Prove that $sumlimits_{n=1}^{infty}a_n< infty $ imply $lim_{x rightarrow infty} frac{1}{x} sum_{n leq x} n...












2












$begingroup$


Let $ {a_n} n in mathbb{N} $ a sequence of non-negative real numbers.



Prove that
$displaystylesum_{n=1}^{infty}a_n< infty $ imply $displaystylelim_{N rightarrow infty} frac{1}{N} sum_{n leq N} n a_n=0$



Would this be accomplished without the condition $a_n geq 0$?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: note that the series $s_N=sum_{n=1}^{N}a_n = a_1+cdot+a_N$ is always an upper bound of $sum_{n=1}^N na_n=1/Na_1+2/Na_2+cdot+a_N$.
    $endgroup$
    – Joseph Zambrano
    Jul 16 '15 at 1:30


















2












$begingroup$


Let $ {a_n} n in mathbb{N} $ a sequence of non-negative real numbers.



Prove that
$displaystylesum_{n=1}^{infty}a_n< infty $ imply $displaystylelim_{N rightarrow infty} frac{1}{N} sum_{n leq N} n a_n=0$



Would this be accomplished without the condition $a_n geq 0$?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: note that the series $s_N=sum_{n=1}^{N}a_n = a_1+cdot+a_N$ is always an upper bound of $sum_{n=1}^N na_n=1/Na_1+2/Na_2+cdot+a_N$.
    $endgroup$
    – Joseph Zambrano
    Jul 16 '15 at 1:30
















2












2








2


4



$begingroup$


Let $ {a_n} n in mathbb{N} $ a sequence of non-negative real numbers.



Prove that
$displaystylesum_{n=1}^{infty}a_n< infty $ imply $displaystylelim_{N rightarrow infty} frac{1}{N} sum_{n leq N} n a_n=0$



Would this be accomplished without the condition $a_n geq 0$?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$




Let $ {a_n} n in mathbb{N} $ a sequence of non-negative real numbers.



Prove that
$displaystylesum_{n=1}^{infty}a_n< infty $ imply $displaystylelim_{N rightarrow infty} frac{1}{N} sum_{n leq N} n a_n=0$



Would this be accomplished without the condition $a_n geq 0$?






real-analysis sequences-and-series




real-analysis sequences-and-series






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share|cite|improve this question













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share|cite|improve this question








edited Dec 29 '18 at 9:56









Martin Sleziak

45k10123277




45k10123277










asked Jul 16 '15 at 1:17









El ChapoEl Chapo

1027




1027












  • $begingroup$
    Hint: note that the series $s_N=sum_{n=1}^{N}a_n = a_1+cdot+a_N$ is always an upper bound of $sum_{n=1}^N na_n=1/Na_1+2/Na_2+cdot+a_N$.
    $endgroup$
    – Joseph Zambrano
    Jul 16 '15 at 1:30




















  • $begingroup$
    Hint: note that the series $s_N=sum_{n=1}^{N}a_n = a_1+cdot+a_N$ is always an upper bound of $sum_{n=1}^N na_n=1/Na_1+2/Na_2+cdot+a_N$.
    $endgroup$
    – Joseph Zambrano
    Jul 16 '15 at 1:30


















$begingroup$
Hint: note that the series $s_N=sum_{n=1}^{N}a_n = a_1+cdot+a_N$ is always an upper bound of $sum_{n=1}^N na_n=1/Na_1+2/Na_2+cdot+a_N$.
$endgroup$
– Joseph Zambrano
Jul 16 '15 at 1:30






$begingroup$
Hint: note that the series $s_N=sum_{n=1}^{N}a_n = a_1+cdot+a_N$ is always an upper bound of $sum_{n=1}^N na_n=1/Na_1+2/Na_2+cdot+a_N$.
$endgroup$
– Joseph Zambrano
Jul 16 '15 at 1:30












3 Answers
3






active

oldest

votes


















4












$begingroup$

You can prove this with Lebesgue's dominated convergence theorem. I'll prove it in a different setting, and I'll let you adapt the proof. Suppose that $gin L^1_+[0,infty)$. Then $$frac 1n int_0^n xg(x)dxto 0$$ as $ntoinfty$.



