Mixing
The sum of an infinite series with integral
$begingroup$
$1+dfrac{1}{9}+dfrac{1}{45}+dfrac{1}{189}+dfrac{1}{729}+dots=sumlimits_{n=1}^infty dfrac{1}{(2n-1)cdot 3^{n-1}}$
I got:
$sumlimits_{n=1}^infty dfrac{1}{(2n-1)cdot 3^{n-1}}=sumlimits_{n=1}^infty dfrac{intlimits_0^1 x^{2n-2},dx}{3^{n-1}}=dots$
And no idea how to impove.
Thx!
sequences-and-series integration
asked Sep 18 '13 at 11:48
Evgeny EgorovEvgeny Egorov
293212
$endgroup$
add a comment |
$begingroup$
$1+dfrac{1}{9}+dfrac{1}{45}+dfrac{1}{189}+dfrac{1}{729}+dots=sumlimits_{n=1}^infty dfrac{1}{(2n-1)cdot 3^{n-1}}$
I got:
$sumlimits_{n=1}^infty dfrac{1}{(2n-1)cdot 3^{n-1}}=sumlimits_{n=1}^infty dfrac{intlimits_0^1 x^{2n-2},dx}{3^{n-1}}=dots$
And no idea how to impove.
Thx!
sequences-and-series integration
asked Sep 18 '13 at 11:48
Evgeny EgorovEvgeny Egorov
293212
$endgroup$
6
$begingroup$
You could try and interchange integration and summation in $$sum_{n=1}^infty int_0^1 left(frac{x^2}{3}right)^{n-1},dx$$
$endgroup$
– Daniel Fischer
Sep 18 '13 at 11:51
$begingroup$
How do you express that series to integral? I've used riemann sum.
$endgroup$
– Chandra Napitupulu
Jan 27 at 2:08
add a comment |
$begingroup$
$1+dfrac{1}{9}+dfrac{1}{45}+dfrac{1}{189}+dfrac{1}{729}+dots=sumlimits_{n=1}^infty dfrac{1}{(2n-1)cdot 3^{n-1}}$
I got:
$sumlimits_{n=1}^infty dfrac{1}{(2n-1)cdot 3^{n-1}}=sumlimits_{n=1}^infty dfrac{intlimits_0^1 x^{2n-2},dx}{3^{n-1}}=dots$
And no idea how to impove.
Thx!
sequences-and-series integration
asked Sep 18 '13 at 11:48
Evgeny EgorovEvgeny Egorov
293212
$endgroup$
$1+dfrac{1}{9}+dfrac{1}{45}+dfrac{1}{189}+dfrac{1}{729}+dots=sumlimits_{n=1}^infty dfrac{1}{(2n-1)cdot 3^{n-1}}$
I got:
$sumlimits_{n=1}^infty dfrac{1}{(2n-1)cdot 3^{n-1}}=sumlimits_{n=1}^infty dfrac{intlimits_0^1 x^{2n-2},dx}{3^{n-1}}=dots$
And no idea how to impove.
Thx!
sequences-and-series integration
sequences-and-series integration
asked Sep 18 '13 at 11:48
Evgeny EgorovEvgeny Egorov
293212
asked Sep 18 '13 at 11:48
Evgeny EgorovEvgeny Egorov
293212
asked Sep 18 '13 at 11:48
Evgeny EgorovEvgeny Egorov
293212
asked Sep 18 '13 at 11:48
Evgeny EgorovEvgeny Egorov
293212
asked Sep 18 '13 at 11:48
Evgeny EgorovEvgeny Egorov
293212
293212
6
$begingroup$
You could try and interchange integration and summation in $$sum_{n=1}^infty int_0^1 left(frac{x^2}{3}right)^{n-1},dx$$
$endgroup$
– Daniel Fischer
Sep 18 '13 at 11:51
$begingroup$
How do you express that series to integral? I've used riemann sum.
$endgroup$
– Chandra Napitupulu
Jan 27 at 2:08
add a comment |
6
$begingroup$
You could try and interchange integration and summation in $$sum_{n=1}^infty int_0^1 left(frac{x^2}{3}right)^{n-1},dx$$
$endgroup$
– Daniel Fischer
Sep 18 '13 at 11:51
$begingroup$
How do you express that series to integral? I've used riemann sum.
