Welcome to our community

Be a part of something great, join today!

strange inequality of infinite series

pantboio

Member
Nov 20, 2012
45
Hi everybody,
while doing a complex analysis exercise, i came to a strange inequality which i don't know how to interpretate. Suppose you have a sequence $\{a_j\}$ of positive real number. Let $\rho$ a positive real number. The inequality i found after some calculation is
$$\sum_{j=1}^{+\infty}\frac{1}{|a_j|^{\rho +\epsilon}}\leq \sum_{j=1}^{+\infty}\frac{1}{|a_j|^{\rho-\epsilon}}$$
for every $\epsilon>0$.
My question is: can i deduce something from this inequality? for example the convergence of the first series (that with $+\epsilon$)? Can i deduce nothing? Is that inequality surely false?
EDIT: the sequence $a_j$ tends to $\infty$
Kind regards
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Hi everybody,
while doing a complex analysis exercise, i came to a strange inequality which i don't know how to interpretate. Suppose you have a sequence $\{a_j\}$ of positive real number. Let $\rho$ a positive real number. The inequality i found after some calculation is
$$\sum_{j=1}^{+\infty}\frac{1}{|a_j|^{\rho +\epsilon}}\leq \sum_{j=1}^{+\infty}\frac{1}{|a_j|^{\rho-\epsilon}}$$
for every $\epsilon>0$.
My question is: can i deduce something from this inequality? for example the convergence of the first series (that with $+\epsilon$)? Can i deduce nothing? Is that inequality surely false?
EDIT: the sequence $a_j$ tends to $\infty$
Kind regards
May be that the inequality holds for every $0<\varepsilon<\rho$ and not for every $\varepsilon >0$...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Another not full clear aspect is that the particular sequence $a_{j} = \ln (2+j)$ satisfies the conditions [all terms are positives and the sequence is unbounded...] , but the series $\displaystyle \sum_{j=1}^{\infty} \frac{1}{a_{j}^{\alpha}}$ diverges for any real $\alpha$. I think that the entire statement of the question should be revised...

Kind regards

$\chi$ $\sigma$
 

pantboio

Member
Nov 20, 2012
45
Another not full clear aspect is that the particular sequence $a_{j} = \ln (2+j)$ satisfies the conditions [all terms are positives and the sequence is unbounded...] , but the series $\displaystyle \sum_{j=1}^{\infty} \frac{1}{a_{j}^{\alpha}}$ diverges for any real $\alpha$. I think that the entire statement of the question should be revised...

Kind regards

$\chi$ $\sigma$
I think so too. As soon as i'll have clearer ideas i'll post a more precise question
 

pantboio

Member
Nov 20, 2012
45
Take $a_j=|z_j|$, where $\{z_j\}$ is the sequence of zeros of an entire function. We suppose $|z_j|\rightarrow +\infty$. Then we introduce two numbers:$b$ which is the exponent of convergence of $\{z_j\}$, defined as the $\inf B$ where $B$ is the set of all $\lambda>0$ such that the series $\sum_{j=1}^{+\infty}\frac{1}{|z_j|^{\lambda}}$ converges. The second quantity is $\rho$, which is
$$\rho=\limsup_{r\rightarrow +\infty}\frac{log\ n(r)}{log\ r}$$
where $n(r)$ is the counting function of $\{z_j\}$, i.e. $n(r)$ is defined to be the number of zeros $z_j$ which satisfy the inequality $|z_j|\leq r$.
CLAIM : $\rho=b$

In order to prove the claim i use the definiton of $\limsup$. We have
$$\rho=\inf\{s>0\ \textrm{s.t.} \frac{log\ n(r)}{log\ r}\leq s\ \forall r\geq r_s\}$$Hence, for every $\epsilon>0$ we have
$$r^{\rho-\epsilon}<n(r)<r^{\rho+\epsilon}$$
definitely for $r$ big enough. From the second inequality we have
$$\sum_{j=1}^{+\infty}\frac{1}{|z_j|^{\rho-\epsilon}}\geq\sum_{j=1}^{+\infty}\frac{1}{n(|z_j|)}=\infty$$
where we assume an enumeration of $z_j$'s such that $|z_1|<|z_2|<\ldots$. This shows that $\rho-\epsilon$ is not in $B$ for every $\epsilon$ and so we have $\rho\leq b$.

The opposite inequality $b\leq \rho$ is what i can't actually prove.
What i can prove is that $\sum_{j=1}^{+\infty}\frac{n(|z_j|)}{|z_j|^{1+\rho+\epsilon}}$ converges $\forall \epsilon$, but i don't know the way to deduce from this what i need, i.e. that $\sum\frac{1}{|z_j|^{\rho+\epsilon}}$ converges
 
Last edited: