If $ sum a_n$ converges, does $sumlimits_{n=2}^{infty} frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$...












4












$begingroup$



If $ sum a_n$ converges and $a_n>0$ for every $n$, does $sumlimits_{n=2}^{infty} frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$ converge as well?




What I did: Define $u_n=frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$. Trying to use the ratio test, I consider:
$$ frac{u_{n+1}}{u_n}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{n^{a_{n+1}}-1}{n^{a_{n}}-1}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}}$$
and now:
$$ frac{ln(n)}{ln(n+1)} rightarrow 1 $$
$$ sqrt{frac{a_{n+1}}{a_n}} rightarrow g in (0;1) $$
but what can I do with the last ratio $$frac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}} ?$$
Or has somebody any other idea for this task?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
    $endgroup$
    – rtybase
    Dec 17 '18 at 17:48
















4












$begingroup$



If $ sum a_n$ converges and $a_n>0$ for every $n$, does $sumlimits_{n=2}^{infty} frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$ converge as well?




What I did: Define $u_n=frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$. Trying to use the ratio test, I consider:
$$ frac{u_{n+1}}{u_n}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{n^{a_{n+1}}-1}{n^{a_{n}}-1}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}}$$
and now:
$$ frac{ln(n)}{ln(n+1)} rightarrow 1 $$
$$ sqrt{frac{a_{n+1}}{a_n}} rightarrow g in (0;1) $$
but what can I do with the last ratio $$frac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}} ?$$
Or has somebody any other idea for this task?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
    $endgroup$
    – rtybase
    Dec 17 '18 at 17:48














4












4








4


1



$begingroup$



If $ sum a_n$ converges and $a_n>0$ for every $n$, does $sumlimits_{n=2}^{infty} frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$ converge as well?




What I did: Define $u_n=frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$. Trying to use the ratio test, I consider:
$$ frac{u_{n+1}}{u_n}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{n^{a_{n+1}}-1}{n^{a_{n}}-1}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}}$$
and now:
$$ frac{ln(n)}{ln(n+1)} rightarrow 1 $$
$$ sqrt{frac{a_{n+1}}{a_n}} rightarrow g in (0;1) $$
but what can I do with the last ratio $$frac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}} ?$$
Or has somebody any other idea for this task?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$





If $ sum a_n$ converges and $a_n>0$ for every $n$, does $sumlimits_{n=2}^{infty} frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$ converge as well?




What I did: Define $u_n=frac{sqrt{a_n}}{ln n}cdot left( n^{a_n}-1 right)$. Trying to use the ratio test, I consider:
$$ frac{u_{n+1}}{u_n}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{n^{a_{n+1}}-1}{n^{a_{n}}-1}=frac{ln(n)}{ln(n+1)} cdot sqrt{frac{a_{n+1}}{a_n}} cdotfrac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}}$$
and now:
$$ frac{ln(n)}{ln(n+1)} rightarrow 1 $$
$$ sqrt{frac{a_{n+1}}{a_n}} rightarrow g in (0;1) $$
but what can I do with the last ratio $$frac{frac{n^{a_{n+1}}}{n^{a_{n}}}-frac{1}{n^{a_{n}}}}{1-frac{1}{n^{a_{n}}}} ?$$
Or has somebody any other idea for this task?






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 18:41









Did

248k23226465




248k23226465










asked Dec 17 '18 at 17:34









VirtualUserVirtualUser

1,096117




1,096117








  • 1




    $begingroup$
    We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
    $endgroup$
    – rtybase
    Dec 17 '18 at 17:48














  • 1




    $begingroup$
    We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
    $endgroup$
    – rtybase
    Dec 17 '18 at 17:48








1




1




$begingroup$
We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
$endgroup$
– rtybase
Dec 17 '18 at 17:48




$begingroup$
We still can have $g=1$ and $sumlimits_{n=1} a_n < infty$ ...
$endgroup$
– rtybase
Dec 17 '18 at 17:48










1 Answer
1






active

oldest

votes


















5












$begingroup$

0. Note that the ratio test here is completely useless, because you do not know anything about $frac{a_{n+1}}{a_n}$ (it could even have a subsequence going to infinity).



