How does this infinite series $1-frac14+frac17-frac110+cdots$ simplify to an integral $int_0^1fracdx1+x^3$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)When can a sum and integral be interchanged?Infinite Series $1+frac12-frac23+frac14+frac15-frac26+cdots$does the infinite series $sum^infty_n=1 (-1)^n frac log(n)n$ converge?Simplify Infinite Series Involving Gamma Function $Gamma$How to simplify this integral?Sum of an infinite series $(1 - frac 12) + (frac 12 - frac 13) + cdots$ - not geometric series?How to find the value of this integral: $int_0^nfrac 1x dx$Does this integral $int_0^infty fracdx(1+e^x)(a+x)$ have a closed form?Does this infinite series converge or diverge?Value of this convergent series: $frac13!+frac25!+frac37!+frac49!+cdots$Calculating a series for$int_0^1 fracx^p-11+x^q dx$ by interchanging sum with integral
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How does this infinite series $1-frac14+frac17-frac110+cdots$ simplify to an integral $int_0^1fracdx1+x^3$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)When can a sum and integral be interchanged?Infinite Series $1+frac12-frac23+frac14+frac15-frac26+cdots$does the infinite series $sum^infty_n=1 (-1)^n frac log(n)n$ converge?Simplify Infinite Series Involving Gamma Function $Gamma$How to simplify this integral?Sum of an infinite series $(1 - frac 12) + (frac 12 - frac 13) + cdots$ - not geometric series?How to find the value of this integral: $int_0^nfrac 1x dx$Does this integral $int_0^infty fracdx(1+e^x)(a+x)$ have a closed form?Does this infinite series converge or diverge?Value of this convergent series: $frac13!+frac25!+frac37!+frac49!+cdots$Calculating a series for$int_0^1 fracx^p-11+x^q dx$ by interchanging sum with integral
$begingroup$
How does the infinite series below simplify to that integral?
$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$
I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.
integration sequences-and-series power-series
edited 3 hours ago
HAMIDINE SOUMARE
2,570417
asked Apr 13 at 16:03
ShreeShree
134
$endgroup$
add a comment |
$begingroup$
How does the infinite series below simplify to that integral?
$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$
I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.
integration sequences-and-series power-series
edited 3 hours ago
HAMIDINE SOUMARE
2,570417
asked Apr 13 at 16:03
ShreeShree
134
$endgroup$
1
$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
Apr 13 at 16:05
$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
Apr 13 at 16:11
add a comment |
$begingroup$
How does the infinite series below simplify to that integral?
$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$
I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.
integration sequences-and-series power-series
edited 3 hours ago
HAMIDINE SOUMARE
2,570417
asked Apr 13 at 16:03
ShreeShree
134
$endgroup$
How does the infinite series below simplify to that integral?
$$1-frac14+frac17-frac110+cdots=int_0^1fracdx1+x^3$$
I thought of simplifying the series to the sum to infinity of $frac16n-5 - frac16n-2$, but this did not help.
integration sequences-and-series power-series
integration sequences-and-series power-series
edited 3 hours ago
HAMIDINE SOUMARE
2,570417
asked Apr 13 at 16:03
ShreeShree
134
edited 3 hours ago
HAMIDINE SOUMARE
2,570417
asked Apr 13 at 16:03
ShreeShree
134
edited 3 hours ago
HAMIDINE SOUMARE
2,570417
edited 3 hours ago
HAMIDINE SOUMARE
2,570417
edited 3 hours ago
HAMIDINE SOUMARE
2,570417
2,570417
asked Apr 13 at 16:03
ShreeShree
134
asked Apr 13 at 16:03
ShreeShree
134
asked Apr 13 at 16:03
ShreeShree
134
134
1
$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
Apr 13 at 16:05
$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
Apr 13 at 16:11
add a comment |
1
$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
Apr 13 at 16:05
$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
Apr 13 at 16:11
1
1
$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
Apr 13 at 16:05
$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
Apr 13 at 16:05
$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
Apr 13 at 16:11
$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
Apr 13 at 16:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
$$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$
answered Apr 13 at 16:15
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,570417
$endgroup$
$begingroup$
For those who aren't clear on the first equality, it's a geometric series formula.
