Answer to Question #334815 in Real Analysis for ABC2468

Question

Show that for any A > 1 and any positive k, $Limit$ as n approaches infinity $n^k/A^n=0$

Answer 1

Choose an arbitrary number $q$ satisfying: $1<q<A$. It is enough to prove the following inequality: $\frac{n^k}{A^n}\leq\frac{q^n}{A^n}$ for sufficiently large $n$ and to show that $\frac{q^n}{A^n}\rightarrow0$, $n\rightarrow\infty$.

Consider the inequality $n^k\leq q^n$. Since the function $ln(x)$ is increasing, it is enough to show the inequality: $k\,ln(n)\leq n\, ln(q)$. Consider the function: $f(x)=x\,ln(q)-k\,ln(x)$. For sufficiently large $x$ the following inequality holds: $x^{\frac12}>ln(x)$ (it follows from the properties of function $x^{\frac12}-ln(x)$ and its derivative). Thus, we obtain: $f(x)=x\,ln(q)-k\,ln(x)>x\,ln(q)-k\,x^{\frac12}=x^{\frac12}(x^{\frac12}ln(q)-k)$. We receive that for $x>(\frac{k}{ln(q)})^2$ $f(x)>0$. Thus, $n\,ln(q)-k\,ln(n)>0$ for $n>(\frac{k}{ln(q)})^2$. Therefore, $n^k\leq q^n$. It remains to prove that $b^n\rightarrow0,n\rightarrow\infty$ for $0<b<1$.Consider the expression $n\,ln(b)$. The latter tends to $-\infty$ as $n\rightarrow\infty$. $b^n=e^{n\,ln(b)}\rightarrow0,n\rightarrow\infty$. It remains to put $b=\frac{q}{A}$ to complete the proof. We obtained that $\frac{n^k}{A^n}\leq\frac{q^n}{A^n}\rightarrow0,n\rightarrow\infty.$