1. Assume $a_n=O(log(n)^{-1})$. Then, $$frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)=(exp(log(n)a_n)-1)frac{a_n^{1/2}}{log(n)} leq Ca_n^{3/2} leq C’a_n$$ so the series converges.



2. Without the assumption that $a_n=O(log(n)^{-1})$, the series may diverge, see example below.



Take $a_{2^{p^4}}=p^{-2}$ (and zero everywhere else). Then, $sum_n{|a_n|}=sum_p{p^{-2}}$ converges.



However, $a_{2^{p^4}}log(2^{p^4}) =log(2) p^2 rightarrow infty$, thus $a_n neq O((log(n)^{-1})$.



Furthermore,
$$sum_n{frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)} = sum_p{frac{p^{-1}}{log(2)p^4}(2^{p^2}-1)}=infty$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I am afraid it is not. See the counter-example at the end.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 18:11










  • $begingroup$
    I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 18:54










  • $begingroup$
    The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:19












  • $begingroup$
    We cannot? But you done this in your proposition?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 20:21






  • 1




    $begingroup$
    I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:34











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

0. Note that the ratio test here is completely useless, because you do not know anything about $frac{a_{n+1}}{a_n}$ (it could even have a subsequence going to infinity).



1. Assume $a_n=O(log(n)^{-1})$. Then, $$frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)=(exp(log(n)a_n)-1)frac{a_n^{1/2}}{log(n)} leq Ca_n^{3/2} leq C’a_n$$ so the series converges.



2. Without the assumption that $a_n=O(log(n)^{-1})$, the series may diverge, see example below.



Take $a_{2^{p^4}}=p^{-2}$ (and zero everywhere else). Then, $sum_n{|a_n|}=sum_p{p^{-2}}$ converges.



However, $a_{2^{p^4}}log(2^{p^4}) =log(2) p^2 rightarrow infty$, thus $a_n neq O((log(n)^{-1})$.



Furthermore,
$$sum_n{frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)} = sum_p{frac{p^{-1}}{log(2)p^4}(2^{p^2}-1)}=infty$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I am afraid it is not. See the counter-example at the end.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 18:11










  • $begingroup$
    I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 18:54










  • $begingroup$
    The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:19












  • $begingroup$
    We cannot? But you done this in your proposition?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 20:21






  • 1




    $begingroup$
    I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:34
















5












$begingroup$

0. Note that the ratio test here is completely useless, because you do not know anything about $frac{a_{n+1}}{a_n}$ (it could even have a subsequence going to infinity).



1. Assume $a_n=O(log(n)^{-1})$. Then, $$frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)=(exp(log(n)a_n)-1)frac{a_n^{1/2}}{log(n)} leq Ca_n^{3/2} leq C’a_n$$ so the series converges.



2. Without the assumption that $a_n=O(log(n)^{-1})$, the series may diverge, see example below.



Take $a_{2^{p^4}}=p^{-2}$ (and zero everywhere else). Then, $sum_n{|a_n|}=sum_p{p^{-2}}$ converges.



However, $a_{2^{p^4}}log(2^{p^4}) =log(2) p^2 rightarrow infty$, thus $a_n neq O((log(n)^{-1})$.



Furthermore,
$$sum_n{frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)} = sum_p{frac{p^{-1}}{log(2)p^4}(2^{p^2}-1)}=infty$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I am afraid it is not. See the counter-example at the end.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 18:11










  • $begingroup$
    I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 18:54










  • $begingroup$
    The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:19












  • $begingroup$
    We cannot? But you done this in your proposition?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 20:21






  • 1




    $begingroup$
    I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:34














5












5








5





$begingroup$

0. Note that the ratio test here is completely useless, because you do not know anything about $frac{a_{n+1}}{a_n}$ (it could even have a subsequence going to infinity).



1. Assume $a_n=O(log(n)^{-1})$. Then, $$frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)=(exp(log(n)a_n)-1)frac{a_n^{1/2}}{log(n)} leq Ca_n^{3/2} leq C’a_n$$ so the series converges.



