Mean value of convergent series Let us in a normed linear space have a sequence $\{a_i\}_{i=1}^\infty$ which converges to some value $b$, how can I show that
$$\lim_{n\rightarrow\infty}\sum_{i=1}^n\frac{a_i}{n}=b$$
My idea is to use a theorem which states that since the series is convergent, for each $\varepsilon>0$ there exist $n\in\mathbf{N}$ such that $d(x_j,x_k) < \varepsilon$ if $k > n$ and $j > n$. Then I think of splitting the sum in two parts and at the same time let $n\rightarrow\infty$ and $\varepsilon\rightarrow0$. The first part of the sum should then tend to $0$ while the second part should thend to $b$.
Does this idea hold? Any better ideas?
 A: I think you just have to do it. Let $m \sigma_{m} = \sum_{i=1}^{m} a_i.$ Choose $\varepsilon > 0.$ For some $N$ we have $\|a_i - b \| < \varepsilon $ for all $i > N.$ Then we have 
$ \| \sigma_{n} - b \| \leq \frac{\sum_{i=1}^{N} \|a_{i} -b\| }{n} +  \sum_{i=N+1}^{n} \frac{\|a_i - b \|}{n} .$  The lefttmost term tends to $0$ as $n \to \infty$ because $N$ is fixed. If you prefer to make it completely formal, it can be made less than $ \varepsilon $ by choosing $n > M$ for some fixed $M.$ The rightmost term is at most $\frac{(n-N) \varepsilon}{n} < \varepsilon . $ Since $\varepsilon $  is arbitrary, the required limit is $b.$ 
A: Since $a_n \rightarrow b$, we have that given $\epsilon > 0$, there exists $N(\epsilon)$ such that for all $n > N(\epsilon)$, we have $\left|a_n -b \right| < \frac{\epsilon}{2}$.
For $n > N(\epsilon)$, we have $\displaystyle \sum_{k=1}^{n} \frac{a_k}n = \displaystyle \sum_{k=1}^{N} \frac{a_k}n + \displaystyle \sum_{k=N+1}^{n} \frac{a_k}n$.
Hence,  for $n>N$, $\displaystyle \left| \sum_{k=1}^{n} \frac{a_k}n -b \right|= \left| \displaystyle \sum_{k=1}^{N} \frac{a_k -b}n + \displaystyle \sum_{k=N+1}^{n} \frac{a_k-b}n \right| \leq \left| \displaystyle \sum_{k=1}^{N} \frac{a_k -b}n \right| + \left| \displaystyle \sum_{k=N+1}^{n} \frac{a_k-b}n \right|$.
The goal is to bound the two terms by $\epsilon/2$.
Let $\displaystyle \sum_{k=1}^{N} \left(a_k -b \right) = f(N)$.
Hence, $\displaystyle \left| \sum_{k=1}^{n} \frac{a_k}n -b \right| \leq \left| \frac{f(N)}{n} \right| + \left(1 - \frac{N}n \right) \frac{\epsilon}{2}$. Now let $M$ such that $\displaystyle \frac{\left|f(N)\right|}M < \frac{\epsilon}{2}$. Hence, for all $n > \max(N,M)$, we have $\displaystyle \left| \sum_{k=1}^{n} \frac{a_k}n -b \right| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.
