Linked Questions

21
votes
3answers
9k views

If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$ [duplicate]

If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$ Since $(a_n)$ is decreasing and bounded below, by Monotonic Convergence ...
9
votes
3answers
791 views

Prove $\lim\limits_{n\to\infty}na_n\ln(n)=0$ [duplicate]

Let $a_n>0$,$\sum a_n$ is convergent,$na_n$ is monotone, Prove: $$\lim\limits_{n\to\infty}na_n\ln(n)=0$$ I try to prove $a_n=o(n\ln(n))$, but it doesn't work.
0
votes
2answers
211 views

Prove that if a sequence converges then $\lim_{k \to \infty} kb_k = 0$ [duplicate]

We have that $(b_n)$ is a sequence of decreasing, non-negative real terms. We wish to show that if $\displaystyle \sum_{i=1}^{\infty} b_n$ converges then it must be the case that $$\lim_{k \to \infty} ...
0
votes
2answers
274 views

Prove that $\lim na_n=0$ [duplicate]

Suppose that the positive series $\sum\limits_{n=1}^{\infty} a_n$ converges and $a_n>a_{n+1}$ for all $ n\in N$. Prove that $\lim na_n=0$ This is my proof. Suppose that $\lim na_n=a$, with $a>0$...
1
vote
0answers
261 views

How to prove that $\lim (n a_n) = 0$? [duplicate]

Possible Duplicate: Series converges implies $\lim{n a_n} = 0$ How do I show that the following: If $ a_1 \geq a_2\geq ... \geq a_n\geq...$ and $\sum a_n$ converges then $\lim (n a_n) = 0$? ...
2
votes
2answers
148 views

A convergent series property [duplicate]

This came up in a friend's exam and it must be one of those ${\epsilon},N(\epsilon)$ arguments I could do in a snapshot in my twenties but now I can't figure out how the proof should go: For a ...
2
votes
0answers
142 views

Terms of a monotone, convergent series must be $o(1/n)$ [duplicate]

Let $(x_n)_{n\in\mathbb N}$ be a non-negative, non-increasing sequence such that $\sum_{n=1}^{\infty} x_n<\infty$. I want to show that $$\lim_{n\to\infty}\{nx_n\}=0.$$ I’ve definitely seen this ...
0
votes
0answers
137 views

Let $\sum_{n=0}^{\infty} a_n$ be a convergent series such that $a_n \downarrow 0$. Prove that, $ \lim_{n\to \infty} na_n = 0 $ [duplicate]

This is my proof so far, Let $\varepsilon > 0$. By CCC, there is $N > 0$, such that, if $n,m>N$, then $|\Sigma_{i=m+1}^n a_i| < \varepsilon$. We can drop the absolute values as $a_i \geq ...
0
votes
1answer
91 views

Question about serie and sequence. [duplicate]

Possible Duplicate: Series converges implies $\lim{n a_n} = 0$ Someone can help me? If $(a_n)$ is a decreasing sequence and $\sum a_n$ converges. Then $\lim {(n.a_n)} = 0$. I don't have idea ...
0
votes
1answer
128 views

Prove: $(na_n)$ converges to 0 if $(a_n)_{n=1}^\infty$ is a monotonously falling sequence of non-negative real numbers [duplicate]

Prove: Let $(a_n)_{n=1}^\infty$ be a monotonously falling sequence of non-negative real numbers, so that $\sum a_n$ is convergent. Then $(na_n)$ is a sequence converging to 0. Well, no idea. All help ...
0
votes
0answers
70 views

Prove limits of sequence using series [duplicate]

I have to prove that if $(a_n)$ is decreasing sequence with positive elements then if series $\sum_{n=1}^{\infty}{a_n}$ converges then $\lim_{n \to \infty}{na_n} = 0$. I know that if $\sum_{n=1}^{\...
2
votes
0answers
46 views

$\sum u_n$ convergent without $u_n = o(1/n)$ [duplicate]

Let $(u_n)$ be a non increasing sequence of reals such that $\sum u_n$ converges. I'm investigating the question: do we always have $u_n = o(1/n)$? I suspect not and I'm trying to build a sequence $(...
0
votes
0answers
39 views

If $a_n \searrow 0$ and $\sum a_n$ converges, is it true that $n a_n \to 0$? [duplicate]

I am posting for a friend: he has a strong feeling that the following should be true but can't prove it: "Let $(a_n)$ be a real sequence. If $a_n \searrow 0$ and $\sum a_n$ converges, then $n a_n \to ...
0
votes
0answers
20 views

convergence of a series that implies the convergence of a sequence [duplicate]

I'm trying to show this: If $ \Sigma_{n=1}^{\infty} {a_n}$ is a convergent series, and $a_n \geq a_{n+1}$ then $\lim\limits_{n \to \infty} {na_n} =0$. I already prove that the sequence $\{ na_n \}$ ...
0
votes
0answers
19 views

A note on the absolutely summable sequence [duplicate]

Let $\{a_n\}$ be a decreasing positive sequence. Suppose that $\sum a_n$ is convergent. Q. Is $na_n$ convergent to $0$.

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