# $\{a_n\}_{n\geq 1}$ be a sequence of strictly decreasing positive numbers then....

Let $\{a_n\}_{n\geq 1}$ be a sequence of strictly decreasing positive numbers i.e., $$a_1>a_2>a_3>\cdots$$

Then which of the following is always true?

• $\lim _{n\rightarrow \infty} a_n=0$
• $\lim _{n\rightarrow \infty} \frac{a_n}{n}=0$
• $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converges.
• $\sum_{n=1}^{\infty}\frac{a_n}{n^2}$ converges.

I guess all options are correct but i could not give a concrete proof.

For third bullet, Take $b_n=\frac{a_n}{n}$ and look for ratio test :

$\lim_{n\rightarrow \infty}\frac{b_{n+1}}{b_n}=\lim_{n\rightarrow \infty}\frac{a_{n+1}}{n+1}.\frac{n}{a_n}\rightarrow 0$ Thus $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converges.

For fourth bullet, I would just use :

As $a_n$ are positive and $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converges with $\frac{a_n}{n^2}<\frac{a_n}{n}$.

we would say $\sum_{n=1}^{\infty}\frac{a_n}{n^2}$ converges.

I can not convince myself that this is a proof.

Thank you :)

• Hint: How does $a_n=42+\frac{1}{n}$ behave? Dec 25, 2013 at 11:10
• Ah... May be that is the reason i could not come up with result $\lim a_n=0$ :D Thank you so much....
– user87543
Dec 25, 2013 at 11:12

The sequence $a_n=1+\frac{1}{n}$ shows that the first and the third point are not true.

the inequalities $$0\leq \frac{ a_n}{ n}\leq \frac {a_1}{ n}$$ prove that the second point is true and the inequalities $$0\leq \frac{ a_n}{ n^2}\leq \frac {a_1}{ n^2}$$ prove that the last point is also true.

• You're welcome.
– user63181
Dec 25, 2013 at 12:29

The correct options are the second and the fourth (Peter’s comment contains a counterexample to both the first and third options). You can deduce the second option from the fourth.

• @SamiBenRomdhane corrected, thanks. Dec 25, 2013 at 11:28

For the third bullet point you can also consider this

$a_n=\frac{n}{2n-1}$

• why would you think this is worth considering? Is there something more interesting than other examples? I would like to know more about this...
– user87543
Dec 25, 2013 at 15:24
• this will disprove bullet point 3 Dec 25, 2013 at 15:43
• Yes Yes i understand that this would disprove third bullet point but i was expecting you want to say that this is more simple/interesting than other examples... Thank you any ways :)
– user87543
Dec 25, 2013 at 15:50
• interesting in the sense that it disproves 1and 3 simultaneosuly Dec 26, 2013 at 3:37
• valid point :) Thank you :)
– user87543
Dec 26, 2013 at 4:38