True or false question regarding the convergence of a series

[songoku]

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]

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.

 

[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 . a_n diverges

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

Or maybe a_n 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]

$\lim_{n \to \infty} na_n=2$ means that n . a_n 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 S_n 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 hintMeaning of convergence:
If the sequence of partial sums S_n 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 . a_n - 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.

 

[Dick]

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]

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 p_n and q_n 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 a_n so let assume a_n = p_n and I need to find suitable q_n so the limit is L.

I do not know what the suitable q_n is.

Thanks

 

[Dick]

Suppose $\sum p_n and \sum q_n$ with p_n and q_n 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 a_n so let assume a_n = p_n and I need to find suitable q_n so the limit is L.

I do not know what the suitable q_n is.

Thanks

How about $q_n=\frac{1}{n}$?

 

[songoku]

How about $q_n=\frac{1}{n}$?

oh my god. I wrote p_n = a_n but what I tried so far was always using p_n = n. a_n, that's why I always got stucked. Got it

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