I have to find and prove the limit of the sequence $X_n=$ $\frac {n^{100}}{1.01^n}$
What is the easier way?
I tried to use Bernoulli's inequality to say lim$\frac {n^{100}}{1.01^n}$ = lim$\frac {n^{100}}{(1+1/10)^n}$ and $ (1+1/10)^n \geq 1+n(1/10).$ But I could get anything.
I think another way is to use the Squeeze Theorem but I have not could find the correct sequences.
Any Ideas?
I only can use the definition, Squeeze Theorem, the Bernoulli's inequality or using operations to reduce the sequence.
Best Answer
We have $$\frac {n^{100}}{1.01^{n}} = \left(\frac{n}{1.01^{n/100}}\right)^{100} $$ By Bernoulli's inequality on the inside quantity, $$\begin{align}\left(\frac{n}{\left(1+\frac{1}{100}\right)^{n/100}}\right)^{100} &\leq \left(\frac{n}{1 + \frac{1}{100}\cdot\frac{n}{100}}\right)^{100} \\ &< \left(\frac{n}{\frac{n}{10000}}\right)^{100} \\ &= 10000^{100} \end{align}$$