Find the sum of $1-frac17+frac19-frac1{15}+frac1{17}-frac1{23}+frac1{25}-dots$
up vote
3
down vote
favorite
Find the sum of $$1-frac17+frac19-frac1{15}+frac1{17}-frac1{23}+frac1{25}-dots$$
a) $dfrac{pi}8(sqrt2-1)$
b) $dfrac{pi}4(sqrt2-1)$
c) $dfrac{pi}8(sqrt2+1)$
d) $dfrac{pi}4(sqrt2+1)$
I have tried a lot.. But i can't find any way to solve it, plz help me.. Advance thanks to u
summation power-series formal-power-series summation-method summation-by-parts
edited Nov 13 at 8:15
user10354138
6,294623
asked Nov 13 at 7:00
user532616
144
|
show 1 more comment
up vote
3
down vote
favorite
Find the sum of $$1-frac17+frac19-frac1{15}+frac1{17}-frac1{23}+frac1{25}-dots$$
a) $dfrac{pi}8(sqrt2-1)$
b) $dfrac{pi}4(sqrt2-1)$
c) $dfrac{pi}8(sqrt2+1)$
d) $dfrac{pi}4(sqrt2+1)$
I have tried a lot.. But i can't find any way to solve it, plz help me.. Advance thanks to u
summation power-series formal-power-series summation-method summation-by-parts
edited Nov 13 at 8:15
user10354138
6,294623
asked Nov 13 at 7:00
user532616
144
How did you narrow it down to those four possibilities?
– Lord Shark the Unknown
Nov 13 at 7:02
1
Maybe you can consider it a sum of two different infinite sums? What would the two expressions be?
– Matti P.
Nov 13 at 7:03
The sum equals $$int_0^1frac{1-x^6}{1-x^8},dx.$$
– Lord Shark the Unknown
Nov 13 at 7:03
But how this sum is equivalent to this integral.. Plz explain clearly
– user532616
Nov 13 at 7:06
expand the integrand as a power series and integrate term-by-term gives the original series
– user10354138
Nov 13 at 7:46
|
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Find the sum of $$1-frac17+frac19-frac1{15}+frac1{17}-frac1{23}+frac1{25}-dots$$
a) $dfrac{pi}8(sqrt2-1)$
b) $dfrac{pi}4(sqrt2-1)$
c) $dfrac{pi}8(sqrt2+1)$
d) $dfrac{pi}4(sqrt2+1)$
I have tried a lot.. But i can't find any way to solve it, plz help me.. Advance thanks to u
summation power-series formal-power-series summation-method summation-by-parts
edited Nov 13 at 8:15
user10354138
6,294623
asked Nov 13 at 7:00
user532616
144
Find the sum of $$1-frac17+frac19-frac1{15}+frac1{17}-frac1{23}+frac1{25}-dots$$
a) $dfrac{pi}8(sqrt2-1)$
b) $dfrac{pi}4(sqrt2-1)$
c) $dfrac{pi}8(sqrt2+1)$
d) $dfrac{pi}4(sqrt2+1)$
I have tried a lot.. But i can't find any way to solve it, plz help me.. Advance thanks to u
summation power-series formal-power-series summation-method summation-by-parts
summation power-series formal-power-series summation-method summation-by-parts
edited Nov 13 at 8:15
user10354138
6,294623
asked Nov 13 at 7:00
user532616
144
edited Nov 13 at 8:15
user10354138
6,294623
asked Nov 13 at 7:00
user532616
144
edited Nov 13 at 8:15
user10354138
6,294623
edited Nov 13 at 8:15
user10354138
6,294623
edited Nov 13 at 8:15
user10354138
6,294623
6,294623
asked Nov 13 at 7:00
user532616
144
asked Nov 13 at 7:00
user532616
144
asked Nov 13 at 7:00
user532616
144
144
How did you narrow it down to those four possibilities?
– Lord Shark the Unknown
Nov 13 at 7:02
1
Maybe you can consider it a sum of two different infinite sums? What would the two expressions be?
– Matti P.
Nov 13 at 7:03
The sum equals $$int_0^1frac{1-x^6}{1-x^8},dx.$$
– Lord Shark the Unknown
Nov 13 at 7:03
But how this sum is equivalent to this integral.. Plz explain clearly
– user532616
Nov 13 at 7:06
expand the integrand as a power series and integrate term-by-term gives the original series
– user10354138
Nov 13 at 7:46
|
show 1 more comment
How did you narrow it down to those four possibilities?
– Lord Shark the Unknown
Nov 13 at 7:02
1
Maybe you can consider it a sum of two different infinite sums? What would the two expressions be?
– Matti P.
