Absolute convergence of series $\sum\limits_{n=2}^\infty{(-1)^{n-1}\over (n+(-1)^n\sqrt{n})^{2\over 3}}$ Exam absolute convergence of series $$\sum\limits_{n=2}^\infty{(-1)^{n-1}\over (n+(-1)^n\sqrt{n})^{2\over 3}}$$
WolframAlpha says it diverges, but I don't know how to show it. As it's not monotonic $\Rightarrow$ I can't estimate it from below. I tried to do Сauchy criteria, but I got mess with radicals and got nothing
$$|u_n|=\frac {1}{n^\frac 23 (1+\frac {(-1)^n}{\sqrt {n}})^\frac 23}$$

$$\sim \frac {1}{n^\frac 23} $$

thus it is not   absolutely convergent.

$$u_n=$$
$$\frac {(-1)^{n-1}}{n^\frac 23}(1-\frac 23\frac {(-1)^n}{\sqrt {n}}(1+\epsilon (n)) )$$

the first series is alternate and the second is absolutely convergent thus $\sum u_n $ is conditionally convergent.
"Amusing" downvote to an otherwise perfectly valid answer.


Consider the function $u(x)=x^a$, then $u'(x)=ax^{a-1}$ hence, if $x_n\sim n\sim y_n$, $$\frac1{u(x_n)}-\frac1{u(y_n)}=\frac{u(y_n)-u(x_n)}{u(x_n)u(y_n)}\sim(y_n-x_n)\frac{u'(n)}{u(n)^2}=(y_n-x_n)\frac{a}{n^{a+1}}$$ The series in the question corresponds to $x_n=n-\sqrt n$, $y_n=n+1+\sqrt{n+1}$ and $a=\frac23$, then $y_n-x_n\sim2\sqrt n$ hence $$\frac1{u(x_n)}-\frac1{u(y_n)}\sim\frac{4/3}{n^{2/3+1/2}}$$ and $\frac23+\frac12&gt;1$ hence the series converges.