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

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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.
 
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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.
ognik said:
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.
 
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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)?
 
Hi ognik,

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

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 $$\sum |U_{n}|$$ converges, then $$\sum U_{n}$$ converges. To prove that $$\sum |U_{n}|$$ 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 $$\lim_{n\rightarrow\infty}n^{p}U_{n}$$ exists, there is a number - call it $$C$$ - such that

$$|n^{p}U_{n}|< C$$ for all $$n$$

Dividing through in this inequality by $$n^{p}$$ we get

$$|U_{n}|<\frac{C}{n^{p}}$$ for all $$n.$$ Thus

$$\sum |U_{n}|<\sum \frac{C}{n^{p}}$$

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

$$\sum |U_{n}|$$

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

$$\sum U_{n}$$

converges too.

To prove that $$\lim_{n\rightarrow\infty}nU_{n}=A>0$$ implies divergence, you can use a similar argument as above, only this time you will want to bound $$U_{n}$$ from below (whereas we bounded $$|U_{n}|$$ 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.
 
ognik said:
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

$\chi$ $\sigma$
 
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Thanks GJA, that's exactly what I needed, but I couldn't justify to myself if I could divide by np, 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.
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...

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