2. Without the assumption that $a_n=O(log(n)^{-1})$, the series may diverge, see example below.



Take $a_{2^{p^4}}=p^{-2}$ (and zero everywhere else). Then, $sum_n{|a_n|}=sum_p{p^{-2}}$ converges.



However, $a_{2^{p^4}}log(2^{p^4}) =log(2) p^2 rightarrow infty$, thus $a_n neq O((log(n)^{-1})$.



Furthermore,
$$sum_n{frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)} = sum_p{frac{p^{-1}}{log(2)p^4}(2^{p^2}-1)}=infty$$






share|cite|improve this answer











$endgroup$



0. Note that the ratio test here is completely useless, because you do not know anything about $frac{a_{n+1}}{a_n}$ (it could even have a subsequence going to infinity).



1. Assume $a_n=O(log(n)^{-1})$. Then, $$frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)=(exp(log(n)a_n)-1)frac{a_n^{1/2}}{log(n)} leq Ca_n^{3/2} leq C’a_n$$ so the series converges.



2. Without the assumption that $a_n=O(log(n)^{-1})$, the series may diverge, see example below.



Take $a_{2^{p^4}}=p^{-2}$ (and zero everywhere else). Then, $sum_n{|a_n|}=sum_p{p^{-2}}$ converges.



However, $a_{2^{p^4}}log(2^{p^4}) =log(2) p^2 rightarrow infty$, thus $a_n neq O((log(n)^{-1})$.



Furthermore,
$$sum_n{frac{a_n^{1/2}}{log(n)}(n^{a_n}-1)} = sum_p{frac{p^{-1}}{log(2)p^4}(2^{p^2}-1)}=infty$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 11:12









Did

248k23226465




248k23226465










answered Dec 17 '18 at 18:08









MindlackMindlack

4,920211




4,920211








  • 1




    $begingroup$
    I am afraid it is not. See the counter-example at the end.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 18:11










  • $begingroup$
    I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 18:54










  • $begingroup$
    The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:19












  • $begingroup$
    We cannot? But you done this in your proposition?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 20:21






  • 1




    $begingroup$
    I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:34














  • 1




    $begingroup$
    I am afraid it is not. See the counter-example at the end.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 18:11










  • $begingroup$
    I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 18:54










  • $begingroup$
    The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:19












  • $begingroup$
    We cannot? But you done this in your proposition?
    $endgroup$
    – VirtualUser
    Dec 17 '18 at 20:21






  • 1




    $begingroup$
    I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
    $endgroup$
    – Mindlack
    Dec 17 '18 at 20:34








1




1




$begingroup$
I am afraid it is not. See the counter-example at the end.
$endgroup$
– Mindlack
Dec 17 '18 at 18:11




$begingroup$
I am afraid it is not. See the counter-example at the end.
$endgroup$
– Mindlack
Dec 17 '18 at 18:11












$begingroup$
I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
$endgroup$
– VirtualUser
Dec 17 '18 at 18:54




$begingroup$
I am looking at this and I am not able to understand why we can assume that $ a_n=O(log(n)^{-1}) $ ?
$endgroup$
– VirtualUser
Dec 17 '18 at 18:54












$begingroup$
The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
$endgroup$
– Mindlack
Dec 17 '18 at 20:19






$begingroup$
The point is that we cannot. If $a_n$ is $O(log(n)^{-1})$, then the second series converges. Else, it is possible that it does not converge.
$endgroup$
– Mindlack
Dec 17 '18 at 20:19














$begingroup$
We cannot? But you done this in your proposition?
$endgroup$
– VirtualUser
Dec 17 '18 at 20:21




$begingroup$
We cannot? But you done this in your proposition?
$endgroup$
– VirtualUser
Dec 17 '18 at 20:21




1




1




$begingroup$
I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
$endgroup$
– Mindlack
Dec 17 '18 at 20:34




$begingroup$
I thought I had made myself clear. I proved two things: 1) in general the series needs not converge. 2) if $a_n$ is $O(log(n)^{-1})$ then the series converges.
$endgroup$
– Mindlack
Dec 17 '18 at 20:34


















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How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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