Leibniz criterion for alternating series Given the series: $\sum_{n=0}^{\infty}(-1)^{n}(\sqrt[n]{n} - 1)^{n}$. Does the series converge?
Attempt to solution (might be incorrect):
$(\sqrt[n]{n} - 1)^{n}> (1+\frac{1}{n})^{n}$
$(1+\frac{1}{n})^{n} \to e \Rightarrow (\sqrt[n]{n} - 1)^{n}$ lower-bounded by $e$. Based on Leibniz Criterion the sequence $\{A_n\}$ (in our case, $(\sqrt[n]{n} - 1)^{n}$) is monotone decreasing, but its limit is not $0$ at infinite $\Rightarrow$ series diverge.
Is it enough to say that since the sequence is lower-bounded, the limit of it at infinite is not $0$, or should I actually calculate the limit of the sequence? 
 A: The alternatingness of the series is something of a red herring as the series converges absolutely. By the root test, to show this it suffices to show that $\lim_{n \rightarrow \infty} |n^{1 \over n} - 1| = 0$. In other words, it suffices to show that $\lim_{n \rightarrow \infty} n^{1 \over n} = 1$. 
There are a few ways to show this limit is in fact $1$. One way is to note that $\ln (n^{1 \over n}) = {\ln(n) \over n}$, and the latter is seen to go to zero as $n$ goes to infinity using L'Hopital's rule. Since the natural log of the $n$th term of the sequence goes to zero, the $n$th term of the sequence goes to $e^0 = 1$. 
A: Your lower bound for $(n^{1/n} -1)^n$ is not correct.
Hint: To show that
$(n^{1/n} -1)^n \rightarrow 0$ write $n^{1/n} = 1 + r_n$ and use the binomial theorem
to show that for $n \ge 2$ we have $r_n < \sqrt{2/(n-1)}.$
Hint2: When you expand $(1+r_n)^n$ the important term in the expansion
is the one in ${r_n}^2 .$
In this way you can show that the series converges absolutely, you don't really need the Leibniz criterion although they will do the job. 
A: Write your series as $\sum_{n = 1}^\infty (-1)^n a_n$ where $a_n = (n^{1/n} - 1)^n$. Note that $(a_n)$ is monotonically decreasing to zero and all terms are positive, hence by the alternating series test (or Leibniz' test) the series converges.
A: I don't use Leibniz criterion.
I put $a_n= (-1)^n \left(\sqrt[n]{n}-1\right)^n\, \forall n\in \mathbb{N}.$
By Cauchy's inequality, I have 
$$\sqrt[n]{\underbrace{1...1}_{(n-4)\times 1} 2. 2 . \sqrt{n}/2.\sqrt{n}/2} \le 1+\frac{1}{\sqrt{n}}, \forall n\ge 4.$$
Hence, $|a_n| \le \frac{1}{\sqrt{n}^n}\, \forall n\ge 4.$
Moreover, since $\sum_{n=4}^{\infty} \frac{1}{\sqrt{n}^n}$ converges, $\sum_{n=1}^{\infty}a_n$ converges.
A: I contribute one idea.
I try to compare $b_n:=\sqrt[n]{n}$ and $\sqrt[n+1]{n+1}$.
Therefore, I compare  $\ln \sqrt[n]{n}=\frac{\ln n}{n}$ and $\ln\sqrt[n+1]{n+1}=\frac{\ln {(n+1)}}{n+1}$.
We consider the function $f(x)= \frac{\ln x}{x}$, where $x\in [3,\infty]$.
It's easy to see that $f$ is a decreasing on $[3,\infty]$.
Hence, $\{b_n\}_{n=3}^{\infty}$ is decreasing.
From that, we have 
$(b_n-1)^n\ge (b_{n+1}-1)^n \ge (b_{n+1}-1)^{n+1}.$
(Because $0<b_{n+1}-1\le 1 \forall n\ge 3.$)