Proof Consider $g_n(x)= dfrac{xg(x)}n {bf 1}_{[0,n]}$. Then $|g_n|leqslant |g|$ and $g_n(x)to 0$. By DCT, we conclude.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'll read about DCT, I I don't heard before
    $endgroup$
    – Sudo su
    Jul 16 '15 at 1:36










  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38



















2












$begingroup$

Let $A_n = a_1 + cdots + a_n$ be the partial sum of ${a_n}$, by summation by parts formula
$$sum_{n = 1}^N na_n = sum_{n = 1}^{N - 1}A_n(n - (n + 1)) + NA_N = NA_N - sum_{n = 1}^{N - 1}A_n.$$
Thus, based on the famous Cesaro's theorem
$$lim_{N to infty} frac{1}{N}sum_{n = 1}^N na_n = lim_{N to infty} A_N - lim_{N to infty} frac{sum_{n = 1}^{N - 1}A_n}{N} = a - a = 0$$
where $a = sum_{n = 1}^infty a_n < infty$.



The above proof shows that as long as $left|sum_{n = 1}^infty a_nright| < infty$, the result always holds. $a_n$ could be negative.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I' ll copy this Proof
    $endgroup$
    – Sudo su
    Jul 16 '15 at 1:36










  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38






  • 1




    $begingroup$
    How do you dervive $lim_{Ntoinfty}(sum_{n=1}^{N-1}A_n)/N=a$? (Not saying that it's not correct, but that that is a non-trivial result in its own right.)
    $endgroup$
    – Steven Stadnicki
    Jul 16 '15 at 1:49












  • $begingroup$
    @StevenStadnicki check en.wikipedia.org/wiki/Ces%C3%A0ro_summation. I would assume this result is so famous that every math student would know that.
    $endgroup$
    – Zhanxiong
    Jul 16 '15 at 1:50












  • $begingroup$
    "The above proof shows that as long as $|sum a_n|<infty$" I think you want to say "as long as $sum a_n$ converges"
    $endgroup$
    – zhw.
    Jul 16 '15 at 2:11





















-1












$begingroup$

$dfrac{displaystyle sum_{n=1}^N na_n}{N}=dfrac{a_1+2a_2+3a_3+cdots +Na_N}{N}=dfrac{a_1+cdots+a_N}{N}+dfrac{N-1}{N}cdotdfrac{a_2+cdots+a_N}{N-1}+cdots + dfrac{a_N}{N}to 0+0+cdots + 0 = 0$ by the well-known Cesaro's theorem as $a_N to 0$ when $N to infty$ since the first series converges imply the general terms tend to $0$. The result still holds without $a_n geq 0$.






share|cite|improve this answer









$endgroup$









  • 5




    $begingroup$
    Each summand tends to zero, but the count of summands is arbitrarily large, so this proof doesn't stand
    $endgroup$
    – Norbert
    Jul 16 '15 at 1:32












  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38










  • $begingroup$
    If that was a valid argument we could also say that $lim_{ntoinfty}frac{1}{n} + frac{1}{n+1} + ldots + frac{1}{2n} = 0 + 0 + ldots + 0 = 0$ while the true result is $log(2)$.
    $endgroup$
    – Winther
    Jul 16 '15 at 2:22












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

You can prove this with Lebesgue's dominated convergence theorem. I'll prove it in a different setting, and I'll let you adapt the proof. Suppose that $gin L^1_+[0,infty)$. Then $$frac 1n int_0^n xg(x)dxto 0$$ as $ntoinfty$.



Proof Consider $g_n(x)= dfrac{xg(x)}n {bf 1}_{[0,n]}$. Then $|g_n|leqslant |g|$ and $g_n(x)to 0$. By DCT, we conclude.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'll read about DCT, I I don't heard before
    $endgroup$
    – Sudo su
    Jul 16 '15 at 1:36










  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38
















4












$begingroup$

You can prove this with Lebesgue's dominated convergence theorem. I'll prove it in a different setting, and I'll let you adapt the proof. Suppose that $gin L^1_+[0,infty)$. Then $$frac 1n int_0^n xg(x)dxto 0$$ as $ntoinfty$.



