Limit of 2 sequences

**mikado** · October 4th 2009, 07:16 AM

Hi,

I'm having some trouble with this question:

Show that if ${a_{n}}$ converges to L, that ${b_{n}}$ where $b_{n}=\frac{1}{n}\left(a_{1}+...+a_{n}\right)$ also converges to L. Look at the case L=0 first.

I got as far as saying that

$|\frac{1}{n}\left(a_{1}+...+a_{n}\right)-0|<\epsilon$

$|\frac{1}{n}\left(a_{1}+...+a_{n}\right)|\leq\frac {1}{n}\left(|a_{1}|+...+|a_{n}|\right)$

and the right part of the inequality can be rewritten as

$\frac{1}{n}\sum^{N}_{i=1}|a_{i}|+\frac{1}{n}\sum^{ n}_{i=N}|a_{i}|$

How should I continue?

Thank you

**Plato** · October 4th 2009, 01:34 PM

Originally Posted by mikado

Show that if ${a_{n}}$ converges to L, that ${b_{n}}$ where $b_{n}=\frac{1}{n}\left(a_{1}+...+a_{n}\right)$ also converges to L. Look at the case L=0 first.

$\frac{1}{n}\left( {\sum\limits_{k = 1}^n {a_k } } \right) - L = \sum\limits_{k = 1}^n {\frac{{a_k - L}}{n}} = \sum\limits_{k = 1}^N {\frac{{a_k - L}}{n}} + \sum\limits_{k = N + 1}^n {\frac{{a_k - L}}{n}}$ .

If $\varepsilon > 0\, \Rightarrow \,\left( {\exists N} \right)\left[ {n \geqslant N\, \Rightarrow \,\left| {a_n - L} \right| < \frac{\varepsilon }{2}} \right]$ .

Now $\,\left( {\exists K > N} \right)\left[ {\frac{1}{K}\left( {\sum\limits_{k = 1}^N {\frac{{a_k - L}}{n}} } \right) < \frac{\varepsilon }{2}} \right]$ .

Carry on.

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