# Limits Verification

1. Jan 23, 2005

Hello all

I encountered the following problems:

(a) Prove that $$\lim_{x\rightarrow \infty}((n+1)^\frac{1}{3} - n^\frac{1}{3}) = 0$$

Using the relation $$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$$ we get $$\frac{((n+1)^\frac{1}{3} - n^\frac{1}{3})((n+1)^\frac{1}{3})^2 + ((n+1)^\frac{1}{3}(n^\frac{1}{3}) + (n^\frac{1}{3})^2}{(n+1)^\frac{1}{3} - n^\frac{1}{3})}$$

Hence $$\frac {1}{((n+1)^\frac{1}{3})^2 + ((n+1)^\frac{1}{3}(n^\frac{1}{3}) + (n^\frac{1}{3})^2}$$

and this approaches 0

Also another problem I encountered was:

Let $$a_n = \frac {10^n}{n!}$$
(a) To what limit does this converge
(b) Is this sequence monotonic
(c) Is it monotonic from a certain n onwards
(d) Give an estimate of the difference between $$a_n$$ and the limit.
(e) From what value of n onwards is this difference less than $$\frac {1}{100}$$?

(a)Is it valid to say that just by looking we can see that the limit converges to 0? How would you do it algebraically? In other words, how do you prove that $$\lim_{x\rightarrow \infty}(\frac {n!}{n^n} = 0)$$
(b&c) I think this sequence is monotonic from a certain n onwards
(d) The estimate in the difference of the limit could be any value $$\epsilon$$. So $$a_n = \frac {10^n}{n!} < \epsilon$$ where $$\epsilon = \frac {1}{100}$$. IS this right? How would you actually estimate the difference?

(e) So $$a_n = \frac {10^n}{n!}< \frac {1}{100}$$. How would you solve
for this?

Thanks a lot

Last edited: Jan 23, 2005
2. Jan 24, 2005

I am not sure if I can use intuituion

3. Jan 24, 2005

### dextercioby

Use the fact that the natural logarithm and the limit commute and the Stirling's approximation

$$\ln n!\sim n\ln n-n$$

for very big "n"...

Daniel.