True or false question regarding the convergence of a series

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  • #1
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Homework Statement:
If ##\lim_{n\rightarrow \infty} (n . a_n) = 2## , then ##\sum_{n=1}^\infty a_n## diverges.

True of false?
Relevant Equations:
not sure
I think ##\lim_{n\rightarrow \infty} a_n = 0## since by direct substitution the value of limit won't be equal to 2 so by direct substitution we must get indeterminate form.

Then how to check for ##\sum_{n=1}^\infty a_n##? I don't think divergence test, integral test, comparison test, limit comparison test, ratio test and root test can be used.

Thanks
 

Answers and Replies

  • #2
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What does it mean, that ##\lim_{n \to \infty} na_n=2\,?## Derive a condition on ##a_n## from that and think of what it means for the series.
 
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  • #3
member 587159
To complement the excellent hint of @fresh_42, recall that ##\sum_{n=1}^\infty 1/n## diverges. One of the test you mentioned can in fact be used.
 
  • #4
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What does it mean, that ##\lim_{n \to \infty} na_n=2\,?## Derive a condition on ##a_n## from that and think of what it means for the series.

To complement the excellent hint of @fresh_42, recall that ##\sum_{n=1}^\infty 1/n## diverges. One of the test you mentioned can in fact be used.
I am not sure I get the hint.

##\lim_{n \to \infty} na_n=2## means that n . an diverges

Is this what you mean? What other information can be obtained besides ##\lim_{n \to \infty} a_n=0\,?##

Or maybe an can only be in form of rational function where the power of denominator is one higher than numerator and the coefficient of highest power in numerator is 2 so when multiplied by n the limit is 2 and by using limit comparison test with 1/n the result is the series always diverges?

Thanks
 
  • #5
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##\lim_{n \to \infty} na_n=2## means that n . an diverges

Is this what you mean? What other information can be obtained besides ##\lim_{n \to \infty} a_n=0\,?##
No, this wasn't what I meant. It is a simple statement ##A \Longrightarrow B##. So all we have is that ##na_n## converges at two. But convergence has a precise meaning which we can write down. It is all we have. You will not solve the problem by talking. What does ##\lim_{n\to \infty} b_n=b## mean in formulas? This will result ín a condition for ##b_n##, especially a lower bound.
 
  • #6
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Sorry still not getting your hint
No, this wasn't what I meant. It is a simple statement ##A \Longrightarrow B##. So all we have is that ##na_n## converges at two. But convergence has a precise meaning which we can write down. It is all we have. You will not solve the problem by talking.

Meaning of convergence:
If the sequence of partial sums Sn has a limit L, the infinite series converges to that
limit, and we write
[tex]\sum_{k=1}^\infty a_k = \lim_n \rightarrow \infty \sum_{k=1}^{n} a_k = \lim_n \rightarrow \infty S_n = L[/tex]

If the sequence of partial sums diverges, the infinite series also diverges.

Is that the one?

What does ##\lim_{n\to \infty} b_n=b## mean in formulas? This will result ín a condition for ##b_n##, especially a lower bound.
I do not know what it means in formulas. Do you mean the definition of limit related to epsilon - delta?

Thanks
 
  • #7
member 587159
Sorry still not getting your hint


Meaning of convergence:
If the sequence of partial sums Sn has a limit L, the infinite series converges to that
limit, and we write
[tex]\sum_{k=1}^\infty a_k = \lim_n \rightarrow \infty \sum_{k=1}^{n} a_k = \lim_n \rightarrow \infty S_n = L[/tex]

If the sequence of partial sums diverges, the infinite series also diverges.

Is that the one?


I do not know what it means in formulas. Do you mean the definition of limit related to epsilon - delta?

Thanks

Yes, use the epsilon-delta (well rather epsilon-N) definition of convergence with a suitable epsilon.

What does ##\lim_n na_n=2## mean in epsilon-delta language?
 
  • #8
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Yes, use the epsilon-delta (well rather epsilon-N) definition of convergence with a suitable epsilon.

What does ##\lim_n na_n=2## mean in epsilon-delta language?

Let me try:

For every ε > 0, there is corresponding number N such that if x > N then |n . an - 2| < ε

## - \varepsilon < n . a_n - 2 < \varepsilon ##

## 2 - \varepsilon < n . a_n < 2 + \varepsilon ##

##\frac{2 - \varepsilon}{n} < a_n < \frac{2 + \varepsilon}{n}##

##\sum_{n=1}^\infty \frac{2 - \varepsilon}{n} < \sum_{n=1}^\infty a_n < \sum_{n=1}^\infty \frac{2 + \varepsilon}{n}##

By comparison test, since the left sum is diverge, then the middle sum also diverges.

Is this correct?
 
  • #9
member 587159
The idea is certainly correct now, but the exposition needs some work. Don't worry, this is usual when learning real analysis.

The "for every ##\epsilon > 0##" part is a bit weird. Just pick one particular epsilon, for example ##\epsilon = 1##. Then for ##n## sufficiently large, say ##n \geq n_0##, we have

##-1 < n a_n - 2 < 1 \implies 1 < n a_n##

In particular, for ##n \geq n_0##

##a_n > 1/n##.

Your inequalities with the series are not entirely correct. We only have this inequality for sufficiently large index values n! So, we should do:

##\sum_{n=1}^\infty a_n = \sum_{n=1}^{n_0-1} a_n + \sum_{n=n_0}^\infty a_n \geq \sum_{n=1}^{n_0-1} a_n + \sum_{n=n_0}^\infty 1/n##

and the right side diverges (if ##\sum_{n=n_0}^\infty 1/n## would converge, then also ##\sum_{n=1}^\infty 1/n##), so we are done.
 
  • #10
Dick
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limit comparison test

Think about how you might apply that test.
 
  • #11
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Think about how you might apply that test.
I do not have any ideas. Where should I start?

Thanks
 
  • #12
Dick
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I do not have any ideas. Where should I start?

Thanks

Start by stating what the test says. It might get you started.
 
  • #13
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Start by stating what the test says. It might get you started.

Suppose ##\sum p_n and \sum q_n ## with pn and qn both > 0 for all n. If ##\lim_{n \rightarrow \infty} \frac{p_n}{q_n} = L## where L > 0 and finite then either both series converge or both diverge.

I want to check an so let assume an = pn and I need to find suitable qn so the limit is L.

I do not know what the suitable qn is.

Thanks
 
  • #14
Dick
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Suppose ##\sum p_n and \sum q_n ## with pn and qn both > 0 for all n. If ##\lim_{n \rightarrow \infty} \frac{p_n}{q_n} = L## where L > 0 and finite then either both series converge or both diverge.

I want to check an so let assume an = pn and I need to find suitable qn so the limit is L.

I do not know what the suitable qn is.

Thanks

How about ##q_n=\frac{1}{n}##?
 
  • #15
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How about ##q_n=\frac{1}{n}##?
oh my god. I wrote pn = an but what I tried so far was always using pn = n. an, that's why I always got stucked. Got it

Thank you very much for all the help fresh_42, math_qed and dick
 
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