$endgroup$
– Chandra Napitupulu
Jan 27 at 2:08
6
6
$begingroup$
You could try and interchange integration and summation in $$sum_{n=1}^infty int_0^1 left(frac{x^2}{3}right)^{n-1},dx$$
$endgroup$
– Daniel Fischer
Sep 18 '13 at 11:51
$begingroup$
You could try and interchange integration and summation in $$sum_{n=1}^infty int_0^1 left(frac{x^2}{3}right)^{n-1},dx$$
$endgroup$
– Daniel Fischer
Sep 18 '13 at 11:51
$begingroup$
How do you express that series to integral? I've used riemann sum.
$endgroup$
– Chandra Napitupulu
Jan 27 at 2:08
$begingroup$
How do you express that series to integral? I've used riemann sum.
$endgroup$
– Chandra Napitupulu
Jan 27 at 2:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
After writing $frac{1}{2n-1}$ as an integral, you have the series
$$sum_{n=1}^infty int_0^1 left(frac{x^2}{3}right)^{n-1},dx.$$
Since the geometric series
$$sum_{k=0}^infty left(frac{x^2}{3}right)^k$$
converges uniformly on the interval $[0,1]$, we can interchange summation and integration, and obtain
$$sum_{n=1}^infty frac{1}{(2n-1)3^{(n-1)}} = int_0^1 frac{3}{3-x^2},dx.$$
The integral can be evaluated in different ways. With partial fraction decomposition
$$frac{3}{3-x^2} = frac{sqrt{3}}{2}left(frac{1}{sqrt{3}-x} + frac{1}{sqrt{3}+x}right)$$
we get
$$int_0^1 frac{3}{3-x^2},dx = frac{sqrt{3}}{2}left[log (sqrt{3}+x) - log (sqrt{3}-x)right]_0^1 = frac{sqrt{3}}{2}log frac{sqrt{3}+1}{sqrt{3}-1} = frac{sqrt{3}}{2}log (2+sqrt{3}).$$
answered Sep 18 '13 at 14:17
Daniel FischerDaniel Fischer
174k17168287
$endgroup$
$begingroup$
(+1) the question was bumped and I wrote an answer that didn't look like yours on the surface, but upon closer inspection, turned out to be pretty close.
$endgroup$
– robjohn♦
Jan 27 at 3:28
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
After writing $frac{1}{2n-1}$ as an integral, you have the series
$$sum_{n=1}^infty int_0^1 left(frac{x^2}{3}right)^{n-1},dx.$$
Since the geometric series
$$sum_{k=0}^infty left(frac{x^2}{3}right)^k$$
converges uniformly on the interval $[0,1]$, we can interchange summation and integration, and obtain
$$sum_{n=1}^infty frac{1}{(2n-1)3^{(n-1)}} = int_0^1 frac{3}{3-x^2},dx.$$
The integral can be evaluated in different ways. With partial fraction decomposition
$$frac{3}{3-x^2} = frac{sqrt{3}}{2}left(frac{1}{sqrt{3}-x} + frac{1}{sqrt{3}+x}right)$$
we get
$$int_0^1 frac{3}{3-x^2},dx = frac{sqrt{3}}{2}left[log (sqrt{3}+x) - log (sqrt{3}-x)right]_0^1 = frac{sqrt{3}}{2}log frac{sqrt{3}+1}{sqrt{3}-1} = frac{sqrt{3}}{2}log (2+sqrt{3}).$$
answered Sep 18 '13 at 14:17
Daniel FischerDaniel Fischer
174k17168287
$endgroup$
$begingroup$
(+1) the question was bumped and I wrote an answer that didn't look like yours on the surface, but upon closer inspection, turned out to be pretty close.
$endgroup$
– robjohn♦
Jan 27 at 3:28
add a comment |
$begingroup$
After writing $frac{1}{2n-1}$ as an integral, you have the series
$$sum_{n=1}^infty int_0^1 left(frac{x^2}{3}right)^{n-1},dx.$$
Since the geometric series
$$sum_{k=0}^infty left(frac{x^2}{3}right)^k$$
converges uniformly on the interval $[0,1]$, we can interchange summation and integration, and obtain
$$sum_{n=1}^infty frac{1}{(2n-1)3^{(n-1)}} = int_0^1 frac{3}{3-x^2},dx.$$
The integral can be evaluated in different ways. With partial fraction decomposition
$$frac{3}{3-x^2} = frac{sqrt{3}}{2}left(frac{1}{sqrt{3}-x} + frac{1}{sqrt{3}+x}right)$$
we get
$$int_0^1 frac{3}{3-x^2},dx = frac{sqrt{3}}{2}left[log (sqrt{3}+x) - log (sqrt{3}-x)right]_0^1 = frac{sqrt{3}}{2}log frac{sqrt{3}+1}{sqrt{3}-1} = frac{sqrt{3}}{2}log (2+sqrt{3}).$$
answered Sep 18 '13 at 14:17
Daniel FischerDaniel Fischer
174k17168287
$endgroup$
$begingroup$
(+1) the question was bumped and I wrote an answer that didn't look like yours on the surface, but upon closer inspection, turned out to be pretty close.
