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Limit of $frac1ln nsum_j=1^nfracx_jj$ in two different cases.
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I have two related problems:
(1) For a sequence $x_n$ with $lim_n to infty x_n = X$, show that $frac1ln nsum_j=1^nfracx_jj$ converges to $X$.
(2) For a sequence $x_n$ with convergent arithmetic means, $lim_n to infty frac1n sum_j=1^n x_j = X$, show that $frac1ln nsum_j=1^nfracx_jj$ converges to $X$.
Attempt:
For (1) I used the Cesaro Stolz theorem:
$$lim_n to inftyfrac1ln nsum_j=1^nfracx_jj = lim_n to inftyfracfracx_n+1n+1ln(n+1)-ln(n) = lim_n to inftyfracx_n+1ln(1+1/n)^n+1 = fracXln(e) = X$$
For (2) I know if all $x_n > 0$ then $frac1ln nsum_j=1^nfracx_jj < fracnln nfrac1nsum_j=1^n x_j$ but this does not seem to help. Also it is not given that all $x_n >0$
Thank you for any assistance.
sequences-and-series convergence
asked Jul 24 at 21:03
user28763
523
add a comment |Â
up vote
4
down vote
favorite
I have two related problems:
(1) For a sequence $x_n$ with $lim_n to infty x_n = X$, show that $frac1ln nsum_j=1^nfracx_jj$ converges to $X$.
(2) For a sequence $x_n$ with convergent arithmetic means, $lim_n to infty frac1n sum_j=1^n x_j = X$, show that $frac1ln nsum_j=1^nfracx_jj$ converges to $X$.
Attempt:
For (1) I used the Cesaro Stolz theorem:
$$lim_n to inftyfrac1ln nsum_j=1^nfracx_jj = lim_n to inftyfracfracx_n+1n+1ln(n+1)-ln(n) = lim_n to inftyfracx_n+1ln(1+1/n)^n+1 = fracXln(e) = X$$
For (2) I know if all $x_n > 0$ then $frac1ln nsum_j=1^nfracx_jj < fracnln nfrac1nsum_j=1^n x_j$ but this does not seem to help. Also it is not given that all $x_n >0$
Thank you for any assistance.
sequences-and-series convergence
asked Jul 24 at 21:03
user28763
523
(1) follows from (2), by the way. To prove (2), do you know summation by parts? Abel's formula converting the sum into an integral is particularly convenient here.
– Daniel Fischer♦
Jul 24 at 21:27
@DanielFischer: Thanks I get sum = $fracS_nn ln n + frac1ln nsum_j=1^n S_j(1/j - 1/(j+1))$ with $S_n$ partial sum of the $sum x_j$ series. I see that the first term goes to $0$. I'll try working with the second sum as you are suggesting.
– user28763
Jul 24 at 21:58
Let $S(t) = sum_j leqslant t x_j$. The arithmetic means converging to $X$ means $S(t) = Xt + o(t)$. Abel's formula gives $$sum_j leqslant t fracx_jj = fracS(t)t + int_1^t fracS(u)u^2,du = X + o(1) + Xlog t + int_1^t fraco(u)u^2,du = Xlog t + o(log t),.$$
– Daniel Fischer♦
Jul 25 at 8:09
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have two related problems:
(1) For a sequence $x_n$ with $lim_n to infty x_n = X$, show that $frac1ln nsum_j=1^nfracx_jj$ converges to $X$.
(2) For a sequence $x_n$ with convergent arithmetic means, $lim_n to infty frac1n sum_j=1^n x_j = X$, show that $frac1ln nsum_j=1^nfracx_jj$ converges to $X$.
Attempt:
For (1) I used the Cesaro Stolz theorem:
$$lim_n to inftyfrac1ln nsum_j=1^nfracx_jj = lim_n to inftyfracfracx_n+1n+1ln(n+1)-ln(n) = lim_n to inftyfracx_n+1ln(1+1/n)^n+1 = fracXln(e) = X$$
For (2) I know if all $x_n > 0$ then $frac1ln nsum_j=1^nfracx_jj < fracnln nfrac1nsum_j=1^n x_j$ but this does not seem to help. Also it is not given that all $x_n >0$
Thank you for any assistance.
sequences-and-series convergence
asked Jul 24 at 21:03
user28763
523
I have two related problems:
(1) For a sequence $x_n$ with $lim_n to infty x_n = X$, show that $frac1ln nsum_j=1^nfracx_jj$ converges to $X$.
