calculate limit of series

Question

While getting limit of infinite series I have came to next expession $$ \lim_{k \to \infty} \frac{1}{\frac{3^k}{k^2}} $$ and do not know how to procede with $$ \lim_{k \to \infty} {3^k} $$?

Answer 1

You can solve this without knowledge of which function grows faster, by using L'Hospital rule twice:

$$ \lim_{k \to \infty} \frac{1}{\frac{3^k}{k^2}} = \lim_{k \to \infty} \frac{k^2}{3^k} = \lim_{k \to \infty} \frac{2k}{3^k \ln 3} = \lim_{k \to \infty} \frac{2}{3^k (\ln 3)^2} = 0 $$

You can read on wikipedia more on L'Hospital rule.

Answer 2

$$ \lim_{k \to \infty} \frac{1}{\frac{3^k}{k^2}}= \lim_{k \to \infty} \frac{k^2}{3^k}=0$$ since $3^k\to\infty$ more fastly than $k^2$

Answer 3

We have $$ \lim_{k \to \infty} \frac{1}{\frac{3^k}{k^2}}= \lim_{k \to \infty} \frac{k^2}{3^k} $$ Now note that ( by binomial theorem ) $$ 3^k=(2+1)^k=2^k+k(2)^{k-1}+\frac{k(k-1)}{2}(2)^{k-2}+\frac{k(k-2)(k-3)}{6}(2)^{k-3}+ \dots +\frac{k(k-2)(k-3)}{6}(2)^{3}+\frac{k(k-1)}{2}(2)^{2}+k(2)^1+1\geq k^3 $$ implies $\frac{k^2}{3^k}\leq \frac{k^2}{k^3}=\frac{1}{k}$. The result then follows from the sandwich theorem applied on inequality: $$ 0\leq\frac{k^2}{3^k}\leq \frac{1}{k}. $$