Given a sequence ${x_n}$ = $\sqrt(1)$ , $-\sqrt(1)$,$\sqrt(2)$,$-\sqrt(2)$…

Given a sequence ${x_n}$ = $\sqrt{1}$ , $-\sqrt{1}$,$\sqrt{2}$,$-\sqrt{2}$...

If $y_n$ = {$x_1$ + $x_2$ + $x_3$ + $x_4$ + ... $x_n$}$1\over n$

Then sequence $y_n$ is

1.$Monotonic$

2.NOT bounded

3.bounded but not convergent (This is correct )

4.convergent

My attempt :I noticed a couple of things here

${x_n}$ is not convergent as two subsequences are convergent to different limits.

The terms of $y_n$ are $1$,$\sqrt(1)$/2,$\sqrt(2)$/3,

So ${y_n}$ seems to go to zero for large n and thus convergent ,but im not sure regd this . Can any1 help it out? Thanks

HINT: $y_{2k}=0$, $y_{2k+1}=\sqrt{k+1}/(2k+1)$.
• Thanks for answer .my carelessness . But i coudnt get expression for $a_{2k+1}$ ..terms are $y_1$ =$\sqrt(1)$/1 ,$y_3$=$\sqrt(2)$/3 .. – Sophie Clad Nov 18 '14 at 6:45
• @SophieClad Have you checked it for initial naturals? If $y_{2k}=0$, $y_{2k+1}$ is the next positive $x_n$ divided by $n$. – Przemysław Scherwentke Nov 18 '14 at 6:49
• @SophieClad Yes it is convergent. It is safer to say: because it is majorizes by $c\sqrt{n}$, but your idea is correct. But I am worried, that 3. is suggestet to be a correct answer. Impossible, if the current formulation of the question is correct. – Przemysław Scherwentke Nov 18 '14 at 7:00