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} $?
calculate limit of series
3 Answers
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.
-
0Yeah, my mistake, I corrected it. – 2012-11-16
$ \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 $k^2$
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}. $