ratio test and root test for a series, but this is a sequence. Just to make sure, you aren't talking about the sum of those terms, are you?

But it does make sense to consider the ratio of successive terms, in combination with what you already did.

 

[BeansNRice]

Stirling's approximation for the factorial quickly shows this expression diverges. Give it a try!

[imath]n! \approx \sqrt{2\pi n}\left(\dfrac n e\right)^n[/imath]

 

[gcc00]

@Dr.Peterson I was taught that sequences and series are equivalent, in the sense that for every serie you can make up a sequence and viceversa (there's a bijection) so it make sense that the techniques are also the same. Anyway, I solved it this way: Doing the ratio test and multiplying for the conjugate:

 

[fresh_42]

Yes, correct. Now you have reduced the problem in post #1 and #8 to the question [imath] \dfrac{\sqrt{n!}}{3^n}\cdot \dfrac{1+a_n}{1+b_n}\longrightarrow \dfrac{\sqrt{n!}}{3^n} [/imath] for [imath] n\to \infty [/imath] because [imath] \displaystyle{\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n =0.}[/imath]

Now use the hint in post #6 and do the same for [imath] \dfrac{\sqrt{n!}}{3^n} [/imath] showing divergence.

 

[Dr.Peterson]

@Dr.Peterson I was taught that sequences and series are equivalent, in the sense that for every serie you can make up a sequence and viceversa (there's a bijection) so it make sense that the techniques are also the same.

No, they are related, but not the same. A series is the sum of a sequence, and the partial sums of a series are a different sequence; you can't do the same things to both.

What must be true of the terms of a series in order for the series to converge is not the same as what must be true of the terms of a sequence for the sequence to converge. Some things are similar, but they are not the same thing.

For example, the sequence [imath]\frac{1}{1},\frac{1}{2},\frac{1}{3},\dots[/imath] converges to zero, but the series [imath]\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots[/imath] diverges.

Be very careful not to confuse the two. What you are doing here, fortunately, is appropriate for sequences.