Jtdylktuy

$lim limits_{n to infty } sqrt[n]{25n+n^3}$

$1leftarrowsqrt[n]{25n}lesqrt[n]{25n+n^3}lesqrt[n]{25n^3+n^3}le sqrt[n]{26n^3}=sqrt[n]{26}cdotsqrt[n]{n^3}to1cdot1=1$

But is it obvious that $ forall _{kin Bbb N} (sqrt[n]n)^k to1 ?$

Since I know $(sqrt[n]n) to1$.

Is it enought to say $(sqrt[n]n)^kto1^k? $

Yes your idea is correct indeed we have that eventually

$$sqrt[n]{25n}le sqrt[n]{25n+n^3}le sqrt[n]{2n^3}$$

and

• $sqrt[n]{25n}=sqrt[n]{25}sqrt[n]{n}to 1$


• $sqrt[n]{2n^3}=(sqrt[n]{2})^3(sqrt[n]{n})^3 to 1$

use $limsup_{n rightarrow infty} |a_n|^{1/n}$ to help you find the answer.

You could also have good approximations for finite values of $n$.

$$a_n=sqrt[n]{kn+n^3}implies log(a_n)=frac 1 n log(kn+n^3)=frac 1 n left(3 log(n)+log left(1+frac{k}{n^2}right) right)$$ Now, using Taylor expansions for large values of $n$
$$log(a_n)=frac{3 log
left({n}right)}{n}+frac{k}{n^3}+Oleft(frac{1}{n^5}right)$$ Continuing with Taylor
$$a_n=e^{log(a_n)}=1+frac{3 log left({n}right)}{n}+frac{9 log
^2left({n}right)}{2 n^2}+frac{6 k+27 log
^3left({n}right)}{6 n^3}+Oleft(frac{1}{n^4}right)$$ which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.