Prove the convergence or divergence of $\sum_{n=0}^\infty \left( (n+1)^\frac1n-(n)^\frac1n\right)$ Let $$u_n:=(n+1)^\frac{1}{n}-(n)^\frac1n$$
How to prove the convergence  or divergence of the series $\sum u_n$ ?
I tried the usual criteria but failed.
 A: With $f(x)=x^{1/n}$, we have $(n+1)^{1/n}-n^{1/n}=f'(\nu)$ for some $\nu\in(n,n+1)$ (Mean Value Theorem).
With $f'(x)=\frac{x^{1/n}}{nx}$, this makes
$$0<u_n<\frac{(n+1)^{1/n}}{n^2}. $$
The numerator tends to $1$ as $n\to \infty$. In fact from $n+1\le2^n$ for $n\ge1$, we find $$0<u_n<\frac2{n^2}$$ so that $\sum u_n$ converges by comparison with a well-known converging series.
A: we have the next inequality is true for $n \geq 6$:
$$(n+1)^{\frac{1}{n}}-n^\frac{1}{n}<(n+1)^{\frac{1}{n}}-(n+2)^{\frac{1}{n+1}}$$
this is a telecopic sum, therefore:
$$\sum_{n=6}^\infty u_n<\sum_{n=6}^\infty\left((n+1)^{\frac{1}{n}}-(n+2)^{\frac{1}{n+1}}\right)=(6+1)^\frac{1}{6}-\lim_{n\to\infty}(n+2)^{\frac{1}{n+1}}=7^\frac{1}{6}-1$$
In the case when $n=0$ we have $u_0=e$
therefore $u_n$ converges
God bless us
A: $$
u_{n} = {(n+1)}^{\frac{1}{n}}-{n}^{\frac{1}{n}}\\
u_{n} = e^{\frac{ln(n+1)}{n}}-e^{\frac{ln(n)}{n}}\\
u_{n} = e^{\frac{ln(n)}{n}}(e^{\frac{ln(n+1)-ln(n)}{n}}-1)\\
u_{n} = e^{\frac{ln(n)}{n}}(e^{\frac{ln(1+\frac{1}{n})}{n}}-1)\\
u_{n} \sim  e^{\frac{ln(n)}{n}}(e^{\frac{1}{n^2}}-1)\\
u_{n} \sim  e^{\frac{ln(n)}{n}}.\frac{1}{n^2}\\
u_{n} \sim  \frac{1}{n^2}
$$
The last line comes from the fact that
$$
e^{\frac{ln(n)}{n}} \sim 1
$$
We conclude, then, that the series converge
A: This is not a solution (so please don't down-vote it)... but merely a graph of the summed series up to $n = 10,000$:

