$$\lim_{n \to \infty}a_n=\dfrac{5^{3\cdot n}}{2^{\left(n+1\right)^2}}$$
I am trying to solve it using the squeeze theorem. I have opened the expression to $$a_n=\dfrac{5^3\cdot 5^n}{2^{n^2}\cdot2^{2n}\cdot2)}$$ I think that the LHS should be $$a_n=\dfrac{2^3\cdot 2^n}{2^{n^2}\cdot2^{2n}\cdot2)}$$ But as for the RHS I do not find a bigger expression, any ideas?
Hint: $0<\dfrac{125^n}{2^{(n+1)^2}} < \dfrac{125^n}{2^{n^2}} < \left(\dfrac{1}{2}\right)^n$
since $\dfrac{125}{2^n} < \dfrac{1}{2}$ when $n > 8$.
$$ \frac{a_{n+1}}{a_{n}}=\\\frac{\frac{5^{3(n+1)}}{2^{(n+2)^2}}}{\frac{5^{3(n)}}{2^{(n+1)^2}}}=\\\frac{5^{3n+3}}{5^{3n}}\frac{2^{(n+1)^2}}{2^{(n+2)^2}}=\\125\frac{2^{n^2+2n+1}}{2^{n^2+4n+4}}=\\125\frac{1}{2^{2n+3}}=\\\frac{125}{8}\frac{1}{4^n}\\$$when n become large $$n\rightarrow \infty\\\frac{a_{n+1}}{a_{n}}=\frac{125}{8}\frac{1}{4^n} \rightarrow 0$$
\dfracin titles. And the title should not consist solely of mathematical symbols. See meta:http://meta.math.stackexchange.com/questions/9687/guidelines-for-good-use-of-latex-in-question-titles/9730#9730 – Martin Sleziak Dec 29 '14 at 10:42