Are the expressions involving the nth root and the natural logarithm equivalent?

Question

Are the following expressions equivalent?

Expression 1:
$1 - \frac{1}{\sqrt[n]{1+n}}, n\in Z^+ $

Expression 2:
$ \frac{ln(1+n)} {n}$

Under what set of conditions is it possible, if at all?

Answer 1

They are not particularly close.

Answer 2

As Richard Ambler suspected by his comment, may be you are considering the case of very large values of $n$ and the limit when $n\to \infty$.

If we use expansions $$A=(n+1)^{\frac{1}{n}}\implies \log(A)={\frac{1}{n}}\log(n+1)={\frac{1}{n}}\left(\log(n)+\log\left(1+\frac 1n\right)\right)$$

$$\log(A)=\frac{\log \left({n}\right)}{n}+\frac{1}{n^2}-\frac{1}{2 n^3}+O\left(\frac{1}{n^4}\right)$$ $$A=e^{\log(A)}=1+\frac{\log \left({n}\right)}{n}+\frac{\log ^2\left({n}\right)+2}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ $$1-\frac 1A=1 - \frac{1}{\sqrt[n]{1+n}}=\frac{\log \left({n}\right)}{n}+\frac{1-\frac{1}{2} \log ^2\left({n}\right)}{n^2}+O\left(\frac{1}{n^3}\right)$$ On the other side, doing the same, $$B=\frac{\log (n+1)}{n}=\frac{\log \left({n}\right)}{n}+\frac{1}{n^2}-\frac{1}{2 n^3}+O\left(\frac{1}{n^4}\right)$$ $$1 - \frac{1}{\sqrt[n]{1+n}}-\frac{\log (n+1)}{n}=-\frac{\log ^2\left({n}\right)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$

So, if you want to know for which $n$, you have $$\left|-(n+1)^{-1/n}-\frac{\log (n+1)}{n}+1\right| \leq \epsilon$$, you need to solve for $n$ $$\frac{\log ^2\left({n}\right)}{2 n^2}=\epsilon\implies n=-\frac{W_{-1}\left(- \sqrt{2\epsilon }\right)}{ \sqrt{2\epsilon }}$$ where appears Lambert function.

For more conveniency, let $\epsilon =10^{-k}$ and compute the corresponding rounded value of $n$ $$\left( \begin{array}{cc} k & n \\ 2 & 22 \\ 3 & 104 \\ 4 & 429 \\ 5 & 1658 \\ 6 & 6171 \\ 7 & 22398 \\ 8 & 79814 \\ 9 & 280500 \\ 10 & 975123 \\ 11 & 3360257 \\ 12 & 11495782 \end{array} \right)$$