Could someone me im & desperate

• mynameisfunk
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.
mynameisfunk

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 $n \rightarrow \infty$.

The Attempt at a Solution

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

mynameisfunk said:

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 $n \rightarrow \infty$.

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.

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mynameisfunk said:

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 $n \rightarrow \infty$.

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.

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

╔(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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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 $a_n> a_0/n[itex]. But the harmonic series [itex]\sum 1/n$ diverges. That leads to a contradiction.

HallsofIvy said:
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 $a_n> a_0/n[itex]. But the harmonic series [itex]\sum 1/n$ 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.

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 !

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...

JG89 said:
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.

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

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.

1. What does "im" stand for in the phrase "Could someone help me im desperate?"

"Im" is short for "I am", which is a common abbreviation used in informal communication.

2. Why do people use the phrase "im desperate" when asking for help?

The phrase "im desperate" is often used to convey a sense of urgency and desperation, indicating that the person is in immediate need of assistance.

3. Is it grammatically correct to use "im" instead of "I am" in a sentence?

While "I am" is the grammatically correct form, "im" is commonly used in informal communication such as text messages and social media posts.

4. How can someone help someone who is "desperate"?

There is no one-size-fits-all answer to this question, as everyone's situation and needs are different. However, some ways to help someone who is desperate may include listening to them, offering emotional support, providing practical assistance, and connecting them with resources or professional help.

5. Is it appropriate to use the phrase "im desperate" in a professional setting?

No, it is not appropriate to use informal language such as "im" and "desperate" in a professional setting. It is important to use proper grammar and professional language when communicating in a professional environment.

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