Evaluting $\lim_{n \to \infty } \sqrt[n]{25n+n^3}$

Question

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

$1\leftarrow\sqrt[n]{25n}\le\sqrt[n]{25n+n^3}\le\sqrt[n]{25n^3+n^3}\le \sqrt[n]{26n^3}=\sqrt[n]{26}\cdot\sqrt[n]{n^3}\to1\cdot1=1$

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

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

Is it enought to say $(\sqrt[n]n)^k\to1^k? $

Answer 1

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

• $\sqrt[n]{25n}=\sqrt[n]{25}\sqrt[n]{n}\to 1$
• $\sqrt[n]{2n^3}=(\sqrt[n]{2})^3(\sqrt[n]{n})^3 \to 1$

Answer 2

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

Answer 3

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 $n$ $$\log(a_n)=\frac{3 \log \left({n}\right)}{n}+\frac{k}{n^3}+O\left(\frac{1}{n^5}\right)$$ Continuing with Taylor $$a_n=e^{\log(a_n)}=1+\frac{3 \log \left({n}\right)}{n}+\frac{9 \log ^2\left({n}\right)}{2 n^2}+\frac{6 k+27 \log ^3\left({n}\right)}{6 n^3}+O\left(\frac{1}{n^4}\right)$$ which, for sure, shows the limit and how "far" is $k$ appearing in the approximation.