(2) For a sequence $x_n$ with convergent arithmetic means, $lim_n to infty frac1n sum_j=1^n x_j = X$, show that $frac1ln nsum_j=1^nfracx_jj$ converges to $X$.
Attempt:
For (1) I used the Cesaro Stolz theorem:
$$lim_n to inftyfrac1ln nsum_j=1^nfracx_jj = lim_n to inftyfracfracx_n+1n+1ln(n+1)-ln(n) = lim_n to inftyfracx_n+1ln(1+1/n)^n+1 = fracXln(e) = X$$
For (2) I know if all $x_n > 0$ then $frac1ln nsum_j=1^nfracx_jj < fracnln nfrac1nsum_j=1^n x_j$ but this does not seem to help. Also it is not given that all $x_n >0$
Thank you for any assistance.
sequences-and-series convergence
asked Jul 24 at 21:03
user28763
523
asked Jul 24 at 21:03
user28763
523
asked Jul 24 at 21:03
user28763
523
asked Jul 24 at 21:03
user28763
523
523
(1) follows from (2), by the way. To prove (2), do you know summation by parts? Abel's formula converting the sum into an integral is particularly convenient here.
– Daniel Fischer♦
Jul 24 at 21:27
@DanielFischer: Thanks I get sum = $fracS_nn ln n + frac1ln nsum_j=1^n S_j(1/j - 1/(j+1))$ with $S_n$ partial sum of the $sum x_j$ series. I see that the first term goes to $0$. I'll try working with the second sum as you are suggesting.
– user28763
Jul 24 at 21:58
Let $S(t) = sum_j leqslant t x_j$. The arithmetic means converging to $X$ means $S(t) = Xt + o(t)$. Abel's formula gives $$sum_j leqslant t fracx_jj = fracS(t)t + int_1^t fracS(u)u^2,du = X + o(1) + Xlog t + int_1^t fraco(u)u^2,du = Xlog t + o(log t),.$$
– Daniel Fischer♦
Jul 25 at 8:09
add a comment |Â
(1) follows from (2), by the way. To prove (2), do you know summation by parts? Abel's formula converting the sum into an integral is particularly convenient here.
– Daniel Fischer♦
Jul 24 at 21:27
@DanielFischer: Thanks I get sum = $fracS_nn ln n + frac1ln nsum_j=1^n S_j(1/j - 1/(j+1))$ with $S_n$ partial sum of the $sum x_j$ series. I see that the first term goes to $0$. I'll try working with the second sum as you are suggesting.
– user28763
Jul 24 at 21:58
Let $S(t) = sum_j leqslant t x_j$. The arithmetic means converging to $X$ means $S(t) = Xt + o(t)$. Abel's formula gives $$sum_j leqslant t fracx_jj = fracS(t)t + int_1^t fracS(u)u^2,du = X + o(1) + Xlog t + int_1^t fraco(u)u^2,du = Xlog t + o(log t),.$$
– Daniel Fischer♦
Jul 25 at 8:09
(1) follows from (2), by the way. To prove (2), do you know summation by parts? Abel's formula converting the sum into an integral is particularly convenient here.
– Daniel Fischer♦
Jul 24 at 21:27
(1) follows from (2), by the way. To prove (2), do you know summation by parts? Abel's formula converting the sum into an integral is particularly convenient here.
– Daniel Fischer♦
Jul 24 at 21:27
@DanielFischer: Thanks I get sum = $fracS_nn ln n + frac1ln nsum_j=1^n S_j(1/j - 1/(j+1))$ with $S_n$ partial sum of the $sum x_j$ series. I see that the first term goes to $0$. I'll try working with the second sum as you are suggesting.
– user28763
Jul 24 at 21:58
@DanielFischer: Thanks I get sum = $fracS_nn ln n + frac1ln nsum_j=1^n S_j(1/j - 1/(j+1))$ with $S_n$ partial sum of the $sum x_j$ series. I see that the first term goes to $0$. I'll try working with the second sum as you are suggesting.