Proof Consider $g_n(x)= dfrac{xg(x)}n {bf 1}_{[0,n]}$. Then $|g_n|leqslant |g|$ and $g_n(x)to 0$. By DCT, we conclude.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'll read about DCT, I I don't heard before
    $endgroup$
    – Sudo su
    Jul 16 '15 at 1:36










  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38














4












4








4





$begingroup$

You can prove this with Lebesgue's dominated convergence theorem. I'll prove it in a different setting, and I'll let you adapt the proof. Suppose that $gin L^1_+[0,infty)$. Then $$frac 1n int_0^n xg(x)dxto 0$$ as $ntoinfty$.



Proof Consider $g_n(x)= dfrac{xg(x)}n {bf 1}_{[0,n]}$. Then $|g_n|leqslant |g|$ and $g_n(x)to 0$. By DCT, we conclude.






share|cite|improve this answer









$endgroup$



You can prove this with Lebesgue's dominated convergence theorem. I'll prove it in a different setting, and I'll let you adapt the proof. Suppose that $gin L^1_+[0,infty)$. Then $$frac 1n int_0^n xg(x)dxto 0$$ as $ntoinfty$.



Proof Consider $g_n(x)= dfrac{xg(x)}n {bf 1}_{[0,n]}$. Then $|g_n|leqslant |g|$ and $g_n(x)to 0$. By DCT, we conclude.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jul 16 '15 at 1:27









Pedro Tamaroff♦Pedro Tamaroff

97.7k10153299




97.7k10153299












  • $begingroup$
    I'll read about DCT, I I don't heard before
    $endgroup$
    – Sudo su
    Jul 16 '15 at 1:36










  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38


















  • $begingroup$
    I'll read about DCT, I I don't heard before
    $endgroup$
    – Sudo su
    Jul 16 '15 at 1:36










  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38
















$begingroup$
I'll read about DCT, I I don't heard before
$endgroup$
– Sudo su
Jul 16 '15 at 1:36




$begingroup$
I'll read about DCT, I I don't heard before
$endgroup$
– Sudo su
Jul 16 '15 at 1:36












$begingroup$
I will try to reproduce. Thanks
$endgroup$
– El Chapo
Jul 16 '15 at 1:38




$begingroup$
I will try to reproduce. Thanks
$endgroup$
– El Chapo
Jul 16 '15 at 1:38











2












$begingroup$

Let $A_n = a_1 + cdots + a_n$ be the partial sum of ${a_n}$, by summation by parts formula
$$sum_{n = 1}^N na_n = sum_{n = 1}^{N - 1}A_n(n - (n + 1)) + NA_N = NA_N - sum_{n = 1}^{N - 1}A_n.$$
Thus, based on the famous Cesaro's theorem
$$lim_{N to infty} frac{1}{N}sum_{n = 1}^N na_n = lim_{N to infty} A_N - lim_{N to infty} frac{sum_{n = 1}^{N - 1}A_n}{N} = a - a = 0$$
where $a = sum_{n = 1}^infty a_n < infty$.



The above proof shows that as long as $left|sum_{n = 1}^infty a_nright| < infty$, the result always holds. $a_n$ could be negative.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I' ll copy this Proof
    $endgroup$
    – Sudo su
    Jul 16 '15 at 1:36










  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38






  • 1




    $begingroup$
    How do you dervive $lim_{Ntoinfty}(sum_{n=1}^{N-1}A_n)/N=a$? (Not saying that it's not correct, but that that is a non-trivial result in its own right.)
    $endgroup$
    – Steven Stadnicki
    Jul 16 '15 at 1:49












  • $begingroup$
    @StevenStadnicki check en.wikipedia.org/wiki/Ces%C3%A0ro_summation. I would assume this result is so famous that every math student would know that.
    $endgroup$
    – Zhanxiong
    Jul 16 '15 at 1:50












  • $begingroup$
    "The above proof shows that as long as $|sum a_n|<infty$" I think you want to say "as long as $sum a_n$ converges"
    $endgroup$
    – zhw.
    Jul 16 '15 at 2:11


















2












$begingroup$

Let $A_n = a_1 + cdots + a_n$ be the partial sum of ${a_n}$, by summation by parts formula
$$sum_{n = 1}^N na_n = sum_{n = 1}^{N - 1}A_n(n - (n + 1)) + NA_N = NA_N - sum_{n = 1}^{N - 1}A_n.$$
Thus, based on the famous Cesaro's theorem
$$lim_{N to infty} frac{1}{N}sum_{n = 1}^N na_n = lim_{N to infty} A_N - lim_{N to infty} frac{sum_{n = 1}^{N - 1}A_n}{N} = a - a = 0$$
where $a = sum_{n = 1}^infty a_n < infty$.