Nov 13 at 7:03
The sum equals $$int_0^1frac{1-x^6}{1-x^8},dx.$$
– Lord Shark the Unknown
Nov 13 at 7:03
But how this sum is equivalent to this integral.. Plz explain clearly
– user532616
Nov 13 at 7:06
expand the integrand as a power series and integrate term-by-term gives the original series
– user10354138
Nov 13 at 7:46
How did you narrow it down to those four possibilities?
– Lord Shark the Unknown
Nov 13 at 7:02
How did you narrow it down to those four possibilities?
– Lord Shark the Unknown
Nov 13 at 7:02
1
1
Maybe you can consider it a sum of two different infinite sums? What would the two expressions be?
– Matti P.
Nov 13 at 7:03
Maybe you can consider it a sum of two different infinite sums? What would the two expressions be?
– Matti P.
Nov 13 at 7:03
The sum equals $$int_0^1frac{1-x^6}{1-x^8},dx.$$
– Lord Shark the Unknown
Nov 13 at 7:03
The sum equals $$int_0^1frac{1-x^6}{1-x^8},dx.$$
– Lord Shark the Unknown
Nov 13 at 7:03
But how this sum is equivalent to this integral.. Plz explain clearly
– user532616
Nov 13 at 7:06
But how this sum is equivalent to this integral.. Plz explain clearly
– user532616
Nov 13 at 7:06
expand the integrand as a power series and integrate term-by-term gives the original series
– user10354138
Nov 13 at 7:46
expand the integrand as a power series and integrate term-by-term gives the original series
– user10354138
Nov 13 at 7:46
|
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Alternating series test gives the sum is convergent, so we can bracket off
begin{align}
1-sum_{k=1}^inftyleft(frac1{8k-1}-frac1{8k+1}right)&=1-sum_{k=1}^inftyfrac{2}{64k^2-1}\
&=1+frac{1}{32}sum_{k=1}^inftyfrac{1}{(frac18)^2-k^2}\
&=frac18left(frac1{1/8}+frac{2}{8}sum_{k=1}^inftyfrac{1}{(frac18)^2-k^2}right)\
&=frac18picot(frac18pi)tag{$dagger$}label{eqn:cot}\
&=frac18pi(1+sqrt2)
end{align}
where we used the Laurent series for cotangent in eqref{eqn:cot}
$$
picot(pi z)=frac1z+2zsum_{k=1}^inftyfrac{1}{z^2-n^2}.
$$
answered Nov 13 at 7:44
user10354138
6,294623
add a comment |
up vote
2
down vote
We are looking for
$$1+sum_{k=1}^infty left(frac1{8k+1}-frac1{8k-1}right)=1-sum_{k=1}^infty frac{2}{64k^2-1}approx 1-frac1{32}zeta(2)approx frac{pi}{8}(sqrt2+1)$$
answered Nov 13 at 7:11
gimusi
86k74292
We need to sum...I add one more step
– gimusi
Nov 13 at 7:13
Sir i have also reached there, but after that i don't know how to go ahead, plz help me by solving this, i shall be very thankful to u
– user532616
Nov 13 at 7:14
What about $$sum_{k=1}^infty frac{2}{64k^2-1}approx (1/32) sum_{k=1}^inftyfrac1{k^2}$$
– gimusi
Nov 13 at 7:19
Since you are given a multiple choice I think that an estimation can suffice to select the correct option (which should be the “a”).
– gimusi
Nov 13 at 7:34
After the editing for the multiple choices the correct aswer should be “d”.
– gimusi
Nov 13 at 8:49
|
show 2 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Alternating series test gives the sum is convergent, so we can bracket off
begin{align}
1-sum_{k=1}^inftyleft(frac1{8k-1}-frac1{8k+1}right)&=1-sum_{k=1}^inftyfrac{2}{64k^2-1}\
&=1+frac{1}{32}sum_{k=1}^inftyfrac{1}{(frac18)^2-k^2}\
&=frac18left(frac1{1/8}+frac{2}{8}sum_{k=1}^inftyfrac{1}{(frac18)^2-k^2}right)\
&=frac18picot(frac18pi)tag{$dagger$}label{eqn:cot}\
&=frac18pi(1+sqrt2)
end{align}
where we used the Laurent series for cotangent in eqref{eqn:cot}
$$
picot(pi z)=frac1z+2zsum_{k=1}^inftyfrac{1}{z^2-n^2}.
