Is Cesaro-Stolz theorem applicable in this case?....

[chisigma]

Recently in the POTW 'graduate' has appeared the following interesting problem ...

Show that if $a_{n}$ is a decreasing sequence of positive real numbers such that $\displaystyle \sum_{n=1}^{\infty} a_{n}$ converges, then $\displaystyle \lim_{n \rightarrow \infty} n\ a_{n} = 0$...

At first I thought is that the easiest way was to use the Cesaro-Stolz theorem , which in simplified version says ...

http://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf

If $b_{n}$ is a sequence of positive real numbers such that $\displaystyle \sum_{n=1}^{\infty} b_{n}$ diverges to infinity and for a sequence $a_{n}$ exists the limit $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}}= l$, then is...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{1} + a_{2} + ...+ a_{n}}{b_{1} + b_{2} + ... + b_{n}} = \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}} = l\ (1)$

In particular if $\displaystyle \sum_{n=1}^{\infty} a_{n}$ converges, the is $l=0$ and setting $b_{n} = \frac{1}{n}$ we arrive directly to $\displaystyle \lim_{n \rightarrow \infty} n\ a_{n} = 0$... but that is correct?... the Cesaro-Stolz theorem holds also in the case $l=0$?...

Kind regards

$\chi$ $\sigma$

 

[Nono713]

Recently in the POTW 'graduate' has appeared the following interesting problem ...

Show that if $a_{n}$ is a decreasing sequence of positive real numbers such that $\displaystyle \sum_{n=1}^{\infty} a_{n}$ converges, then $\displaystyle \lim_{n \rightarrow \infty} n\ a_{n} = 0$...

At first I thought is that the easiest way was to use the Cesaro-Stolz theorem , which in simplified version says ...

http://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf

If $b_{n}$ is a sequence of positive real numbers such that $\displaystyle \sum_{n=1}^{\infty}$ diverges and for a sequence $a_{n}$ exists the limit $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}}= l$, then is...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{1} + a_{2} + ...+ a_{n}}{b_{1} + b_{2} + ... + b_{n}} = \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}} = l\ (1)$

In particular if $\displaystyle \sum_{n=1}^{\infty} a_{n}$ converges, the is $l=0$ and setting $b_{n} = \frac{1}{n}$ we arrive directly to $\displaystyle \lim_{n \rightarrow \infty} n\ a_{n} = 0$... but that is correct?... the Cesaro-Stolz theorem holds also in the case $l=0$?...

Kind regards

$\chi$ $\sigma$

I'm not sure that is correct. Aren't you vacuously assuming that $\lim_{n \to \infty} n a_n = 0$ already when you assume $l = 0$ to begin with? It doesn't seem like you are using the Cesaro-Stolz theorem at all here, as you are not using the equality to:
$$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{1} + a_{2} + ...+ a_{n}}{b_{1} + b_{2} + ... + b_{n}}$$
which is the only thing the theorem states. So the reasoning doesn't seem correct to me. I could be wrong, though. In any case, of course, you don't need such a theorem for the problem, since you can do it the way Euge shows or I believe also by comparison with the harmonic series, that is, suppose the sequence grew asymptotically as fast as or slower than $1 / n$, then the series does not converge. hence we must have $a_n \in O(1 / n^{1 + \epsilon})$ for some $\epsilon > 0$ and so $n a_n \in O(1 / n^\epsilon) \to 0$ (this isn't rigorous but I think the general idea works).​

 

[Fallen Angel]

Hi,

I also thought on it, but it's not a good idea, at least not as good as it could seem.

If $\displaystyle\sum a_{n}$ converges you have not guaranteed that $lim \dfrac{a_{n}}{b_{n}}=0$, you need to prove it, In particular if $b_{n}=\dfrac{1}{n}$ you need to prove the same statement that you are required.

Maybe this is not clearly explained but I don't know how to explain it better, I mean it's a circular argument.

What I tried it's more or less what Bacterius said, comparing it with a series of the form $\displaystyle\sum\dfrac{\epsilon}{n}$ with $\epsilon > 0$ that diverges for any $\epsilon$, but I think my fail was thinking there wasn't any convergent subseries of that

 

[Opalg]

My solution, not as elegant as the neat proof given by Euge, goes back to the epsilon-N definition of convergence.

