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Let $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$ with $b_n = (a_1 + ... + a_n) / n$

I've proved that $b_n \rightarrow 0$ for $n \rightarrow \infty$ when $\lim_{n \rightarrow \infty} a_n = 0$

However I'm having problems proving the following:

Assume $a_n \rightarrow a$ for $n \rightarrow \infty$. Use the sequence $x_n = a_n - a$ to show that $b_n \rightarrow a$ for $n \rightarrow \infty$.

Could someone help me out ?

Thanks.

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  • $\begingroup$ What is $b_n - a$ in terms of $x_n$? $\endgroup$ – Daniel Fischer Nov 14 '13 at 9:59
  • $\begingroup$ $b_n - a = (x_1 + x_2 + ... + x_n)/n$ $\endgroup$ – Shuzheng Nov 14 '13 at 10:09
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    $\begingroup$ Yep, so that takes you to the situation you already proved. $\endgroup$ – Daniel Fischer Nov 14 '13 at 10:19
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Take sequence $y_n=(x_1+...+x_n)/n=b_n-a$. Now if $a_n\rightarrow a$ then $x_n\rightarrow 0$ and by theorem u proved you get $a=lim(y_n)+lim(a)=lim(b_n-a)+lim(a)=lim(b_n)$.

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  • $\begingroup$ Thanks that helped me out. However I don't think you can write $\lim_{n \rightarrow \infty}(b_n - a) = \lim_{n \rightarrow \infty}(b_n) - lim_{n \rightarrow \infty}(a)$ because this equality is valid if the two sequences are convergent, but we are trying to prove $\{b_n\}$ is convergent. $\endgroup$ – Shuzheng Nov 14 '13 at 10:20
  • $\begingroup$ Yeah, because $\forall \epsilon \exists N : |(b_n - a) - 0| \le \epsilon \Rightarrow |b_n - a| \le \epsilon$ when $n \ge N$. Are there any formula proving your equality ? $\endgroup$ – Shuzheng Nov 14 '13 at 10:29
  • $\begingroup$ Of course u were right i edited now it should be ok. $\endgroup$ – user52045 Nov 14 '13 at 10:31
  • $\begingroup$ No because you still use $\lim_{n \rightarrow \infty}(b_n - a) = \lim_{n \rightarrow \infty}(b_n) - \lim_{n \rightarrow \infty}(a)$ before proving convergence of $\{b_n\}$ $\endgroup$ – Shuzheng Nov 14 '13 at 10:34
  • $\begingroup$ Notice that last equality shows that limit $b_n$ exists as sum of 2 sequences with has limits. $\endgroup$ – user52045 Nov 14 '13 at 10:39

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