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?

  • $\begingroup$ There are inequalities that you need to use such as $\lfloor a_n \rfloor \lfloor t \rfloor \leq \lfloor ta_n \rfloor $ $\endgroup$ – Paul Jul 27 '18 at 12:28
  • $\begingroup$ Is this right now? $\endgroup$ – J.Dane Jul 27 '18 at 12:49
  • $\begingroup$ What level class is this? If this is an elementary calculus class, your answer is probably sufficient. If this is an analysis class (or if you want to be 100% correct), you will need to justify moving limits in and out of the summation. If interchanging limits is not a topic you've discussed in your class thus far, then your solution should be sufficient. $\endgroup$ – JavaMan Jul 27 '18 at 13:04
  • $\begingroup$ We've discussed it but only on finite terms. Can I move the limits when I am working with infinite terms? $\endgroup$ – J.Dane Jul 27 '18 at 13:11
  • $\begingroup$ Something is missing. When you sum $a_n-\frac{1}{t}$ you end up with $\sum_{n=1}^{\infty}a_n-\frac{\infty}{t}$. How did you get rid of $\frac{\infty}{t}$, since the limit is $\frac{\infty}{\infty}$? $\endgroup$ – herb steinberg Jul 27 '18 at 15:16

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

$$\frac1{t}\sum_{n=1}^\infty\lfloor ta_{n}\rfloor \ge \frac1{t}\sum_{n=1}^N\lfloor ta_{n} \rfloor \ge \frac1{t}\sum_{n=1}^N (ta_{n}-1) = \sum_{n=1}^N a_{n} -N/t > S-\epsilon/2 - \epsilon/2 = S-\epsilon.$$

  • $\begingroup$ Thank you very useful. So should I prove the other side of the inequality ie $\frac1{t}\sum_{n=1}^\infty\lfloor ta_{n}\rfloor \le\frac1{t}\sum_{n=1}^\infty ta_{n}= \sum_{n=1}^\infty a_{n}=S$ or is this enough? $\endgroup$ – J.Dane Jul 27 '18 at 16:24
  • $\begingroup$ Yes, that's the easy side. And you still have the $S=\infty$ case to do. $\endgroup$ – zhw. Jul 27 '18 at 16:25
  • $\begingroup$ Ok thank you again $\endgroup$ – J.Dane Jul 27 '18 at 16:26

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