If $a_n$ is decreasing, positive and tends towards $0$ then is true that $\frac{1}{N}\sum_{n=1}^N a_n\to 0$? 
If $a_n$ is decreasing, positive and tends towards $0$ then is true that $\frac{1}{N}\sum_{n=1}^N a_n\to 0$?

I believe this is true because you're taking the average of smaller and smaller things. This is almost like a law of large numbers.
 A: Use the Stolz–Cesàro theorem (an analogy of the L'hospital theorem for sequences):
$$
\lim_{N\to\infty}\frac{\sum_{n=1}^Na_n}{N}=\lim_{N\to\infty}\frac{\sum_{n=1}^{N+1}a_n-\sum_{n=1}^Na_n}{(N+1)-N}=\lim_{N\to\infty}a_{N+1}=0
$$
A: This is one of my favorite proofs about Cesaro means. There is a quicker proof here but for some reason I prefer the proof I give below, and I thought I'd share it just in case.
Proposition
Let $(a_n)_{n\in\mathbb{N}}$ be a sequence of real numbers converging to $a\in\mathbb{R}$. Define $$C_n=\frac{1}{n}\sum_{k=1}^{n}a_k$$ Then $$\lim_{n\to\infty}C_n=a$$
Proof
Since $(a_n)_{n\in\mathbb{N}}$ converges to $a$, then the sequence $$ b_n=|a_n-a|$$ converges to $0$, and is thus bounded, so there exists an $N_1\in\mathbb{N}$ such that for all $n\in\mathbb{N}$, we have $|a_n-a|\leq N_1$.
Let $\varepsilon>0$. By the Archimedian Property, there exists an $N_2\in\mathbb{N}$ such that $1/N_2<\varepsilon$.
As $(a_n)_{n\in\mathbb{N}}$ converges to $a$, then there exists an $N_3\in\mathbb{N}$ such that for all $n\in\mathbb{N}$ with $n\geq N_3$, we have $$|a_n-a|<\frac{1}{N_2}$$
Now, let $N=\max\{N_1,N_2,N_3\}$ and $n\geq N^2$. Then
$$\begin{align*}
|C_n-a|
&=\left|\frac{a_1+a_2+\dots+a_n}{n}-a\right|\\
&=\left|\frac{a_1+a_2+\dots+a_n-an}{n}\right|\\
&=\left|\frac{(a_1-a)+(a_2-a)+\dots+(a_N-a)+\dots+(a_n-a)}{n}\right|\\
&\leq\frac{|a_1-a|}{n}+\frac{|a_2-a|}{n}+\dots+\frac{|a_{N-1}-a|}{n}+\frac{|a_N-a|}{n}+\dots+\frac{|a_n-a|}{n}\\
&\leq\underbrace{\frac{N+N+\dots+N}{n}}_{\text{$N-1$ times}}+\underbrace{\frac{\varepsilon}{n}+\frac{\varepsilon}{n}+\dots+\frac{\varepsilon}{n}}_{\text{$n-N+1$ times}}\\
&\leq\frac{N(N-1)+\varepsilon(n-N+1)}{n}\\
&=\frac{N^2}{n}-\frac{N}{n}+\varepsilon-\varepsilon\frac{N-1}{n}\\
&<\frac{N^2}{n}+\varepsilon
\end{align*}$$
If we then let some $\varepsilon'>0$, we can resize $\varepsilon$ (which was hinted at by using the Archimedean Property) so that $N^2/n +\varepsilon<\varepsilon'$ for a large enough $n$ in order to achieve the result. Thus for any $\varepsilon'>0$, we have $|C_n-a|<\varepsilon'$ which completes the proof.
A: Since the $a_n$ are decreasing, notice that $na_{n+1} \leq a_1 + \cdots + a_n$, from which we may use some fairly simple calculations to get that $S_{n+1} \leq S_n$ (the $S_i$ being the $i$th averages in question). So the $S_i$ are positive, decreasing as well so by Weierstrass they do converge at some point, and for any $\epsilon > 0$, pick $N$ such that $a_n < \epsilon/2$ for all $n > N$.
Then pick $M > N$ such that $a_1 + \cdots + a_N/M < \epsilon/2$. Then $S_M < \epsilon$.
As stated in the comments, this is kind of a suboptimal proof in that you can show the same for much weaker conditions but I haven't had coffee today.
