Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Does not contain a proof, but the values $b_n=\frac{1}{n}\sum_{i=1}^n a_n$ is called the Cesaro mean. – Thomas Andrews Mar 14 '12 at 17:52
up vote 3 down vote accepted

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 $\displaystyle \left| \sum_{k=1}^{n} \frac{a_k}n -b \right| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.

share|cite|improve this answer

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 $b.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.