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Let $\{a_n\}_{n\geq0}$ is a sequence of complex numbers such that $\frac{a_0+\cdots+a_n}{n+1}$ converges to $a$. Suppose also that $\exists M>0$ such that $n|a_{n}-a_{n-1}|<M$ for all $n\geq 1$. Then show that $a_n$ converges to $a$.

N.B. I started with denoting $b_{n}=a_{n}-a_{n-1}$ then $n|b_n|<M$ and $s_n=\frac{a_0+\cdots+a_n}{n+1}$ then $a_{n}-s_{n}=\frac{(n+1)a_{n}-a_{n}-\cdots-a_0}{n+1}=\frac{na_n-a_{n-1}-\cdots-a_0}{n+1}=\frac{nb_n+(n-1)a_{n-1}-a_{n-2}-\cdots-a_0}{n+1}=\cdots=\frac{nb_n+(n-1)b_{n-1}+\cdots+b_1}{n+1}$

which implies $|a_n-s_n|\leq\frac{Mn}{n+1}\to M$. But this only shows $a_n$ is bounded. Now employing Bolzano-Weierstrass Theorem we see $\exists \{a_{n_k}\}_{k\geq 1}$ which converges . From here I have no idea. How to show $a_n\to a$.

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  • $\begingroup$ Compare math.stackexchange.com/questions/736480/…. $\endgroup$ – Martin R Oct 1 '17 at 17:51
  • $\begingroup$ You have already proved that the sequence $(a_n)_{n\geq 0}$ is bounded. Let $(a_{n_k})_{k\geq 0}$ a subsequence. Using Bolzano-Weierstrass we get that there exists a convergent subsequence $(a_{n_{k_l}})_{l\geq 0}$. As soon as we have proved that $$ \lim a_{n_{k_l}} =a $$ then we are done. $\endgroup$ – Severin Schraven Oct 1 '17 at 19:03
  • $\begingroup$ In fact the subsequence may not converge to $a$. Since we can choose $a_n=(-1)^n$. Then we see $a=0$. But $a_{n_k}=1$ converges to 1.although this does not satisfy the hypothesis that $n|a_n-a_{n-1}|$ is not bounded. @severin can you please elaborate on how we can show the result that you claimed? $\endgroup$ – mudok Oct 2 '17 at 6:40
  • $\begingroup$ @MartinR the link you referred is not helpful since there also the problem is unresolved. I have now come to know that this problem is from rudin and I saw the hint that was provided in the book yet I am not completely convinced with the approach. $\endgroup$ – mudok Oct 2 '17 at 15:12
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Assume not. Then there exists $\epsilon>0$ such that for infinitely many $n$, $|a_n-a|>\epsilon$. Say, we have $a_n>a+\epsilon$ (the case $a_n<a-\epsilon$ is similar). Then we still have $a_{n+k}-a>\frac\epsilon 2$ for $0\le k\le m:=\lfloor \frac{\epsilon n}{2M}\rfloor $. Therefore, $$ |a_1+\ldots+ a_{n+m}-(n+m)a|\ge m\epsilon-|a_1+\ldots+a_n-na|$$ or after division by $n+m$, $$ |s_{n+m}-a|\ge \frac{m}{n+m}\epsilon -\frac n{n+m}|s_n-a|$$ so that $$ |s_{n+m}-a|+|s_n-a|\ge \frac m{m+n}(\epsilon-|s_n-a|).$$ As $n\to \infty$, the left hand side tends $\to 0$ and the right hand side $\to \frac{\epsilon^2}{2M+\epsilon}$, contradiction.

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  • $\begingroup$ Much Thanks Sir. $\endgroup$ – mudok Oct 5 '17 at 12:19

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