# Limit of the sequence $\frac{n}{a^n}$ for a>1

In general, the sequence defined by $$a_n=\frac{n}{c^n}$$ Diverges if $\,c\lt1$, and converges to $0$ if $\,c\gt1$. The proof of divergence for $\,c\lt1$ is quite straight forward, but I'm trying to proof that the sequence $\{a_n\}$ converges to $0$ when $\,c\gt1$ by only using the limit definition for convergence. Other results from limits are allowed, as long as they can be proved with algebra and by using some functions (differentiation or integration is not allowed, while trascendental functions are).

I have been trying to prove that $$\frac{n}{c^n}$$ is less than some other expresion involving $n$ so a direct relationship with some $\epsilon\gt0$ can be obtained, in order to use the definition of limit. Other ways involve to prove that ${a_n}<b_n$ for some sequence $\{b_n\}$ that is known to be convergent to $0$ but I haven't found the way.

• $a_{n+1}/a_n=\frac{n+1}{nc}$, which becomes, eventually, $\le \frac12\left(1+\frac1c\right)<1$. So, the tail of the sequence is dominated by $K\alpha^n$ for some constants $\alpha<1$ and $K$. – user228113 Nov 5 '17 at 23:10

Write $c=1+p$ for some $p>0$. Then $(1+p)^{n}\geq 1+np+\dfrac{n(n-1)}{2}p^{2}$ and $\dfrac{n}{c^{n}}\leq\dfrac{n}{1+np+\dfrac{n(n-1)}{2}p^{2}}$. Now taking $n\rightarrow\infty$.

It suffices to show that for $n$ large enough, $c^n>n^2$. (Because then $\frac{n}{c^n}<\frac{n}{n^2}=\frac{1}{n}\to0$.)

This is equivalent to showing $\sqrt{c}>\sqrt[n]{n}$ for $n$ large enough.

Since $c>1$, it suffices to show $\lim_{n\to\infty}\sqrt[n]{n}=1$. Here is one proof of that.

Hint:

Prove that $\;\dfrac{a_{n+1}}{a_n}\le k$ for some $k<1$ if $n$ is large enough, then deduce that$(a_n)$ is eventually bounded from above by a geometric series with ratio $k$.