Is $\sum_{n=2}^{\infty}\frac{(-1)^n}{\sqrt n+(-1)^n}$ convergent?

Question

Is the following series convergent?

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{\sqrt n+(-1)^n}$$

I think, the above series is divergent, since

$$\sum_{n=2}^{\infty}\frac{(-1)^n}{\sqrt n+(-1)^n}=\frac{1}{\sqrt 2+1}-\frac{1}{\sqrt 3-1}+\frac{1}{\sqrt 4+1}-\frac{1}{\sqrt 5-1}+\dots\geq$$ $$\frac{-2}{(\sqrt 3-1)^2}+\frac{-2}{(\sqrt 5-1)^2}+\dots=\sum_{n=2}^{\infty}\frac{-2}{(\sqrt n-1)^2}$$

And the last series is divergent. IS my argument correct? Thanks.

Answer 1

The series diverges since

$$\sum_{n=2}^m \frac{(-1)^n}{\sqrt{n} + (-1)^n} = \sum_{n=2}^m \frac{(-1)^n(\sqrt{n} - (-1)^n)}{n - 1} \\ = \sum_{n=2}^m \frac{(-1)^n\sqrt{n}}{n - 1} - \sum_{n=2}^m \frac{1}{n - 1}, $$

with the first series on the RHS convergent by Dirichlet and the second a divergent harmonic series.

Answer 2

No !. when the sign of general term changes, we cannot use comparison test.

Taylor expansion gives,

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

$$=\frac {(-1)^n}{\sqrt {n}}-\frac {1}{n}+\frac {(-1)^n}{n\sqrt {n}}(1+\epsilon (n)) $$ $$=v_n+w_n+t_n $$

with

$\sum v_n$ convergent as alernate.

$\sum w_n $ divergent, and $\sum t_n$ absolutely convergent cause $$|t_n|\sim \frac {1}{n^\frac 32} $$

We conclude that $\sum u_n $ is divergent.