# does this series converge / diverge

does this series converge/diverge conditionally or absolutly

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

i can use the facts that:

$\lim_{x\to0}\frac{\sin x}{x} =1$;

$\sin(\frac{1}{\sqrt{n}})$ is monotone decreasing

my problem is the alternating $-1$'s in the denominator, i dont know how to deal with that, though the fraction looks similar to an $e$ based limit

• What is the limit of $\frac{\sqrt{n}}{(-1)^n+\sqrt{n}}$ ? You will find this term is not important to determine the convergence of the series. – Xoff Dec 7 '13 at 16:44
• the limit of this term is ofcourse 1, but $\frac{\sqrt{n}}{(-1)^n + \sqrt{n}} \cdot \sin(\frac{1}{\sqrt{n}})$ is going to $0$, but i cannot prove that it is monotone decending – guynaa Dec 7 '13 at 17:08

In fact, this is a bit tricky, since the terms are NOT monotonically decreasing. Absolute convergence (or lack thereof) is quite easy, as indicated in my other answer, but for conditional convergence, the only simple way seems to be:

1. Write $\sin(1/\sqrt{n})$ as $1/\sqrt{n} + O(1/n^{3/2}).$

Since $\sum n^{-3/2}$ converges (absolutely), we can throw away that part, so the absolute convergence of the series is equivalent to the convergence of

$$\sum (-1)^n\frac{1}{((-1)^n + \sqrt{n}}.$$

Now in this series, pair terms (the $n$-th with the $n+1$-st). the sum of two such equals:

$$\frac{2 + \sqrt{n} - \sqrt{n+1}}{(-1)^n + \sqrt{n})((-1)^{n+1} + \sqrt{n+1})}.$$

Now, the numerator converges to $2$ as $n$ grows large, but the numerator approaches $n,$ so the series DOES NOT converge conditionally.

Since you know the limit of $\sin(x)/x,$ you know that the terms of your series are like $1/\sqrt{n}$ in absolute value. What does this tells you about absolute convergence? As for conditional convergence, if you could prove that the terms of your series are monotonically decreasing in absolute value, you would be done. Can you?

• i can show that $\lim_{n\to\infty} \frac{\sqrt{n}}{(-1)^n + \sqrt{n}} \cdot \sin(\frac{1}{\sqrt{n}}) = 0$, but i cant show that its monotone decending so i can leibniz test – guynaa Dec 7 '13 at 17:30