$endgroup$
– Alexis Olson
Apr 13 at 19:35
$begingroup$
Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
$endgroup$
– StammeringMathematician
16 hours ago
1
$begingroup$
@StammeringMathematician Check out this question.
$endgroup$
– Alexis Olson
8 hours ago
$begingroup$
@AlexisOlson Thanks a lot.
$endgroup$
– StammeringMathematician
6 hours ago
add a comment |
$begingroup$
for $x$ real, $ngeq 0$ integer
beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
endalign
For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign
Therefore,
beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
endalign
For $xin[0;1],ngeq 0$, integer,
beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
and,
beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
Therefore,
beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
Therefore,
beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign
answered Apr 13 at 16:47
FDPFDP
6,16211929
$endgroup$
$begingroup$
Why can we interchange $sum$ and $int$
$endgroup$
– StammeringMathematician
16 hours ago
$begingroup$
Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
$endgroup$
– FDP
15 hours ago
$begingroup$
Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
$endgroup$
– StammeringMathematician
15 hours ago
$begingroup$
My computation shows that series converges slowly.
$endgroup$
– FDP
14 hours ago
add a comment |
$begingroup$
If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign
answered Apr 13 at 16:15
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
$$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$
answered Apr 13 at 16:15
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,570417
$endgroup$
$begingroup$
For those who aren't clear on the first equality, it's a geometric series formula.
$endgroup$
– Alexis Olson
Apr 13 at 19:35
$begingroup$
Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
$endgroup$
– StammeringMathematician
16 hours ago
1
$begingroup$
@StammeringMathematician Check out this question.
$endgroup$
– Alexis Olson
8 hours ago
$begingroup$
@AlexisOlson Thanks a lot.
$endgroup$
– StammeringMathematician
6 hours ago
add a comment |
$begingroup$
$$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
$$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$
answered Apr 13 at 16:15
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,570417
$endgroup$
$begingroup$
For those who aren't clear on the first equality, it's a geometric series formula.
$endgroup$
– Alexis Olson
Apr 13 at 19:35
$begingroup$
Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
$endgroup$
– StammeringMathematician
16 hours ago
1
$begingroup$
@StammeringMathematician Check out this question.
$endgroup$
– Alexis Olson
8 hours ago
$begingroup$
@AlexisOlson Thanks a lot.
$endgroup$
– StammeringMathematician
6 hours ago
add a comment |
$begingroup$
$$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
$$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$
answered Apr 13 at 16:15
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,570417
$endgroup$
$$int_0^1fracdx1-(-x)^3=int_0^1sum_n=0^infty(-x)^3ndx=sum_n=0^infty(-1)^3nint_0^1x^3ndx$$
$$=sum_n=0^inftyfrac(-1)^3n3n+1= 1-frac14+frac17-frac110+cdots $$
answered Apr 13 at 16:15
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,570417
edited 18 hours ago
answered Apr 13 at 16:15
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,570417
answered Apr 13 at 16:15
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,570417
answered Apr 13 at 16:15
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,570417
2,570417
$begingroup$
For those who aren't clear on the first equality, it's a geometric series formula.
$endgroup$
– Alexis Olson
Apr 13 at 19:35
$begingroup$
Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
$endgroup$
– StammeringMathematician
16 hours ago
1
$begingroup$
@StammeringMathematician Check out this question.
$endgroup$
– Alexis Olson
8 hours ago
$begingroup$
@AlexisOlson Thanks a lot.
$endgroup$
– StammeringMathematician
6 hours ago
add a comment |
$begingroup$
For those who aren't clear on the first equality, it's a geometric series formula.
$endgroup$
– Alexis Olson
Apr 13 at 19:35
$begingroup$
Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
$endgroup$
– StammeringMathematician
16 hours ago
1
$begingroup$
@StammeringMathematician Check out this question.