$endgroup$
– robjohn♦
Jan 27 at 3:28
add a comment |
$begingroup$
After writing $frac{1}{2n-1}$ as an integral, you have the series
$$sum_{n=1}^infty int_0^1 left(frac{x^2}{3}right)^{n-1},dx.$$
Since the geometric series
$$sum_{k=0}^infty left(frac{x^2}{3}right)^k$$
converges uniformly on the interval $[0,1]$, we can interchange summation and integration, and obtain
$$sum_{n=1}^infty frac{1}{(2n-1)3^{(n-1)}} = int_0^1 frac{3}{3-x^2},dx.$$
The integral can be evaluated in different ways. With partial fraction decomposition
$$frac{3}{3-x^2} = frac{sqrt{3}}{2}left(frac{1}{sqrt{3}-x} + frac{1}{sqrt{3}+x}right)$$
we get
$$int_0^1 frac{3}{3-x^2},dx = frac{sqrt{3}}{2}left[log (sqrt{3}+x) - log (sqrt{3}-x)right]_0^1 = frac{sqrt{3}}{2}log frac{sqrt{3}+1}{sqrt{3}-1} = frac{sqrt{3}}{2}log (2+sqrt{3}).$$
answered Sep 18 '13 at 14:17
Daniel FischerDaniel Fischer
174k17168287
$endgroup$
After writing $frac{1}{2n-1}$ as an integral, you have the series
$$sum_{n=1}^infty int_0^1 left(frac{x^2}{3}right)^{n-1},dx.$$
Since the geometric series
$$sum_{k=0}^infty left(frac{x^2}{3}right)^k$$
converges uniformly on the interval $[0,1]$, we can interchange summation and integration, and obtain
$$sum_{n=1}^infty frac{1}{(2n-1)3^{(n-1)}} = int_0^1 frac{3}{3-x^2},dx.$$
The integral can be evaluated in different ways. With partial fraction decomposition
$$frac{3}{3-x^2} = frac{sqrt{3}}{2}left(frac{1}{sqrt{3}-x} + frac{1}{sqrt{3}+x}right)$$
we get
$$int_0^1 frac{3}{3-x^2},dx = frac{sqrt{3}}{2}left[log (sqrt{3}+x) - log (sqrt{3}-x)right]_0^1 = frac{sqrt{3}}{2}log frac{sqrt{3}+1}{sqrt{3}-1} = frac{sqrt{3}}{2}log (2+sqrt{3}).$$
answered Sep 18 '13 at 14:17
Daniel FischerDaniel Fischer
174k17168287
answered Sep 18 '13 at 14:17
Daniel FischerDaniel Fischer
174k17168287
answered Sep 18 '13 at 14:17
Daniel FischerDaniel Fischer
174k17168287
answered Sep 18 '13 at 14:17
Daniel FischerDaniel Fischer
174k17168287
174k17168287
$begingroup$
(+1) the question was bumped and I wrote an answer that didn't look like yours on the surface, but upon closer inspection, turned out to be pretty close.
$endgroup$
– robjohn♦
Jan 27 at 3:28
add a comment |
$begingroup$
(+1) the question was bumped and I wrote an answer that didn't look like yours on the surface, but upon closer inspection, turned out to be pretty close.
$endgroup$
– robjohn♦
Jan 27 at 3:28
$begingroup$
(+1) the question was bumped and I wrote an answer that didn't look like yours on the surface, but upon closer inspection, turned out to be pretty close.
$endgroup$
– robjohn♦
Jan 27 at 3:28
$begingroup$
(+1) the question was bumped and I wrote an answer that didn't look like yours on the surface, but upon closer inspection, turned out to be pretty close.
$endgroup$
– robjohn♦
Jan 27 at 3:28
add a comment |
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$begingroup$
You could try and interchange integration and summation in $$sum_{n=1}^infty int_0^1 left(frac{x^2}{3}right)^{n-1},dx$$
$endgroup$
– Daniel Fischer
Sep 18 '13 at 11:51
$begingroup$
How do you express that series to integral? I've used riemann sum.
$endgroup$
– Chandra Napitupulu
Jan 27 at 2:08