How to solve this limit $lim_{nto infty}(frac{2^n}{n^k})$?

Answer :

$$lim_{nto+infty} frac{2^n}{n^k}=lim_{nto+infty}frac{e^{nln(2)}}{e^{kln(n)}}=  lim_{nto+infty}e^{n(ln(2)-kfrac{ln(n)}{n}) } $$ Note that $$ lim _{nto+infty} kfrac{ln(n)} {n} =0$$

So :

$$lim_{nto+infty} frac{2^n}{n^k}=lim_{nto+infty} e^{nln(2)}=+infty $$

Because $ln(2)>0$.

Suppose $k>0$. Let $m=[k]+1$. Then, for $n>2m+2$ $$ frac{2^n}{n^k}=frac{(1+1)^n}{n^k}ge frac{binom{n}{m}}{n^{m+1}}=frac{n(n-1)cdots(n-m+1)}{m!n^{m+1}}ge frac{(n-m+1)^{n-m}}{m!n^{m+1}}=frac{1}{m!}(1-frac{m}{n})^{m+1}(n-m)^{n-2m-1} $$ Since $$lim_{nto infty}frac{1}{m!}(1-frac{m}{n})^{m+1}(n-m)^{n-2m-1}=infty $$ hence one has $$ lim_{nto infty}frac{2^n}{n^k}=infty. $$

Consider the sequence $a_n=frac{2^n}{n^k}$.

$$frac {a_{n+1}}{a_n}= 2frac{n^k}{(n+1)^k}=2left(frac{n}{n+1} right)^k =2left(1-frac{1}{n+1} right )^k.$$

Since $k$ is fixed, for large $n$ this ratio exceeds, say $frac 32$, so $lim a_n = infty$.

Take the logarithm of the expression (assuming $k ge 1$)

$$ lnleft(frac{2^n}{n^k}right) =ln(2^n)-ln(n^k) = nln(2) +lnleft(frac{1}{n^k} right) $$ It follows begin{align} lim_{nto infty}lnleft(frac{2^n}{n^k}right) &= lim_{nto infty}left[nln(2) +lnleft(frac{1}{n^k} right)right] \ &=lim_{nto infty}(nln(2)) + lim_{nto infty}lnleft(frac{1}{n^k} right) \ &=+infty + 0 = +infty end{align} So the limit of the logarithm is $+infty$ and so is the limit $$ lim_{nto infty}left(frac{2^n}{n^k}right) = +infty $$ simply explained, $ln(x) to +infty implies x to +infty$