If a positive decreasing sequence $a_n$ converges, then $a_n=o(1/n)$

Question

How can I prove that if a positive decreasing sequence $a_n$ converges, then $a_n=o\left (\dfrac{1}{n}\right )$?

If I suppose $a_n$ is not $o\left (\dfrac{1}{n}\right )$, then I can find an $\varepsilon >0$ such that there are infinitely many $n$ such that $a_n>\dfrac{\varepsilon}{n}$. So if I construct $S=\{n_1,n_2,\ldots \}$, the set of all such $n$, then I can say that$$\sum a_n>\varepsilon \left (1+\frac{n_2-n_1}{n_2}+\frac{n_3-n_2}{n_3}+\cdots \right ),$$but I'm blocked at here. Would you please help me?

I don't know the exact term but when I studied it , it was under the heading Abel Pringsheim Theorem — Mr.Gandalf Sauron, Jul 31 '22 at 14:20
This has been asked billions of times on the site. See for example : https://math.stackexchange.com/questions/4603/if-a-n-subset0-infty-is-non-increasing-and-sum-n-1-infty-a-n-infty — TheSilverDoe, Jul 31 '22 at 14:22

Salcio · Answer 1 · 2022-07-31T14:30:36.287

-2

If $\sum a_n$ is finite, then for any $\epsilon$ there is an index $m$ such that $\sum_{n=m}^{\infty} a_n <\epsilon$ .
Hence, $\sum_{n=m}^{2m} a_n < \epsilon$. Since the sequence is monotonic it means that $ma_{2m} <\epsilon$.
So, $ma_m$ tends to $0$.

edited Jul 31 '22 at 14:30

answered Jul 31 '22 at 14:25

Salcio

2,495

Please strive not to post more (dupe) answers to dupes of FAQs. – TheSilverDoe Jul 31 '22 at 14:31
@TheSilverDoe ? – Salcio Jul 31 '22 at 14:32

score -2 · Answer 2 · answered Jul 31 '22 at 14:25

-2

This is true. Because you can prove from here

$$\lim_{n\to\infty} na_n=0 \Rightarrow \lim_{n\to\infty} \frac{a_n}{\left(\frac{1}n\right)}=0$$

Therefore,

$$a_n=o\left(\frac{1}n\right)$$

answered Jul 31 '22 at 14:25

MathFail

21,128

Please strive not to post more (dupe) answers to dupes of FAQs. – TheSilverDoe Jul 31 '22 at 14:31

If a positive decreasing sequence $a_n$ converges, then $a_n=o(1/n)$

2 Answers2