$endgroup$
– Alexis Olson
8 hours ago
$begingroup$
@AlexisOlson Thanks a lot.
$endgroup$
– StammeringMathematician
6 hours ago
$begingroup$
For those who aren't clear on the first equality, it's a geometric series formula.
$endgroup$
– Alexis Olson
Apr 13 at 19:35
$begingroup$
For those who aren't clear on the first equality, it's a geometric series formula.
$endgroup$
– Alexis Olson
Apr 13 at 19:35
$begingroup$
Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
$endgroup$
– StammeringMathematician
16 hours ago
$begingroup$
Why can we interchange the sum and integration? what is the idea? thanks. @AlexisOlson If you are free, please respond. Thanks.
$endgroup$
– StammeringMathematician
16 hours ago
1
1
$begingroup$
@StammeringMathematician Check out this question.
$endgroup$
– Alexis Olson
8 hours ago
$begingroup$
@StammeringMathematician Check out this question.
$endgroup$
– Alexis Olson
8 hours ago
$begingroup$
@AlexisOlson Thanks a lot.
$endgroup$
– StammeringMathematician
6 hours ago
$begingroup$
@AlexisOlson Thanks a lot.
$endgroup$
– StammeringMathematician
6 hours ago
add a comment |
$begingroup$
for $x$ real, $ngeq 0$ integer
beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
endalign
For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign
Therefore,
beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
endalign
For $xin[0;1],ngeq 0$, integer,
beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
and,
beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
Therefore,
beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
Therefore,
beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign
answered Apr 13 at 16:47
FDPFDP
6,16211929
$endgroup$
$begingroup$
Why can we interchange $sum$ and $int$
$endgroup$
– StammeringMathematician
16 hours ago
$begingroup$
Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
$endgroup$
– FDP
15 hours ago
$begingroup$
Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
$endgroup$
– StammeringMathematician
15 hours ago
$begingroup$
My computation shows that series converges slowly.
$endgroup$
– FDP
14 hours ago
add a comment |
$begingroup$
for $x$ real, $ngeq 0$ integer
beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
endalign
For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign
Therefore,
beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
endalign
For $xin[0;1],ngeq 0$, integer,
beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
and,
beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
Therefore,
beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
Therefore,
beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign
answered Apr 13 at 16:47
FDPFDP
6,16211929
$endgroup$
$begingroup$
Why can we interchange $sum$ and $int$
$endgroup$
– StammeringMathematician
16 hours ago
$begingroup$
Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
$endgroup$
– FDP
15 hours ago
$begingroup$
Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
$endgroup$
– StammeringMathematician
15 hours ago
$begingroup$
My computation shows that series converges slowly.
$endgroup$
– FDP
14 hours ago
add a comment |
$begingroup$
for $x$ real, $ngeq 0$ integer
beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
endalign
For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign
Therefore,
beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
endalign
For $xin[0;1],ngeq 0$, integer,
beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
and,
beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
Therefore,
beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
Therefore,
beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign
answered Apr 13 at 16:47
FDPFDP
6,16211929
$endgroup$
for $x$ real, $ngeq 0$ integer
beginalignfrac11+x^3&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11-(-x^3)\
&=frac1-(-x^3)^n+11-(-x^3)+frac(-x^3)^n+11+x^3\
endalign
For $xneq 1$, $ngeq 0$ integer, beginalignsum_k=0^n x^k=frac1-x^n+11-xendalign
Therefore,
beginalignint_0^1 frac11+x^3,dx&=int_0^1 left(sum_k=0^n (-x^3)^kright),dx+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n left(int_0^1 (-x^3)^k,dxright)+int_0^1 frac(-x^3)^n+11+x^3,dx\
&=sum_k=0^n frac(-1)^k3k+1+int_0^1 frac(-x^3)^n+11+x^3,dx\
endalign
For $xin[0;1],ngeq 0$, integer,
beginalignfracx^3(n+1)1+x^3leq x^3(n+1)endalign
and,
beginalignint_0^1 x^3(n+1),dx=frac13n+4endalign
Therefore,
beginalignleft|int_0^1 frac(-x^3)^n+11+x^3,dxright|leq frac13n+4endalign
beginalignleft|int_0^1 frac11+x^3,dx-sum_k=0^n frac(-1)^k3k+1right|leq frac13n+4endalign
Therefore,
beginalignboxedint_0^1 frac11+x^3,dx=sum_k=0^infty frac(-1)^k3k+1endalign
answered Apr 13 at 16:47
FDPFDP
6,16211929
answered Apr 13 at 16:47
FDPFDP
6,16211929
answered Apr 13 at 16:47
FDPFDP
6,16211929
answered Apr 13 at 16:47
FDPFDP
6,16211929
6,16211929
$begingroup$
Why can we interchange $sum$ and $int$
$endgroup$
– StammeringMathematician
16 hours ago
$begingroup$
Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
$endgroup$
– FDP
15 hours ago
$begingroup$
Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
$endgroup$
– StammeringMathematician
15 hours ago
$begingroup$
My computation shows that series converges slowly.
