Convergence and Monotonicity in Limits Verification

[courtrigrad]

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

 

[courtrigrad]

I am not sure if I can use intuituion

 

[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.