Are the expressions involving the nth root and the natural logarithm equivalent? 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?
 A: 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 $n$
$$\left(
\begin{array}{cc}
 k & n \\
 2 & 22 \\
 3 & 104 \\
 4 & 429 \\
 5 & 1658 \\
 6 & 6171 \\
 7 & 22398 \\
 8 & 79814 \\
 9 & 280500 \\
 10 & 975123 \\
 11 & 3360257 \\
 12 & 11495782
\end{array}
\right)$$
A: They are not particularly close.