$endgroup$
– FDP
14 hours ago
add a comment |
$begingroup$
Why can we interchange $sum$ and $int$
$endgroup$
– StammeringMathematician
16 hours ago
$begingroup$
Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
$endgroup$
– FDP
15 hours ago
$begingroup$
Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
$endgroup$
– StammeringMathematician
15 hours ago
$begingroup$
My computation shows that series converges slowly.
$endgroup$
– FDP
14 hours ago
$begingroup$
Why can we interchange $sum$ and $int$
$endgroup$
– StammeringMathematician
16 hours ago
$begingroup$
Why can we interchange $sum$ and $int$
$endgroup$
– StammeringMathematician
16 hours ago
$begingroup$
Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
$endgroup$
– FDP
15 hours ago
$begingroup$
Only finite sums (that is, not series) are used. You can interchange sum and integral signs when the sum involves finite number of terms.
$endgroup$
– FDP
15 hours ago
$begingroup$
Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
$endgroup$
– StammeringMathematician
15 hours ago
$begingroup$
Thanks, I got it. Actually I was confused with the answer above yours one as there summation is over infinite terms.
$endgroup$
– StammeringMathematician
15 hours ago
$begingroup$
My computation shows that series converges slowly.
$endgroup$
– FDP
14 hours ago
$begingroup$
My computation shows that series converges slowly.
$endgroup$
– FDP
14 hours ago
add a comment |
$begingroup$
If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign
answered Apr 13 at 16:15
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
add a comment |
$begingroup$
If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign
answered Apr 13 at 16:15
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
add a comment |
$begingroup$
If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign
answered Apr 13 at 16:15
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
If $lvert xrvert<1$, let$$f(x)=sum_n=0^inftyfracx^3n+13n+1.$$Then $$f'(x)=sum_n=0^infty x^3n=frac11-x^3.$$Thereforebeginalign1-frac14+frac17-frac110+cdots&=lim_xto1f(x)\&=int_0^1f'(x),mathrm dx\&=int_0^1frac11-x^3,mathrm dx.endalign
answered Apr 13 at 16:15
José Carlos SantosJosé Carlos Santos
175k24134243
answered Apr 13 at 16:15
José Carlos SantosJosé Carlos Santos
175k24134243
answered Apr 13 at 16:15
José Carlos SantosJosé Carlos Santos
175k24134243
answered Apr 13 at 16:15
José Carlos SantosJosé Carlos Santos
175k24134243
175k24134243
add a comment |
add a comment |
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$begingroup$
en.wikipedia.org/wiki/Binomial_series $$(1+x^3)^-1$$
$endgroup$
– lab bhattacharjee
Apr 13 at 16:05
$begingroup$
It can be obtained by integrating term by term the infinite series expansion of $(1+x^3)^-1$.
$endgroup$
– StubbornAtom
Apr 13 at 16:11