The above proof shows that as long as $left|sum_{n = 1}^infty a_nright| < infty$, the result always holds. $a_n$ could be negative.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I' ll copy this Proof
    $endgroup$
    – Sudo su
    Jul 16 '15 at 1:36










  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38






  • 1




    $begingroup$
    How do you dervive $lim_{Ntoinfty}(sum_{n=1}^{N-1}A_n)/N=a$? (Not saying that it's not correct, but that that is a non-trivial result in its own right.)
    $endgroup$
    – Steven Stadnicki
    Jul 16 '15 at 1:49












  • $begingroup$
    @StevenStadnicki check en.wikipedia.org/wiki/Ces%C3%A0ro_summation. I would assume this result is so famous that every math student would know that.
    $endgroup$
    – Zhanxiong
    Jul 16 '15 at 1:50












  • $begingroup$
    "The above proof shows that as long as $|sum a_n|<infty$" I think you want to say "as long as $sum a_n$ converges"
    $endgroup$
    – zhw.
    Jul 16 '15 at 2:11
















2












2








2





$begingroup$

Let $A_n = a_1 + cdots + a_n$ be the partial sum of ${a_n}$, by summation by parts formula
$$sum_{n = 1}^N na_n = sum_{n = 1}^{N - 1}A_n(n - (n + 1)) + NA_N = NA_N - sum_{n = 1}^{N - 1}A_n.$$
Thus, based on the famous Cesaro's theorem
$$lim_{N to infty} frac{1}{N}sum_{n = 1}^N na_n = lim_{N to infty} A_N - lim_{N to infty} frac{sum_{n = 1}^{N - 1}A_n}{N} = a - a = 0$$
where $a = sum_{n = 1}^infty a_n < infty$.



The above proof shows that as long as $left|sum_{n = 1}^infty a_nright| < infty$, the result always holds. $a_n$ could be negative.






share|cite|improve this answer











$endgroup$



Let $A_n = a_1 + cdots + a_n$ be the partial sum of ${a_n}$, by summation by parts formula
$$sum_{n = 1}^N na_n = sum_{n = 1}^{N - 1}A_n(n - (n + 1)) + NA_N = NA_N - sum_{n = 1}^{N - 1}A_n.$$
Thus, based on the famous Cesaro's theorem
$$lim_{N to infty} frac{1}{N}sum_{n = 1}^N na_n = lim_{N to infty} A_N - lim_{N to infty} frac{sum_{n = 1}^{N - 1}A_n}{N} = a - a = 0$$
where $a = sum_{n = 1}^infty a_n < infty$.



The above proof shows that as long as $left|sum_{n = 1}^infty a_nright| < infty$, the result always holds. $a_n$ could be negative.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 3 '15 at 17:04

























answered Jul 16 '15 at 1:32









ZhanxiongZhanxiong

8,94911033




8,94911033












  • $begingroup$
    I' ll copy this Proof
    $endgroup$
    – Sudo su
    Jul 16 '15 at 1:36










  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38






  • 1




    $begingroup$
    How do you dervive $lim_{Ntoinfty}(sum_{n=1}^{N-1}A_n)/N=a$? (Not saying that it's not correct, but that that is a non-trivial result in its own right.)
    $endgroup$
    – Steven Stadnicki
    Jul 16 '15 at 1:49












  • $begingroup$
    @StevenStadnicki check en.wikipedia.org/wiki/Ces%C3%A0ro_summation. I would assume this result is so famous that every math student would know that.
    $endgroup$
    – Zhanxiong
    Jul 16 '15 at 1:50












  • $begingroup$
    "The above proof shows that as long as $|sum a_n|<infty$" I think you want to say "as long as $sum a_n$ converges"
    $endgroup$
    – zhw.
    Jul 16 '15 at 2:11




















  • $begingroup$
    I' ll copy this Proof
    $endgroup$
    – Sudo su
    Jul 16 '15 at 1:36










  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38






  • 1




    $begingroup$
    How do you dervive $lim_{Ntoinfty}(sum_{n=1}^{N-1}A_n)/N=a$? (Not saying that it's not correct, but that that is a non-trivial result in its own right.)
    $endgroup$
    – Steven Stadnicki
    Jul 16 '15 at 1:49












