True or false question regarding the convergence of a series

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

fresh_42
Mentor
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.

songoku and member 587159
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.

songoku
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

fresh_42
Mentor
##\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.

songoku
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
$$\sum_{k=1}^\infty a_k = \lim_n \rightarrow \infty \sum_{k=1}^{n} a_k = \lim_n \rightarrow \infty S_n = L$$

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

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
$$\sum_{k=1}^\infty a_k = \lim_n \rightarrow \infty \sum_{k=1}^{n} a_k = \lim_n \rightarrow \infty S_n = L$$

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?

songoku
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?

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.

songoku
Dick
Homework Helper
limit comparison test

Think about how you might apply that test.

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

Thanks

Dick
Homework Helper
I do not have any ideas. Where should I start?

Thanks

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

songoku
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

Dick
Homework Helper
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