Find this limit: $\lim_{n \rightarrow \infty } n \cdot c^{n} \ \ \ $

Question

Find this limit: $\lim_{n \rightarrow \infty } n \cdot c^{n} \ \ \ $ with $|c|<1$

My atempt
We can define $c=\frac{1}{a}\ \ \ $ with $a>1$

Then, \begin{align} \lim_{n \rightarrow \infty } n \cdot c^{n} &= \lim_{n \rightarrow \infty } \frac{n}{a^{n}}\\ &=\lim_{n \rightarrow \infty } \frac{n}{a^{n}} \cdot \frac{(\frac{1}{n})}{(\frac{1}{n})}\\&=\lim_{n \rightarrow \infty } \frac{1}{\frac{a^{n}}{n}} \end{align}

And, $\lim_{n \rightarrow \infty } \frac{a^{n}}{n}$ goes to $\infty$

So,

\begin{align} \lim_{n \rightarrow \infty } n \cdot c^{n} &=0 \end{align}

My question

• I know this is a very easy question, but how can I prove the fact that $\lim_{n \rightarrow \infty } \frac{a^{n}}{n}$ goes to $\infty$?
• Am I correct? Is there any other way to find this limit?

Thank you very much for your support!

Answer 1

Best to write $\lvert c\rvert=\frac{1}{1+a}$ with $a>0$, then use the binomial theorem. You likely can't assume the limit to infinity if you need to prove this result.

For $\lvert c \rvert\lt1$ and $n\gt 2$,

$$ \begin{align} \lvert nc^n\rvert&=\frac{n}{(1+a)^n}\\&=\frac{n}{1+an+\frac{1}{2}n(n+1)a^2+\cdots}\\&\leq\frac{n}{\frac{1}{2}n(n+1)a^2}\\&\leq\frac{2}{na^2} \end{align} $$

which converges to zero straightforwardly. Then the results follow by sandwiching (squeeze theorem).

In case you are interested $\frac{a^n}{n}$ does tend to $\infty$ for $a\gt1$ by a similar method:

Write $a=1+b, b\gt 0$. Then for $n\gt2$, $$ \begin{align} \frac{a^n}{n}&=\frac{(1+b)^n}{n}\\&=\frac{1+bn+\frac{1}{2}n(n+1)b^2+\cdots}{n}\\&\geq \frac{\frac{1}{2}n(n+1)b^2}{n}\\&\geq \frac{nb^2}{2} \end{align} $$ and the last term of course tends to infinity.