  • $begingroup$
    @StevenStadnicki check en.wikipedia.org/wiki/Ces%C3%A0ro_summation. I would assume this result is so famous that every math student would know that.
    $endgroup$
    – Zhanxiong
    Jul 16 '15 at 1:50












  • $begingroup$
    "The above proof shows that as long as $|sum a_n|<infty$" I think you want to say "as long as $sum a_n$ converges"
    $endgroup$
    – zhw.
    Jul 16 '15 at 2:11


















$begingroup$
I' ll copy this Proof
$endgroup$
– Sudo su
Jul 16 '15 at 1:36




$begingroup$
I' ll copy this Proof
$endgroup$
– Sudo su
Jul 16 '15 at 1:36












$begingroup$
I will try to reproduce. Thanks
$endgroup$
– El Chapo
Jul 16 '15 at 1:38




$begingroup$
I will try to reproduce. Thanks
$endgroup$
– El Chapo
Jul 16 '15 at 1:38




1




1




$begingroup$
How do you dervive $lim_{Ntoinfty}(sum_{n=1}^{N-1}A_n)/N=a$? (Not saying that it's not correct, but that that is a non-trivial result in its own right.)
$endgroup$
– Steven Stadnicki
Jul 16 '15 at 1:49






$begingroup$
How do you dervive $lim_{Ntoinfty}(sum_{n=1}^{N-1}A_n)/N=a$? (Not saying that it's not correct, but that that is a non-trivial result in its own right.)
$endgroup$
– Steven Stadnicki
Jul 16 '15 at 1:49














$begingroup$
@StevenStadnicki check en.wikipedia.org/wiki/Ces%C3%A0ro_summation. I would assume this result is so famous that every math student would know that.
$endgroup$
– Zhanxiong
Jul 16 '15 at 1:50






$begingroup$
@StevenStadnicki check en.wikipedia.org/wiki/Ces%C3%A0ro_summation. I would assume this result is so famous that every math student would know that.
$endgroup$
– Zhanxiong
Jul 16 '15 at 1:50














$begingroup$
"The above proof shows that as long as $|sum a_n|<infty$" I think you want to say "as long as $sum a_n$ converges"
$endgroup$
– zhw.
Jul 16 '15 at 2:11






$begingroup$
"The above proof shows that as long as $|sum a_n|<infty$" I think you want to say "as long as $sum a_n$ converges"
$endgroup$
– zhw.
Jul 16 '15 at 2:11













-1












$begingroup$

$dfrac{displaystyle sum_{n=1}^N na_n}{N}=dfrac{a_1+2a_2+3a_3+cdots +Na_N}{N}=dfrac{a_1+cdots+a_N}{N}+dfrac{N-1}{N}cdotdfrac{a_2+cdots+a_N}{N-1}+cdots + dfrac{a_N}{N}to 0+0+cdots + 0 = 0$ by the well-known Cesaro's theorem as $a_N to 0$ when $N to infty$ since the first series converges imply the general terms tend to $0$. The result still holds without $a_n geq 0$.






share|cite|improve this answer









$endgroup$









  • 5




    $begingroup$
    Each summand tends to zero, but the count of summands is arbitrarily large, so this proof doesn't stand
    $endgroup$
    – Norbert
    Jul 16 '15 at 1:32












  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38










  • $begingroup$
    If that was a valid argument we could also say that $lim_{ntoinfty}frac{1}{n} + frac{1}{n+1} + ldots + frac{1}{2n} = 0 + 0 + ldots + 0 = 0$ while the true result is $log(2)$.
    $endgroup$
    – Winther
    Jul 16 '15 at 2:22
















-1












$begingroup$

$dfrac{displaystyle sum_{n=1}^N na_n}{N}=dfrac{a_1+2a_2+3a_3+cdots +Na_N}{N}=dfrac{a_1+cdots+a_N}{N}+dfrac{N-1}{N}cdotdfrac{a_2+cdots+a_N}{N-1}+cdots + dfrac{a_N}{N}to 0+0+cdots + 0 = 0$ by the well-known Cesaro's theorem as $a_N to 0$ when $N to infty$ since the first series converges imply the general terms tend to $0$. The result still holds without $a_n geq 0$.






