Convergence of $\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n^{1+1/n}}$ Does the series $$\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n^{1+1/n}}$$
converge absolutely, converge conditionally, or diverge?
I've tried applying the ratio test and the root test, and in both cases the limit is $1$, so I cannot conclude anything. 
 A: This series does not converge absolutely, by the limit comparison test
$$ \lim_{n\to\infty} \frac{1/n^{1+1/n}}{1/n} = \lim_{n\to\infty} \frac{1}{n^{1/n}} = 1 $$
with the divergent harmonic series $\sum \frac{1}{n}$.
But since
$$ \frac{1}{n^{1+1/n}} = \frac{1}{n} \exp\left( - \frac{\log n}{n} \right) = \frac{1}{n} + O\left(\frac{\log n}{n^2} \right), $$
the series converges conditionally by noting that
$$ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{1+1/n}} = \sum_{n=1}^{\infty} (-1)^{n} \left( \frac{1}{n}  + O\left(\frac{\log n}{n^2} \right) \right), $$
which is a sum of two convergent series. This is my favorite style of argument for proving that an alternating series with complicated term converges.
If you feel uncomfortable with the Big-Oh notation, then consider the function
$$ f(x) = x^{-1-\frac{1}{x}} = \exp\left( - \frac{x+1}{x}\log x \right). $$
Then by the logarithmic differentiation,
$$\frac{f'(x)}{f(x)} = -\frac{x+1-\log x}{x^2} < 0 $$
for large $x$ and thus $f(x)$ is a non-negative decreasing function. Since it is immediate that $f(x) \to 0$ s $x \to \infty$, the conclusion follows from the alternating series test that
$$ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{1+1/n}} = \sum_{n=1}^{\infty} (-1)^{n} f(n) $$
converges.
A: The series converges conditionally. Because
$$
\frac1{n^{1+1/n}}\leq\frac1n\to0,
$$
and so we have an alternating series with the general term going to zero (Leibnitz Criterion). Edit: as correctly pointed out in the comments, Leibnitz criterion requires us to check for monotonicity. This can be done by looking at the derivative of $f(t)=1/(t^{1+1/t})$. We have
$$
f'(t)=\frac1{t^3}\,e^{-\frac1t\log t}\,(-t+\log t -1).
$$
As the factors out of the brackets are positive for $t>0$, and the factor in brackets is negative for all $t>0$, we get that $f$ is decreasing, and so is the sequence. 
The series does not converge absolutely. Because
$$
\sum_n\frac1{n^{1+1/n}}=\sum_n\frac1n\,\frac1{n^{1/n}}>e^{-1/2}\,\sum_{n\geq n_0}\frac1n=\infty
$$
Note that $n^{-1/n}=e^{-\frac1n\,\log n}\to1$. Fix $n_0$ such that $\frac1n\,\log n<1/2$ for all $n\geq n_0$ (such $n_0$ is very small, but that doesn't matter). Then, for all $n\geq n_0$,
$$
\frac1{n^{1/n}}=n^{-1/n}=e^{-\frac1n\log n}>e^{-1/2}
$$
A: by alternating series test , $$ \lim_{n\to\infty} \frac{1}{n^{1+\frac{1}{n}}} = 0 $$
so, given series converges
