# Does a converge series has a sum that growing not slower than a linear function?

Consider a series $a_n \in {\mathbb R}_+$ ($n \in {\mathbb Z}_+$) with $$\lim_{n\rightarrow \infty} a_n = 0.$$

It is easy to find an example that $$\lim_{n\rightarrow \infty} \sum_{i=1}^{n}a_i = \infty.$$ For example $a_n = 1/n$ or $a_n = 1/\log(n)$.

However, does such a $a_n$ exist that $\lim_{n\rightarrow \infty} a_n = 0$ and

$$\lim_{n\rightarrow \infty} \frac {1}{n}\sum_{i=1}^{n}a_i = c,$$ where $c > 0$?

PS: For $a_n = 1/n$ or $a_n = 1/\log(n)$, it can be verified that $c=0$.

If a sequence $(a_n)$ is convergence and $a_n\to L$, it can be shown that $$\lim_{n\rightarrow \infty} \frac {1}{n}\sum_{i=1}^{n}a_i = L$$ also. So basically, you cannot find a sequence $(a_n)$ such that $lim_{n\to\infty} a_n=0$ while
$\lim_{n\rightarrow \infty} \frac {1}{n}\sum_{i=1}^{n}a_i > 0$

• I've edited my answer to (hopefully) match your question :) – BigbearZzz Oct 7 '15 at 3:29
• I'm glad I could help. – BigbearZzz Oct 7 '15 at 3:35

I believe that this is true with each $a_1 = 1$. We note:

$$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^n 1 =\lim_{n\rightarrow \infty} \frac{1}{n} n = 1$$

• Thanks, but $\lim_{n\rightarrow \infty} a_n$ seems not $0$ – Ryan Oct 7 '15 at 3:24
• sorry, my question initially confusing, and I have re-edited it – Ryan Oct 7 '15 at 3:26
• Sorry, I didn't know you were looking for that condition. Bigbear's answer implies that no such sequence exists. – Frank Wang Oct 7 '15 at 3:27