share|cite|improve this answer









$endgroup$









  • 5




    $begingroup$
    Each summand tends to zero, but the count of summands is arbitrarily large, so this proof doesn't stand
    $endgroup$
    – Norbert
    Jul 16 '15 at 1:32












  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38










  • $begingroup$
    If that was a valid argument we could also say that $lim_{ntoinfty}frac{1}{n} + frac{1}{n+1} + ldots + frac{1}{2n} = 0 + 0 + ldots + 0 = 0$ while the true result is $log(2)$.
    $endgroup$
    – Winther
    Jul 16 '15 at 2:22














-1












-1








-1





$begingroup$

$dfrac{displaystyle sum_{n=1}^N na_n}{N}=dfrac{a_1+2a_2+3a_3+cdots +Na_N}{N}=dfrac{a_1+cdots+a_N}{N}+dfrac{N-1}{N}cdotdfrac{a_2+cdots+a_N}{N-1}+cdots + dfrac{a_N}{N}to 0+0+cdots + 0 = 0$ by the well-known Cesaro's theorem as $a_N to 0$ when $N to infty$ since the first series converges imply the general terms tend to $0$. The result still holds without $a_n geq 0$.






share|cite|improve this answer









$endgroup$



$dfrac{displaystyle sum_{n=1}^N na_n}{N}=dfrac{a_1+2a_2+3a_3+cdots +Na_N}{N}=dfrac{a_1+cdots+a_N}{N}+dfrac{N-1}{N}cdotdfrac{a_2+cdots+a_N}{N-1}+cdots + dfrac{a_N}{N}to 0+0+cdots + 0 = 0$ by the well-known Cesaro's theorem as $a_N to 0$ when $N to infty$ since the first series converges imply the general terms tend to $0$. The result still holds without $a_n geq 0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jul 16 '15 at 1:28









DeepSeaDeepSea

71.4k54488




71.4k54488








  • 5




    $begingroup$
    Each summand tends to zero, but the count of summands is arbitrarily large, so this proof doesn't stand
    $endgroup$
    – Norbert
    Jul 16 '15 at 1:32












  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38










  • $begingroup$
    If that was a valid argument we could also say that $lim_{ntoinfty}frac{1}{n} + frac{1}{n+1} + ldots + frac{1}{2n} = 0 + 0 + ldots + 0 = 0$ while the true result is $log(2)$.
    $endgroup$
    – Winther
    Jul 16 '15 at 2:22














  • 5




    $begingroup$
    Each summand tends to zero, but the count of summands is arbitrarily large, so this proof doesn't stand
    $endgroup$
    – Norbert
    Jul 16 '15 at 1:32












  • $begingroup$
    I will try to reproduce. Thanks
    $endgroup$
    – El Chapo
    Jul 16 '15 at 1:38










  • $begingroup$
    If that was a valid argument we could also say that $lim_{ntoinfty}frac{1}{n} + frac{1}{n+1} + ldots + frac{1}{2n} = 0 + 0 + ldots + 0 = 0$ while the true result is $log(2)$.
    $endgroup$
    – Winther
    Jul 16 '15 at 2:22








5




5




$begingroup$
Each summand tends to zero, but the count of summands is arbitrarily large, so this proof doesn't stand
$endgroup$
– Norbert
Jul 16 '15 at 1:32






$begingroup$
Each summand tends to zero, but the count of summands is arbitrarily large, so this proof doesn't stand
$endgroup$
– Norbert
Jul 16 '15 at 1:32














$begingroup$
I will try to reproduce. Thanks
$endgroup$
– El Chapo
Jul 16 '15 at 1:38




$begingroup$
I will try to reproduce. Thanks
$endgroup$
– El Chapo
Jul 16 '15 at 1:38












$begingroup$
If that was a valid argument we could also say that $lim_{ntoinfty}frac{1}{n} + frac{1}{n+1} + ldots + frac{1}{2n} = 0 + 0 + ldots + 0 = 0$ while the true result is $log(2)$.
$endgroup$
– Winther
Jul 16 '15 at 2:22




$begingroup$
If that was a valid argument we could also say that $lim_{ntoinfty}frac{1}{n} + frac{1}{n+1} + ldots + frac{1}{2n} = 0 + 0 + ldots + 0 = 0$ while the true result is $log(2)$.
$endgroup$
– Winther
Jul 16 '15 at 2:22


















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