Could someone me im & desperate

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In summary: But I can't seem to find the paper or notes I used. Thanks for the help!In summary, the statements are true.
  • #1
mynameisfunk
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could someone please help me I am urgent & desperate

Homework Statement



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

Homework Equations





The Attempt at a Solution



I thought i had it solved by setting [itex]a_n[/itex] to [itex]1/(n^2)[/itex]
 
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  • #2


mynameisfunk said:

Homework Statement



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

Homework Equations





The Attempt at a Solution



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

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.

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


Thus

[itex] na_n < \epsilon [/itex] for [itex] n > n_0[/itex] . That completes the proof.

The second one is similar since the sequence is cauchy.

EDIT

So both statements are true.
 
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  • #3


mynameisfunk said:

Homework Statement



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

Homework Equations





The Attempt at a Solution



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

I would think about alternating series.
 
  • #4
Dick said:
I would think about alternating series.
Isn't a_n >= 0 .
 
  • #5


╔(σ_σ)╝ said:
Isn't a_n >= 0 .

True. I missed that. Then think about series with LOTS of zero terms.
 
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  • #6


Prove the contrapositive. Since [itex]a_n\ge 0[/itex] for all n, so is [itex]na_n[/itex]. If [itex]na_n[/itex] does NOT go to 0, there must be some [itex]a_0> 0[/itex] such that [itex]na_n> a_0[/itex]. That means that [itex]a_n> a_0/n[itex]. But the harmonic series [itex]\sum 1/n[/itex] diverges. That leads to a contradiction.
 
  • #7


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

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.
 
  • #8
Dick said:
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 !
 
  • #9


For part a), it isn't too difficult to prove if the [tex] a_n [/tex] are monotonically decreasing, then [tex] n a_n \rightarrow 0 [/tex], so you can prove that a sub-sequence of [tex] n a_n [/tex] converges to 0, but without the monotonicity I don't know...
 
  • #10


JG89 said:
For part a), it isn't too difficult to prove if the [tex] a_n [/tex] are monotonically decreasing, then [tex] n a_n \rightarrow 0 [/tex], so you can prove that a sub-sequence of [tex] n a_n [/tex] 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.
 
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  • #11


Good idea Dick. I've spent too much time trying to prove it correct :redface:
 
  • #12


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.
 

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