What are limit tests and how do they establish convergence of a series?

[ognik]

I've just studied integral tests for convergence, 1st timer, but some detail is escaping me.
The text reads:

1. Show that if $ \lim_{{n}\to{\infty}} {n}^{p}\: {U}_{n}\implies A \lt \infty\: (p \gt 1) $
Then $ \sum_{n=1}^{\infty} {U}_{n}\: $ converges

2. Show that if $ \lim_{{n}\to{\infty}} n {U}_{n}\implies A \gt 0 $ the series diverges

"These two tests, known as limit tests, are often convenient for establishing
the convergence of a series. They may be treated as comparison
tests"
--------------------
I can 'see' these are true, but formal proof escapes me. I tried 'limit of products=product of limits':

$ \lim_{{n}\to{\infty}}{n}^{p}{U}_{n} = \lim_{{n}\to{\infty}} {n}^{p}\lim_{{n}\to{\infty}}{U}_{n} \implies A $
$ \therefore \lim_{{n}\to{\infty}}{U}_{n} \implies A/ \lim_{{n}\to{\infty}} {n}^{p} $

Clearly $ \lim_{{n}\to{\infty}} {n}^{p} \implies\infty\: for\: p \gt 1 $
$ \therefore \lim_{{n}\to{\infty}}{U}_{n} \implies A/\infty = 0 $

But I'm stuck here, how do I relate the above result to $ \sum_{n=1}^{\infty} {U}_{n}\: $ ?
Appreciate the help, thanks.

 

[Dethrone]

1) Can be done using contradiction. Suppose $\lim_{{n}\to{\infty}} {n}^{p}U_n =A$ and $\sum_{n=1}^{\infty}U_n$ divergent, that is, $\lim_{{n}\to{\infty}}U_n = B$ ($\ne 0$). Therefore, $\lim_{{n}\to{\infty}}n^p U_n=\lim_{{n}\to{\infty}}n^p B \rightarrow \infty$, which is a contradiction.

But I'm stuck here, how do I relate the above result to $ \sum_{n=1}^{\infty} {U}_{n}\: $ ?
Appreciate the help, thanks.

I do not think they're relatable, because $\lim_{{n}\to{\infty}}a_n=0$ does not imply $\sum a_n$ convergent.

 

[ognik]

Thanks, that works well, however from the context I suspect I am supposed to relate this to the integral test.

Does 1) imply that $ {U}_{n} < A/{n}^{p} $ ? In which case for p > 1, $ A/{n}^{p} $ converges (p series)?

 

[GJA]

Hi ognik,

We need to be a little careful using a proof by contradiction here. In supposing \(\displaystyle \sum U_{n}\) diverges, we cannot conclude \(\displaystyle \lim _{n\rightarrow\infty}U_{n}=B\neq 0.\) For example, the harmonic series \(\displaystyle \sum\frac{1}{n}\) diverges and yet \(\displaystyle \lim_{n\rightarrow\infty}\frac{1}{n}=0.\)

I think the approach we want to take in the first case is to use the fact that absolute convergence implies convergence; i.e. if \(\displaystyle \sum |U_{n}|\) converges, then \(\displaystyle \sum U_{n}\) converges. To prove that \(\displaystyle \sum |U_{n}|\) converges we want to use the following idea:

If a sequence converges, then it must be bounded.

Have you seen a statement along these lines yet?

What it tells us is that since \(\displaystyle \lim_{n\rightarrow\infty}n^{p}U_{n}\) exists, there is a number - call it \(\displaystyle C\) - such that

\(\displaystyle |n^{p}U_{n}|< C\) for all \(\displaystyle n\)

Dividing through in this inequality by \(\displaystyle n^{p}\) we get

\(\displaystyle |U_{n}|<\frac{C}{n^{p}}\) for all \(\displaystyle n.\) Thus

\(\displaystyle \sum |U_{n}|<\sum \frac{C}{n^{p}}\)

Since the sum on the right converges using the p-test (or you could use the integral test with \(\displaystyle \int_{1}^{\infty}\frac{C}{x^{p}}dx\) if you wish), the sum

\(\displaystyle \sum |U_{n}|\)

also converges. Thus, we have absolute convergence, and so it follows that

\(\displaystyle \sum U_{n}\)

converges too.

To prove that \(\displaystyle \lim_{n\rightarrow\infty}nU_{n}=A>0\) implies divergence, you can use a similar argument as above, only this time you will want to bound \(\displaystyle U_{n}\) from below (whereas we bounded \(\displaystyle |U_{n}|\) from above in the first argument). Take a shot at trying this for yourself, and if you get stuck, we can revisit the problem. Let me know if anything is unclear/not quite right.

 

[chisigma]

I've just studied integral tests for convergence, 1st timer, but some detail is escaping me.
The text reads:

1. Show that if $ \lim_{{n}\to{\infty}} {n}^{p}\: {U}_{n}\implies A \lt \infty\: (p \gt 1) $
Then $ \sum_{n=1}^{\infty} {U}_{n}\: $ converges

2. Show that if $ \lim_{{n}\to{\infty}} n {U}_{n}\implies A \gt 0 $ the series diverges

"These two tests, known as limit tests, are often convenient for establishing
the convergence of a series. They may be treated as comparison
tests"
--------------------
I can 'see' these are true, but formal proof escapes me. I tried 'limit of products=product of limits':

$ \lim_{{n}\to{\infty}}{n}^{p}{U}_{n} = \lim_{{n}\to{\infty}} {n}^{p}\lim_{{n}\to{\infty}}{U}_{n} \implies A $
$ \therefore \lim_{{n}\to{\infty}}{U}_{n} \implies A/ \lim_{{n}\to{\infty}} {n}^{p} $

Clearly $ \lim_{{n}\to{\infty}} {n}^{p} \implies\infty\: for\: p \gt 1 $
$ \therefore \lim_{{n}\to{\infty}}{U}_{n} \implies A/\infty = 0 $

But I'm stuck here, how do I relate the above result to $ \sum_{n=1}^{\infty} {U}_{n}\: $ ?
Appreciate the help, thanks.

In my opinion the Cesaro-Stolz Theorem is the most simple way to answer to the second questions [divergence]... this theorem etablishes that, given two strictly increasing and divergent sequences $a_{n}$ and $b_{n}$, if ...

$\displaystyle \lim_{n \rightarrow\ \infty} \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}= l\ (1)$

... then is also...

$\displaystyle \lim_{n \rightarrow\ \infty} \frac{a_{n}}{b_{n}}= l\ (2)$

Setting $\displaystyle a_{n} = \sum_{k=1}^{n} U_{k}$ and $\displaystyle b_{n}= \sum_{k=1}^{n} \frac{1}{k}$ and taking into account that the series $\displaystyle \sum_{k=1}^{\infty} \frac{1}{k}$ diverges, You can conclude that $\displaystyle \sum_{k=1}^{\infty} U_{k}$ diverges...

Kind regards

$\chi$ $\sigma$

 

[ognik]

Thanks GJA, that's exactly what I needed, but I couldn't justify to myself if I could divide by n^p, the way you have presented it is clear.

Thanks also chisigma, I haven't encountered that test yet (looked it up on Wiki) - but it is very similar to the limit comparison test, I suspect I'll come across that CS theorum later in the book.