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


2.NOT bounded

3.bounded but not convergent (This is correct )


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: $y_{2k}=0$, $y_{2k+1}=\sqrt{k+1}/(2k+1)$.

  • $\begingroup$ 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 .. $\endgroup$ – Sophie Clad Nov 18 '14 at 6:45
  • $\begingroup$ @SophieClad Have you checked it for initial naturals? If $y_{2k}=0$, $y_{2k+1}$ is the next positive $x_n$ divided by $n$. $\endgroup$ – Przemysław Scherwentke Nov 18 '14 at 6:49
  • $\begingroup$ Yes i have seen pattern of odd terms but coudnt get to general expression . Also as per ur formulae if i take limit of odd terms subsequence it goes to 0 .so both subsequences are converging to zero . Am i right ? $\endgroup$ – Sophie Clad Nov 18 '14 at 6:52
  • $\begingroup$ @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. $\endgroup$ – Przemysław Scherwentke Nov 18 '14 at 7:00
  • $\begingroup$ can u tell me how u have come upto expression of odd terms ? and u say it s majorizes by c sqrtn .wat do u mean by this ?. i just took limit of odd term general expression and it ges to zero ,isn't that good enough ? $\endgroup$ – Sophie Clad Nov 18 '14 at 7:06

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