– user28763
Jul 24 at 21:58
Let $S(t) = sum_j leqslant t x_j$. The arithmetic means converging to $X$ means $S(t) = Xt + o(t)$. Abel's formula gives $$sum_j leqslant t fracx_jj = fracS(t)t + int_1^t fracS(u)u^2,du = X + o(1) + Xlog t + int_1^t fraco(u)u^2,du = Xlog t + o(log t),.$$
– Daniel Fischer♦
Jul 25 at 8:09
Let $S(t) = sum_j leqslant t x_j$. The arithmetic means converging to $X$ means $S(t) = Xt + o(t)$. Abel's formula gives $$sum_j leqslant t fracx_jj = fracS(t)t + int_1^t fracS(u)u^2,du = X + o(1) + Xlog t + int_1^t fraco(u)u^2,du = Xlog t + o(log t),.$$
– Daniel Fischer♦
Jul 25 at 8:09
add a comment |Â
2 Answers
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Summation by parts gives:
beginalign
sum_j=1^n fracx_jj &= frac1n sum_j=1^n x_j - sum_j=0^n-1 left(sum_i=1^jx_iright)left(frac1j+1 - frac1jright) \
&= frac1n sum_j=1^n x_j - sum_j=0^n-1 left(frac1jsum_i=1^jx_iright)left(fracjj+1 - 1right)\
&= sum_j=0^n-1 frac1j+1left(frac1jsum_i=1^jx_iright) + frac1n sum_j=1^n x_j \
endalign
The assumption is $lim_ntoinfty frac1n sum_i=1^n x_i = X$ so also $lim_ntoinfty frac1n-1 sum_i=1^n-1 x_i = X$. The first part gives
$$X = lim_ntoinfty frac1ln nsum_j=1^n fracfrac1j-1 sum_i=1^j-1 x_ij = lim_ntoinfty frac1ln nsum_j=0^n-1 fracfrac1j sum_i=1^j x_ij+1 = lim_ntoinfty frac1ln nleft[sum_j=1^n fracx_jj - frac1n sum_j=1^n x_jright]$$
so
$$lim_ntoinfty frac1ln n sum_j=1^n fracx_jj = X + lim_ntoinfty frac1ln nleft(frac1n sum_j=1^n x_jright) = X$$
answered Jul 24 at 22:08
mechanodroid
22.2k52041
Thank you. This is what I was looking for.
– user28763
Aug 3 at 17:21
add a comment |Â
up vote
2
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Replacing $x_n$ by $x_n-X$ if necessary, we can assume that $X=0$. Let $s_n:=sum_i=1^nx_i$. Then there exists a sequence $left(delta_nright)_ngeqslant 1$ converging to $0$ such that $x_n=s_n-s_n-1=ndelta_n-left(n-1right)delta_n-1$ hence
$$
frac 1ln nsum_j=1^nfracx_jj=frac 1ln nsum_j=1^nfracjdelta_j-left(j-1right)delta_j-1j=frac 1ln nsum_j=1^nleft(delta_j-delta_j-1right)+frac 1ln nsum_j=1^nfracvarepsilon_j-1j.
$$
The first sum in the last expression is telescopic and goes to $0$ as $n$ goes to infinity; for the second one, fix an integer $N$ and observe that for all $ngeqslant N$,
$$
leftlvert frac 1ln nsum_j=1^nfracvarepsilon_j-1jrightrvert leqslant frac Nln n+frac 1ln nsum_j=1^nfrac1jsup_igeqslant Nleftlvert delta_irightrvert.
$$
Since the quantity $frac 1ln nsum_j=1^nfrac1j$ can be bounded independently on $n$ (say by $c$, we get
$$
limsup_nto +inftyleftlvert frac 1ln nsum_j=1^nfracvarepsilon_j-1jrightrvert leqslant csup_igeqslant Nleftlvert delta_irightrvert
$$
and since $N$ is arbitrary, we can conclude.
