I want to evaluate this sum $\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt {n+\sqrt{n+\sqrt{\cdots}}}+(-1)^n}$ such that I want to know what is the value of that series if it is converge , I have started by the evaluation of the series of the term :$\frac{(-1)^n}{\sqrt n+(-1)^n}$ using the following idea:
\begin{eqnarray*} \frac{(-1)^n}{\sqrt n+(-1)^n}&=&\frac{(-1)^n}{\sqrt n}\times\frac{1}{1+\frac{(-1)^n}{\sqrt n}}\\ &=&\frac{(-1)^n}{\sqrt n}\left(1-\frac{(-1)^n}{\sqrt n}+\frac{1}{n}+o\left(\frac 1n\right)\right)\\ &=&\frac{(-1)^n}{\sqrt n}-\frac{1}{n}+\frac{(-1)^n}{n\sqrt n}+o\left(\frac{1}{n\sqrt n}\right). \end{eqnarray*} The series of general term $\frac{(-1)^n}{\sqrt n+(-1)^n}$ is sum of two convergents series (Altern series ) with harmonic series which it is diverge which gives a diverge series , but really I can't use the same idea to check whethere $\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt {n+\sqrt{n+\sqrt{\cdots}}}+(-1)^n}$ converge or not ? and what is its value however the nested radical which montioned in denominator of that series follow the quadratic equation $x^2=n+x$ which means $x=\frac{1+\sqrt{1+4n}}{2}$ or $x=\frac{1-\sqrt{1+4n}}{2}$ then we have two possibility for the titled sum for $x\geq 0$ probably yield to divergent series , but rea really its difficult for me to assure that is a convergent series ?and what about its bounds ? any help ?
As you noticed, $$y_n = \sqrt{n + \sqrt{n+\sqrt{n+...}}} = \frac{1+\sqrt{1+4n}}{2}$$ by using the fact that $y_n = \sqrt{n+y_n}$.
Thus the sum simplifies to $$\sum_{n=1}^{\infty} \frac{(-1)^n}{\frac{1+\sqrt{1+4n}}{2} + (-1)^n}$$
This can be split up into even $n$ and odd $n$ as $$\sum_{n=1}^{\infty} \left(\frac{1}{\frac{1+\sqrt{1+8n}}{2} + 1} + \frac{-1}{\frac{1+\sqrt{-3+8n}}{2} -1}\right)$$
Expanding this yields $$\frac{1}{3}-\frac{2}{-1+\sqrt{5}}+\sum_{n=2}^{\infty}\frac{5+2\sqrt{-3+8n}-\sqrt{1+8n}+2n\sqrt{1+8n}-2n\sqrt{-3+8n}}{\left(2-2n\right)\left(2-4n\right)}-\sum_{n=2}^{\infty}\frac{8n}{\left(2-2n\right)\left(2-4n\right)}$$
Although the first sum converges, the second sum diverges by the limit comparison test with $\frac{1}{n}$.