Let $\sum_{n=1}^\infty a_{n}$ be a series of positive terms. Show that $$\lim_{t\to \infty}\frac1{t}\sum_{n=1}^\infty\lfloor ta_{n}\rfloor =\sum_{n=1}^\infty a_{n}$$
So this is what I could come up with. I hope it's right
$$ ta_{1}-1 \lt \lfloor ta_{1}\rfloor \le ta_{1}$$ $$ ta_{2}-1 \lt \lfloor ta_{2}\rfloor \le ta_{2}$$ $$\vdots$$ $$ ta_{n}-1 \lt \lfloor ta_{n}\rfloor \le ta_{n}$$ $$\vdots$$ Dividing with $t$ we get $$a_{1}-\frac 1{t} \lt \frac{\lfloor ta_{1}\rfloor}{t} \le a_{1} $$ $$a_{2}-\frac 1{t} \lt \frac{\lfloor ta_{2}\rfloor}{t} \le a_{2} $$ $$\vdots$$ $$a_{n}-\frac 1{t} \lt \frac{\lfloor ta_{n}\rfloor}{t} \le a_{n} $$ $$\vdots$$
If we use the limit when $t \to \infty$ in every inequality and then by summing all of them we get $$\sum_{n=1}^\infty a_{n} \le \lim_{t\to \infty}\frac1{t}\sum_{n=1}^\infty\lfloor ta_{n}\rfloor \le \sum_{n=1}^\infty a_{n} $$ So the limit is $$\lim_{t\to \infty}\frac1{t}\sum_{n=1}^\infty\lfloor ta_{n}\rfloor =\sum_{n=1}^\infty a_{n} $$
I'm not very sure if this is right. Can somebody tell me what to do or does someone has another method?
This should help: Let $S=\sum_{n=1}^{\infty}a_n$ and assume $S<\infty.$ Let $\epsilon>0.$ Then there exists $N$ such that $\sum_{n=1}^{N}a_n > S-\epsilon/2.$ Choose $t_0$ such that $N/t<\epsilon/2$ for $t\ge t_0.$ Then $t\ge t_0$ implies
$$\frac1{t}\sum_{n=1}^\infty\lfloor ta_{n}\rfloor \ge \frac1{t}\sum_{n=1}^N\lfloor ta_{n} \rfloor \ge \frac1{t}\sum_{n=1}^N (ta_{n}-1) = \sum_{n=1}^N a_{n} -N/t > S-\epsilon/2 - \epsilon/2 = S-\epsilon.$$