answered Jul 24 at 21:31
Davide Giraudo
121k15146249
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Summation by parts gives:
beginalign
sum_j=1^n fracx_jj &= frac1n sum_j=1^n x_j - sum_j=0^n-1 left(sum_i=1^jx_iright)left(frac1j+1 - frac1jright) \
&= frac1n sum_j=1^n x_j - sum_j=0^n-1 left(frac1jsum_i=1^jx_iright)left(fracjj+1 - 1right)\
&= sum_j=0^n-1 frac1j+1left(frac1jsum_i=1^jx_iright) + frac1n sum_j=1^n x_j \
endalign
The assumption is $lim_ntoinfty frac1n sum_i=1^n x_i = X$ so also $lim_ntoinfty frac1n-1 sum_i=1^n-1 x_i = X$. The first part gives
$$X = lim_ntoinfty frac1ln nsum_j=1^n fracfrac1j-1 sum_i=1^j-1 x_ij = lim_ntoinfty frac1ln nsum_j=0^n-1 fracfrac1j sum_i=1^j x_ij+1 = lim_ntoinfty frac1ln nleft[sum_j=1^n fracx_jj - frac1n sum_j=1^n x_jright]$$
so
$$lim_ntoinfty frac1ln n sum_j=1^n fracx_jj = X + lim_ntoinfty frac1ln nleft(frac1n sum_j=1^n x_jright) = X$$
answered Jul 24 at 22:08
mechanodroid
22.2k52041
Thank you. This is what I was looking for.
– user28763
Aug 3 at 17:21
add a comment |Â
up vote
1
down vote
accepted
Summation by parts gives:
beginalign
sum_j=1^n fracx_jj &= frac1n sum_j=1^n x_j - sum_j=0^n-1 left(sum_i=1^jx_iright)left(frac1j+1 - frac1jright) \
&= frac1n sum_j=1^n x_j - sum_j=0^n-1 left(frac1jsum_i=1^jx_iright)left(fracjj+1 - 1right)\
&= sum_j=0^n-1 frac1j+1left(frac1jsum_i=1^jx_iright) + frac1n sum_j=1^n x_j \
endalign
The assumption is $lim_ntoinfty frac1n sum_i=1^n x_i = X$ so also $lim_ntoinfty frac1n-1 sum_i=1^n-1 x_i = X$. The first part gives
$$X = lim_ntoinfty frac1ln nsum_j=1^n fracfrac1j-1 sum_i=1^j-1 x_ij = lim_ntoinfty frac1ln nsum_j=0^n-1 fracfrac1j sum_i=1^j x_ij+1 = lim_ntoinfty frac1ln nleft[sum_j=1^n fracx_jj - frac1n sum_j=1^n x_jright]$$
so
$$lim_ntoinfty frac1ln n sum_j=1^n fracx_jj = X + lim_ntoinfty frac1ln nleft(frac1n sum_j=1^n x_jright) = X$$
answered Jul 24 at 22:08
mechanodroid
22.2k52041
Thank you. This is what I was looking for.
– user28763
Aug 3 at 17:21
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Summation by parts gives:
beginalign
sum_j=1^n fracx_jj &= frac1n sum_j=1^n x_j - sum_j=0^n-1 left(sum_i=1^jx_iright)left(frac1j+1 - frac1jright) \
&= frac1n sum_j=1^n x_j - sum_j=0^n-1 left(frac1jsum_i=1^jx_iright)left(fracjj+1 - 1right)\
&= sum_j=0^n-1 frac1j+1left(frac1jsum_i=1^jx_iright) + frac1n sum_j=1^n x_j \
endalign
The assumption is $lim_ntoinfty frac1n sum_i=1^n x_i = X$ so also $lim_ntoinfty frac1n-1 sum_i=1^n-1 x_i = X$. The first part gives
$$X = lim_ntoinfty frac1ln nsum_j=1^n fracfrac1j-1 sum_i=1^j-1 x_ij = lim_ntoinfty frac1ln nsum_j=0^n-1 fracfrac1j sum_i=1^j x_ij+1 = lim_ntoinfty frac1ln nleft[sum_j=1^n fracx_jj - frac1n sum_j=1^n x_jright]$$
so
$$lim_ntoinfty frac1ln n sum_j=1^n fracx_jj = X + lim_ntoinfty frac1ln nleft(frac1n sum_j=1^n x_jright) = X$$
answered Jul 24 at 22:08
mechanodroid
22.2k52041
Summation by parts gives:
beginalign
sum_j=1^n fracx_jj &= frac1n sum_j=1^n x_j - sum_j=0^n-1 left(sum_i=1^jx_iright)left(frac1j+1 - frac1jright) \
&= frac1n sum_j=1^n x_j - sum_j=0^n-1 left(frac1jsum_i=1^jx_iright)left(fracjj+1 - 1right)\
&= sum_j=0^n-1 frac1j+1left(frac1jsum_i=1^jx_iright) + frac1n sum_j=1^n x_j \
endalign
The assumption is $lim_ntoinfty frac1n sum_i=1^n x_i = X$ so also $lim_ntoinfty frac1n-1 sum_i=1^n-1 x_i = X$. The first part gives
$$X = lim_ntoinfty frac1ln nsum_j=1^n fracfrac1j-1 sum_i=1^j-1 x_ij = lim_ntoinfty frac1ln nsum_j=0^n-1 fracfrac1j sum_i=1^j x_ij+1 = lim_ntoinfty frac1ln nleft[sum_j=1^n fracx_jj - frac1n sum_j=1^n x_jright]$$
so
$$lim_ntoinfty frac1ln n sum_j=1^n fracx_jj = X + lim_ntoinfty frac1ln nleft(frac1n sum_j=1^n x_jright) = X$$
answered Jul 24 at 22:08
mechanodroid
22.2k52041
answered Jul 24 at 22:08
mechanodroid
22.2k52041
answered Jul 24 at 22:08
mechanodroid
22.2k52041
answered Jul 24 at 22:08
mechanodroid
22.2k52041
22.2k52041
Thank you. This is what I was looking for.