Given $\varepsilon>0$, choose $M$ so that $\sum_{n=M}^\infty a_n <\varepsilon/2$. Then choose $N>M$ so that $n>N\;\Rightarrow\; a_n < \varepsilon/(2M)$. If $n>N$ then $$\begin{aligned} na_n &= Ma_n + (n-M)a_n \\ &< Ma_n + \sum_{r=M+1}^n a_r \\ &< \varepsilon/2 + \varepsilon/2 = \varepsilon . \end{aligned}$$ Since $\varepsilon$ is arbitrary, it follows that $\lim_{n\to\infty}na_n = 0.$

 

[chisigma]

I fear that my post was misinterpreted ... the problem is not privilege one solution over another in a particular problem, the problem is whether or not a theorem is applicable to that particular problem ...

I think we all agree that the theorem of Cesaro-Stolz is valid ... what I asked of you is simply if the theorem is valid also in the case, defining $\displaystyle \lim_{n \rightarrow \infty} \frac {a_ {1} + a_ {2} + ... + a_ {n}} {b_ {1} + b_ {2} + ... + b_ {n}} = l$ , is $ l = 0 $ ...

The question is not irrelevant, because if the theorem is applicable to the case l = 0, then for every convergent series $\displaystyle \sum_{n=1}^{\infty} a_{n}$, You get to more stringent conclusions of the type...

$\displaystyle \lim_{n \rightarrow \infty} n\ \ln n\ a_{n} = 0$

$\displaystyle \lim_{n \rightarrow \infty} n\ \ln n\ \ln (\ln n)\ a_{n} = 0$

... etc...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Looking at the problem with more attention and found this excellent work ...

https://books.google.it/books?id=YNP1fcTjtBcC&pg=PA266&lpg=PA266&dq=stolz+cesaro+proof&source=bl&ots=yEqW0FAu3R&sig=vOBbtq6doA4C813qhYj9HVRXSSo&hl=en&sa=X&ei=sxa1VOHDAof8ywOViYGgCA&ved=0CEYQ6AEwBjgK#v=onepage&q=stolz%20cesaro%20proof&f=false

Here I learned that the road from me undertaken passing through the so-called 'Reciprocal of the Cesaro Stolz lemma' that says ...

Let be $a_{n}$ and $b_{n}$ two sequences of real positive numbers such that...

a) $0 < b_{1} < b_{2} < ... < b_{n} < ...$ and $\displaystyle \lim _{n \rightarrow \infty} b_{n} = \infty$...

b) $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}} = l \in \mathbb{R}$

c) $\displaystyle \lim_{n \rightarrow \infty} \frac{b_{n}}{b_{n+1}} = L \in \mathbb{R}$ \ $\{1\}$...

Then the limit...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}- a_{n}}{b_{n+1}- b_{n}}$

... exists and is equal to l...

Therefore it seems to be no problem with the value of l [any real number and therefore also l = 0 ...], as with the value of L that is supposed $ L \ne 1 $ ... however the script ends with the following lines ...

Theorem B.3 shows that il limit $\displaystyle \lim_{n \rightarrow \infty} \frac{b_{n}}{b_{n+1}}$ exists and it is not equal to 1, the reciprocal of Cesaro-Stolz lemma is valid. We stop our line of investigation here and invite the reader to study further the additional conditions required such that the reciprocal of the Cesaro-Stolz lemma remains valid in the trouble case $\displaystyle \lim_{n \rightarrow \infty} \frac{b_{n}}{b_{n+1}}= 1$...

It seems that an extra hard work I am requested (Thinking)...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Looking at the problem with more attention and found this excellent work ...

https://books.google.it/books?id=YNP1fcTjtBcC&pg=PA266&lpg=PA266&dq=stolz+cesaro+proof&source=bl&ots=yEqW0FAu3R&sig=vOBbtq6doA4C813qhYj9HVRXSSo&hl=en&sa=X&ei=sxa1VOHDAof8ywOViYGgCA&ved=0CEYQ6AEwBjgK#v=onepage&q=stolz%20cesaro%20proof&f=false

Here I learned that the road from me undertaken passing through the so-called 'Reciprocal of the Cesaro Stolz lemma' that says ...

Let be $a_{n}$ and $b_{n}$ two sequences of real positive numbers such that...

a) $0 < b_{1} < b_{2} < ... < b_{n} < ...$ and $\displaystyle \lim _{n \rightarrow \infty} b_{n} = \infty$...

b) $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}} = l \in \mathbb{R}$

c) $\displaystyle \lim_{n \rightarrow \infty} \frac{b_{n}}{b_{n+1}} = L \in \mathbb{R}$ \ $\{1\}$...

Then the limit...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}- a_{n}}{b_{n+1}- b_{n}}$

... exists and is equal to l...

Therefore it seems to be no problem with the value of l [any real number and therefore also l = 0 ...], as with the value of L that is supposed $ L \ne 1 $ ... however the script ends with the following lines ...

