• Give an example of a sequence $(a_n)$ such that $(a_n)$ does not converge but $(b_n)=\dfrac{a_1+a_2+\ldots a_n}{n}$ converges.

I took $(a_n)=(-1)^n$ which is divergent and then took $(b_{k})$ where $b_{k}=\dfrac{a_1+a_2+\ldots a_{k}}{k}\forall n$.Hence $b_{k}$ is convergent.

Now $b_k=-\dfrac{1}{k} ;k=$ odd ;$b_k=0;k= $ even

Is this example okay?

  • Give an example to show there exists sequences $a_n,b_n$ which converge to $0$ but $c_n=\dfrac{a_n^2}{b_n}$ converges to $5$.

Here I took $a_n=\sqrt \frac{5}{n},b_n=\frac{1}{n}$.

Is this example okay?

  • $\begingroup$ The 2nd is ok, but your explanation for the first example isn't ok: you have to check the convergence of the whole sequence $(b_n)$, not only its subsequence $(b_{2n})$. Look at $(b_{2n+1})$. $\endgroup$ – paf Aug 17 '16 at 1:15
  • $\begingroup$ I don't have any $b_{2n+1} $ sequence here.I have defined $b_k=\dfrac{a_1+a_2+ \ldots +a_{2k}}{2k}$ $\endgroup$ – Learnmore Aug 17 '16 at 1:56
  • $\begingroup$ Is it okay now @paf $\endgroup$ – Learnmore Aug 17 '16 at 1:58
  • $\begingroup$ No. You are only allowed to choose the sequence $(a_n)$. You have to define $b_n=\dfrac{a_1+a_2+\cdots +a_n}{n}$ for all $n$. $\endgroup$ – Zoe H Aug 17 '16 at 2:06
  • 1
    $\begingroup$ Note: $b_n =0$ if n is even. $b_n = 1/n$ if n is is odd. So $0 \le b_n \le 1/n $. That's good. For all $\epsilon >0$ if $n > 1/\epsilon $ then $|b_n -0| < \epsilon$. (Or use sandwich theorem) $\endgroup$ – fleablood Aug 17 '16 at 5:18

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