– user28763
Aug 3 at 17:21
add a comment |Â
Thank you. This is what I was looking for.
– user28763
Aug 3 at 17:21
Thank you. This is what I was looking for.
– user28763
Aug 3 at 17:21
Thank you. This is what I was looking for.
– user28763
Aug 3 at 17:21
add a comment |Â
up vote
2
down vote
Replacing $x_n$ by $x_n-X$ if necessary, we can assume that $X=0$. Let $s_n:=sum_i=1^nx_i$. Then there exists a sequence $left(delta_nright)_ngeqslant 1$ converging to $0$ such that $x_n=s_n-s_n-1=ndelta_n-left(n-1right)delta_n-1$ hence
$$
frac 1ln nsum_j=1^nfracx_jj=frac 1ln nsum_j=1^nfracjdelta_j-left(j-1right)delta_j-1j=frac 1ln nsum_j=1^nleft(delta_j-delta_j-1right)+frac 1ln nsum_j=1^nfracvarepsilon_j-1j.
$$
The first sum in the last expression is telescopic and goes to $0$ as $n$ goes to infinity; for the second one, fix an integer $N$ and observe that for all $ngeqslant N$,
$$
leftlvert frac 1ln nsum_j=1^nfracvarepsilon_j-1jrightrvert leqslant frac Nln n+frac 1ln nsum_j=1^nfrac1jsup_igeqslant Nleftlvert delta_irightrvert.
$$
Since the quantity $frac 1ln nsum_j=1^nfrac1j$ can be bounded independently on $n$ (say by $c$, we get
$$
limsup_nto +inftyleftlvert frac 1ln nsum_j=1^nfracvarepsilon_j-1jrightrvert leqslant csup_igeqslant Nleftlvert delta_irightrvert
$$
and since $N$ is arbitrary, we can conclude.
answered Jul 24 at 21:31
Davide Giraudo
121k15146249
add a comment |Â
up vote
2
down vote
Replacing $x_n$ by $x_n-X$ if necessary, we can assume that $X=0$. Let $s_n:=sum_i=1^nx_i$. Then there exists a sequence $left(delta_nright)_ngeqslant 1$ converging to $0$ such that $x_n=s_n-s_n-1=ndelta_n-left(n-1right)delta_n-1$ hence
$$
frac 1ln nsum_j=1^nfracx_jj=frac 1ln nsum_j=1^nfracjdelta_j-left(j-1right)delta_j-1j=frac 1ln nsum_j=1^nleft(delta_j-delta_j-1right)+frac 1ln nsum_j=1^nfracvarepsilon_j-1j.
$$
The first sum in the last expression is telescopic and goes to $0$ as $n$ goes to infinity; for the second one, fix an integer $N$ and observe that for all $ngeqslant N$,
$$
leftlvert frac 1ln nsum_j=1^nfracvarepsilon_j-1jrightrvert leqslant frac Nln n+frac 1ln nsum_j=1^nfrac1jsup_igeqslant Nleftlvert delta_irightrvert.
$$
Since the quantity $frac 1ln nsum_j=1^nfrac1j$ can be bounded independently on $n$ (say by $c$, we get
$$
limsup_nto +inftyleftlvert frac 1ln nsum_j=1^nfracvarepsilon_j-1jrightrvert leqslant csup_igeqslant Nleftlvert delta_irightrvert
$$
and since $N$ is arbitrary, we can conclude.
