How would you prove this converges? $\sum_{1}^{\infty} \frac{(-1)^n}{[n-(-1)^n]^{\frac{2}{3}}}$ How would you check if this converges or not?
$$\sum_{1}^{\infty} \frac{(-1)^n}{[n-(-1)^n]^{\frac{2}{3}}}$$
It looks like a telescopic sequence so I thought I'd first write the beginning values:
$$S_n=\frac{-1}{2^{\frac{2}{3}}} + \frac{1}{1^{\frac{2}{3}}} + \frac{-1}{4^{\frac{2}{3}}} ... $$ 
But I got no conclusions from it...
What am I missing here?
Thanks.
 A: We have:
$$ \frac{1}{\left[n+(-1)^n\right]^{\frac{2}{3}}} = \frac{1}{n^{\frac{2}{3}}}+O\left(\frac{1}{n^{\frac{5}{3}}}\right)$$
where:
$$ \sum_{n\geq 1}\frac{(-1)^n}{n^{5/3}}$$
is an absolutely convergent series and
$$ \sum_{n\geq 1}\frac{(-1)^n}{n^{2/3}}$$
is conditionally convergent by Leibniz' test.
A: First, note that your series is not absolutely convergent: placing a modulus inside the sum you get $\sum \frac 1 {\big( n - (-1)^n \big)^{\frac 2 3}}$ which, keeping only the dominant term and applying the limit comparison test, behaves like $\sum \frac 1 {n^{\frac 2 3}}$ which is clearly divergent.
Now, is your series convergent? One cannot apply Leibniz's test because the sequence $\frac 1 {\big( n - (-1)^n \big)^{\frac 2 3}}$ does not decrease, even though it tends to $0$.
Let us examine once more the definition of the concept of convergent series: $x_1 + x_2 + \cdots + x_n + \cdots$ converges if and only if the sequences $(x_1 + x_2) + (x_3 + x_4) + \cdots + (x_{2N-1} + x_{2N})$ and $(x_1 + x_2) + (x_3 + x_4) + \cdots + (x_{2N-1} + x_{2N}) + x_{2N+1}$ converge, both, to the same limit, when $N \to \infty$.
In our case, $x_{2n+1} \to 0$, so we just have to show that $(x_1 + x_2) + (x_3 + x_4) + \cdots + (x_{2N-1} + x_{2N}) = \sum \limits _{n=1} ^N (x_{2n-1} + x_{2n})$ converges. In order to study this convergence we shall transform the summand as follows:
$$x_{2n-1} + x_{2n} = \\ -\frac 1 { {\sqrt[3] {2n-1 +1}} ^2} + \frac 1 { {\sqrt[3] {2n-1}} ^2} = \\ \Big( \frac 1 {\sqrt[3] {2n-1}} + \frac 1 {\sqrt[3] {2n}} \Big) \Big( \frac 1 {\sqrt[3] {2n-1}} - \frac 1 {\sqrt[3] {2n}} \Big) \sim \\ \frac 1 {\sqrt[3] n} \frac {\sqrt[3] {2n} - \sqrt[3] {2n-1}} {\sqrt[3] {2n} \sqrt[3] {2n-1}} \sim \\ \frac 1 n \frac {\sqrt[3] {2n} - \sqrt[3] {2n-1}} 1 = \\ \frac 1 n \frac {2n - (2n-1)} { {\sqrt[3] {2n}}^2 + {\sqrt[3] {2n}} {\sqrt[3] {2n-1}} + {\sqrt[3] {2n-1}}^2 } \sim \\ \frac 1 {n^{\frac 5 3}} ,$$
where the notation $a_n \sim b_n$ means $a_n, b_n >0$ and $\lim \limits _{n \to \infty} \frac {a_n} {b_n} \in (0, \infty)$.
Since $\sum \frac 1 {n^{\frac 5 3}}$ converges, the convergence of your series follows now easily.
A: You need two things:
(1) The terms tend to zero (though not monotonically).
(2) If you consider pairs of terms
for $2n-1$ and $2n$,
if $f(n) = \frac1{n^{\frac{2}{3}}}$
$-\frac1{(2n)^{\frac{2}{3}}}
+\frac1{(2n-1)^{\frac{2}{3}}}
=-f(2n)+f(2n-1)
=f'(x)$
where
$2n-1 < x < 2n$
by the mean value theorem.
But
$f'(x)
= \frac23\frac1{x^{5/3}}
$,
so
$\sum_{n=2}^{\infty} (-f(2n)+f(2n-1))
$
converges.
A: Let $(S_n)$ be the sequence of partial sums for the given series, 
and let $(T_n)$ be the sequence of partial sums for  $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2/3}}$.
Since this series converges by the Alternating Series Test, $\;T_n\to l$ for some number $l$
and therefore $S_{2n}=T_{2n}\to l$.  $\;\;$Then $S_{2n+1}=S_{2n}-\frac{1}{(2n+2)^{2/3}}\to l$ also, 
so $S_n\to l$ and the given series converges.