Theorem B.3 shows that il limit $\displaystyle \lim_{n \rightarrow \infty} \frac{b_{n}}{b_{n+1}}$ exists and it is not equal to 1, the reciprocal of Cesaro-Stolz lemma is valid. We stop our line of investigation here and invite the reader to study further the additional conditions required such that the reciprocal of the Cesaro-Stolz lemma remains valid in the trouble case $\displaystyle \lim_{n \rightarrow \infty} \frac{b_{n}}{b_{n+1}}= 1$...

It seems that an extra hard work I am requested (Thinking)...

Meeting the challenge in the general case in which it is $\displaystyle \lim_{n \rightarrow \infty } \frac{b_{n}}{b_{n+1}} = 1$ seems a very difficult task... however, as part of the original problem the following special case of the theorem of mutual Cesaro Stolz can be treated ...

Let be $a_{n}$ and $b_{n}$ two increasing sequences of real positive numbers such that...

a) $0 < b_{1} < b_{2} < ... < b_{n} < ...$, $0 < a_{1} < a_{2} < ... < a_{n} < ...$ and $\displaystyle \lim _{n \rightarrow \infty} b_{n} = \infty$...

b) $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}} = 0 $

c) $\displaystyle \lim_{n \rightarrow \infty} \frac{b_{n}}{b_{n+1}} = 1$...

d) $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{a_{n+1}} = 1$...

e) for any n is $\displaystyle \frac{a_{n}}{a_{n+1}} \ge \frac{b_{n}}{b_{n+1}}$...

Then the limit...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1}- a_{n}}{b_{n+1}- b_{n}}$

... exists and is equal to 0...

The proof is comfortable and it is based on the identity...

$\displaystyle \frac{a_{n+1} - a_{n}}{b_{n+1}- b_{n}} = \frac{a_{n+1}}{b_{n+1}}\ \frac{1 - \frac{a_{n}}{a_{n+1}}}{1 - \frac{b_{n}}{b_{n+1}}}\ (1)$

If n tends to infinity, the first term of the second member of (1) tends to 0 and for the fractional term is $\displaystyle \frac{1 - \frac{a_{n}}{a_{n+1}}}{1 - \frac{b_{n}}{b_{n+1}}} \le 1$ so that the demonstration is performed...

My conclusion therefore is that the original problem theorem inverse of Cesaro Stolz, although in modified form by me in a particular case. could be applied and indeed would have led to an extension of the results ...

... of course further investigation of the case $\displaystyle \lim_{n \rightarrow \infty} \frac{b_{n}}{b_{n+1}} = 1$ remains open and of gratest importance ​...

Kind regards

$\chi$ $\sigma$

P.S. ... of course the condition e) is not really necessary and we only need that $\displaystyle \frac{1 - \frac{a_{n}}{a_{n+1}}}{1 - \frac{b_{n}}{b_{n+1}}} $ tends for $n \rightarrow \infty$ to some finite value $\lambda$... this concept will be better explained in my next post...

 

[chisigma]

We come now to the more general case in which it is $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}} = l \ne 0$ and and we try to answer the question...

Let be $a_{n}$ and $b_{n}$ two positive real numbers sequences such that...

a) $0 < b_{1} < b_{2} < ... < b_{n} < ...$, $0 < a_{1} < a_{2} < ... < a_{n} < ...$ and $\displaystyle \lim_{ n \rightarrow \infty} b_{n} = \infty$...

b) $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}} = l \ne 0$...

c) $\displaystyle \lim_{n \rightarrow \infty} \frac{b_{n}}{b_{n+1}} = 1$ ...

d) $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{a_{n+1}} = 1$ ...

... under what conditions is...

$\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1} - a_{n}}{b_{n+1} - b_{n}} = l$?... Even now the answer is given by the identity...

$\displaystyle \frac{a_{n+1} - a_{n}}{b_{n+1} - b_{n}} = \frac {a_{n+1}}{b_{n+1}}\ \frac{1 - \frac{a_{n}}{a_{n+1}}}{1- \frac{b_{n}}{b_{n+1}}}\ (1)$

... if $\displaystyle \lim_{n \rightarrow \infty} \frac{1 - \frac{a_{n}}{a_{n+1}}}{1- \frac{b_{n}}{b_{n+1}}} = \lambda$ then $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1} - a_{n}}{b_{n+1} - b_{n}} = l\ \lambda$, so that the requirements are satisfied only if $\lambda = 1$ ...

Kind regards

$\chi$ $\sigma$
Je suis Dieudonné! … the freedom of thought applies to all... for him too!...