answered Jul 24 at 21:31
Davide Giraudo
121k15146249
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Replacing $x_n$ by $x_n-X$ if necessary, we can assume that $X=0$. Let $s_n:=sum_i=1^nx_i$. Then there exists a sequence $left(delta_nright)_ngeqslant 1$ converging to $0$ such that $x_n=s_n-s_n-1=ndelta_n-left(n-1right)delta_n-1$ hence
$$
frac 1ln nsum_j=1^nfracx_jj=frac 1ln nsum_j=1^nfracjdelta_j-left(j-1right)delta_j-1j=frac 1ln nsum_j=1^nleft(delta_j-delta_j-1right)+frac 1ln nsum_j=1^nfracvarepsilon_j-1j.
$$
The first sum in the last expression is telescopic and goes to $0$ as $n$ goes to infinity; for the second one, fix an integer $N$ and observe that for all $ngeqslant N$,
$$
leftlvert frac 1ln nsum_j=1^nfracvarepsilon_j-1jrightrvert leqslant frac Nln n+frac 1ln nsum_j=1^nfrac1jsup_igeqslant Nleftlvert delta_irightrvert.
$$
Since the quantity $frac 1ln nsum_j=1^nfrac1j$ can be bounded independently on $n$ (say by $c$, we get
$$
limsup_nto +inftyleftlvert frac 1ln nsum_j=1^nfracvarepsilon_j-1jrightrvert leqslant csup_igeqslant Nleftlvert delta_irightrvert
$$
and since $N$ is arbitrary, we can conclude.
answered Jul 24 at 21:31
Davide Giraudo
121k15146249
Replacing $x_n$ by $x_n-X$ if necessary, we can assume that $X=0$. Let $s_n:=sum_i=1^nx_i$. Then there exists a sequence $left(delta_nright)_ngeqslant 1$ converging to $0$ such that $x_n=s_n-s_n-1=ndelta_n-left(n-1right)delta_n-1$ hence
$$
frac 1ln nsum_j=1^nfracx_jj=frac 1ln nsum_j=1^nfracjdelta_j-left(j-1right)delta_j-1j=frac 1ln nsum_j=1^nleft(delta_j-delta_j-1right)+frac 1ln nsum_j=1^nfracvarepsilon_j-1j.
$$
The first sum in the last expression is telescopic and goes to $0$ as $n$ goes to infinity; for the second one, fix an integer $N$ and observe that for all $ngeqslant N$,
$$
leftlvert frac 1ln nsum_j=1^nfracvarepsilon_j-1jrightrvert leqslant frac Nln n+frac 1ln nsum_j=1^nfrac1jsup_igeqslant Nleftlvert delta_irightrvert.
$$
Since the quantity $frac 1ln nsum_j=1^nfrac1j$ can be bounded independently on $n$ (say by $c$, we get
$$
limsup_nto +inftyleftlvert frac 1ln nsum_j=1^nfracvarepsilon_j-1jrightrvert leqslant csup_igeqslant Nleftlvert delta_irightrvert
$$
and since $N$ is arbitrary, we can conclude.
answered Jul 24 at 21:31
Davide Giraudo
121k15146249
answered Jul 24 at 21:31
Davide Giraudo
121k15146249
answered Jul 24 at 21:31
Davide Giraudo
121k15146249
answered Jul 24 at 21:31
Davide Giraudo
121k15146249
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(1) follows from (2), by the way. To prove (2), do you know summation by parts? Abel's formula converting the sum into an integral is particularly convenient here.
– Daniel Fischer♦
Jul 24 at 21:27
@DanielFischer: Thanks I get sum = $fracS_nn ln n + frac1ln nsum_j=1^n S_j(1/j - 1/(j+1))$ with $S_n$ partial sum of the $sum x_j$ series. I see that the first term goes to $0$. I'll try working with the second sum as you are suggesting.
– user28763
Jul 24 at 21:58
Let $S(t) = sum_j leqslant t x_j$. The arithmetic means converging to $X$ means $S(t) = Xt + o(t)$. Abel's formula gives $$sum_j leqslant t fracx_jj = fracS(t)t + int_1^t fracS(u)u^2,du = X + o(1) + Xlog t + int_1^t fraco(u)u^2,du = Xlog t + o(log t),.$$
– Daniel Fischer♦
Jul 25 at 8:09