Could someone me im & desperate

[mynameisfunk]

could someone please help me I am urgent & desperate

Homework Statement

True or False, give proof or counterexample
(1) If a_n \geq 0 for all n and \sum a_n converges then na_n \rightarrow 0 as n \rightarrow \infty.
(2) if a_n \geq 0 for all n and \sum a_n converges, then n(a_n-a_{n-1}) \rightarrrow 0 as n \rightarrow \infty.

Homework Equations

The Attempt at a Solution

I thought i had it solved by setting a_n to 1/(n^2)

 

[╔(σ_σ)╝]

Homework Statement

True or False, give proof or counterexample
(1) If a_n \geq 0 for all n and \sum a_n converges then na_n \rightarrow 0 as n \rightarrow \infty.
(2) if a_n \geq 0 for all n and \sum a_n converges, then n(a_n-a_{n-1}) \rightarrrow 0 as n \rightarrow \infty.

Homework Equations

The Attempt at a Solution

I thought i had it solved by setting a_n to 1/(n^2)

I don't have much knowledge on infinite series from an analysis point of view. What I have is a limited knowledge from first year engineering calculus so don't take my solution for gold. However, I believe the reasoning is sound. Since you are "desperate" hopefully this will suffice.

a_n < \frac{\epsilon}{n} for every n > n_0 since \epsilon >0 and the harmonic series diverges.


Thus

na_n < \epsilon for n > n_0 . That completes the proof.

The second one is similar since the sequence is cauchy.

EDIT

So both statements are true.

 

[Dick]

Homework Statement

True or False, give proof or counterexample
(1) If a_n \geq 0 for all n and \sum a_n converges then na_n \rightarrow 0 as n \rightarrow \infty.
(2) if a_n \geq 0 for all n and \sum a_n converges, then n(a_n-a_{n-1}) \rightarrrow 0 as n \rightarrow \infty.

Homework Equations

The Attempt at a Solution

I thought i had it solved by setting a_n to 1/(n^2)

I would think about alternating series.

 

[╔(σ_σ)╝]

I would think about alternating series.

Isn't a_n >= 0 .

 

[Dick]

Isn't a_n >= 0 .

True. I missed that. Then think about series with LOTS of zero terms.

 

[HallsofIvy]

Prove the contrapositive. Since a_n\ge 0 for all n, so is na_n. If na_n does NOT go to 0, there must be some a_0> 0 such that na_n> a_0. That means that a_n> a_0/n. But the harmonic series \sum 1/n <b>diverges</b>. That leads to a contradiction.

 

[Dick]

Prove the contrapositive. Since a_n\ge 0 for all n, so is na_n. If na_n does NOT go to 0, there must be some a_0> 0 such that na_n> a_0. That means that a_n> a_0/n. But the harmonic series \sum 1/n <b>diverges</b>. That leads to a contradiction.

<br /> <br /> That's not true. n*a_n doesn't have to be bounded away from zero. It just has to "NOT go to 0". Those are two different things.

 

[╔(σ_σ)╝]

That's not true. n*a_n doesn't have to be bounded away from zero. It just has to "NOT go to 0". Those are two different things.

There is a theorem and states that of a_n does not go to 0 then a_n is eventually bounded away from 0. If not a_n would be in every epsilon neighbourhood of 0. :-$. I am confused !

 

[JG89]

For part a), it isn't too difficult to prove if the a_n are monotonically decreasing, then n a_n \rightarrow 0, so you can prove that a sub-sequence of n a_n converges to 0, but without the monotonicity I don't know...

 

[Dick]

For part a), it isn't too difficult to prove if the a_n are monotonically decreasing, then n a_n \rightarrow 0, so you can prove that a sub-sequence of n a_n converges to 0, but without the monotonicity I don't know...

Right! If a_n=1/n if n is even and a_n=0 if n is odd then n*a_n does not converge to zero. Of course, in this example the series a_n doesn't converge. But it's not hard to fix it so a_n does converge. Put more 0's in! Just making the odd ones zero isn't enough.

 

[JG89]

Good idea Dick. I've spent too much time trying to prove it correct

 

[l'Hôpital]

Interesting enough, I had a very similar question (well, part i), not ii) ) in my Analysis midterm that was due Wednesday. I believe I used the Cauchy-Condensation Test for i), and made some argument about na_n being bounded by 2^k a_2^k, and since the latter went to zero, so did na_n.