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Given a sequence of real numbers $a_n$ ($n \in \Bbb{N}$) that converges to $a \in \Bbb{R}$, show that the sequence $b_n$ ($n \in \Bbb{N}$) also converges to $a$.

$$b_n=\frac{1}{n}(a_1+a_2+a_3+...+a_n)$$

I would be grateful for any tips, I have no idea where to look for sequences converging to sequences(?) - or is only $a_n$ a sequence and $a$ is not? I thought about the opposite, but couldn't help myself that way either ("if $b_n$ converges to $a$, does $a_n$ converge to $a$?"). I'd be grateful for any solutions to similar problems, that won't give away the solution - which I have in my book. I'd still like to understand it without looking that up though.

user1994928
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    Also, the standard terminology is that $a_n$ converges to $a$, not against $a$. –  May 22 '18 at 21:24
  • Thanks, English is not my first language, the direct German translation would be as I wrote it above - so that help alot for further questions. – user1994928 May 22 '18 at 21:26

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