$$
answered Nov 13 at 7:44
user10354138
6,294623
add a comment |
up vote
2
down vote
accepted
Alternating series test gives the sum is convergent, so we can bracket off
begin{align}
1-sum_{k=1}^inftyleft(frac1{8k-1}-frac1{8k+1}right)&=1-sum_{k=1}^inftyfrac{2}{64k^2-1}\
&=1+frac{1}{32}sum_{k=1}^inftyfrac{1}{(frac18)^2-k^2}\
&=frac18left(frac1{1/8}+frac{2}{8}sum_{k=1}^inftyfrac{1}{(frac18)^2-k^2}right)\
&=frac18picot(frac18pi)tag{$dagger$}label{eqn:cot}\
&=frac18pi(1+sqrt2)
end{align}
where we used the Laurent series for cotangent in eqref{eqn:cot}
$$
picot(pi z)=frac1z+2zsum_{k=1}^inftyfrac{1}{z^2-n^2}.
$$
answered Nov 13 at 7:44
user10354138
6,294623
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Alternating series test gives the sum is convergent, so we can bracket off
begin{align}
1-sum_{k=1}^inftyleft(frac1{8k-1}-frac1{8k+1}right)&=1-sum_{k=1}^inftyfrac{2}{64k^2-1}\
&=1+frac{1}{32}sum_{k=1}^inftyfrac{1}{(frac18)^2-k^2}\
&=frac18left(frac1{1/8}+frac{2}{8}sum_{k=1}^inftyfrac{1}{(frac18)^2-k^2}right)\
&=frac18picot(frac18pi)tag{$dagger$}label{eqn:cot}\
&=frac18pi(1+sqrt2)
end{align}
where we used the Laurent series for cotangent in eqref{eqn:cot}
$$
picot(pi z)=frac1z+2zsum_{k=1}^inftyfrac{1}{z^2-n^2}.
$$
answered Nov 13 at 7:44
user10354138
6,294623
Alternating series test gives the sum is convergent, so we can bracket off
begin{align}
1-sum_{k=1}^inftyleft(frac1{8k-1}-frac1{8k+1}right)&=1-sum_{k=1}^inftyfrac{2}{64k^2-1}\
&=1+frac{1}{32}sum_{k=1}^inftyfrac{1}{(frac18)^2-k^2}\
&=frac18left(frac1{1/8}+frac{2}{8}sum_{k=1}^inftyfrac{1}{(frac18)^2-k^2}right)\
&=frac18picot(frac18pi)tag{$dagger$}label{eqn:cot}\
&=frac18pi(1+sqrt2)
end{align}
where we used the Laurent series for cotangent in eqref{eqn:cot}
$$
picot(pi z)=frac1z+2zsum_{k=1}^inftyfrac{1}{z^2-n^2}.
$$
answered Nov 13 at 7:44
user10354138
6,294623
edited Nov 13 at 8:17
answered Nov 13 at 7:44
user10354138
6,294623
answered Nov 13 at 7:44
user10354138
6,294623
answered Nov 13 at 7:44
user10354138
6,294623
6,294623
add a comment |
add a comment |
up vote
2
down vote
We are looking for
$$1+sum_{k=1}^infty left(frac1{8k+1}-frac1{8k-1}right)=1-sum_{k=1}^infty frac{2}{64k^2-1}approx 1-frac1{32}zeta(2)approx frac{pi}{8}(sqrt2+1)$$
answered Nov 13 at 7:11
gimusi
86k74292
We need to sum...I add one more step
– gimusi
Nov 13 at 7:13
Sir i have also reached there, but after that i don't know how to go ahead, plz help me by solving this, i shall be very thankful to u
– user532616
Nov 13 at 7:14
What about $$sum_{k=1}^infty frac{2}{64k^2-1}approx (1/32) sum_{k=1}^inftyfrac1{k^2}$$
– gimusi
Nov 13 at 7:19
Since you are given a multiple choice I think that an estimation can suffice to select the correct option (which should be the “a”).
– gimusi
Nov 13 at 7:34
After the editing for the multiple choices the correct aswer should be “d”.
– gimusi
Nov 13 at 8:49
|
show 2 more comments
up vote
2
down vote
We are looking for
$$1+sum_{k=1}^infty left(frac1{8k+1}-frac1{8k-1}right)=1-sum_{k=1}^infty frac{2}{64k^2-1}approx 1-frac1{32}zeta(2)approx frac{pi}{8}(sqrt2+1)$$
answered Nov 13 at 7:11
gimusi
86k74292
We need to sum...I add one more step
– gimusi
Nov 13 at 7:13
Sir i have also reached there, but after that i don't know how to go ahead, plz help me by solving this, i shall be very thankful to u
– user532616
Nov 13 at 7:14
What about $$sum_{k=1}^infty frac{2}{64k^2-1}approx (1/32) sum_{k=1}^inftyfrac1{k^2}$$
– gimusi
Nov 13 at 7:19
Since you are given a multiple choice I think that an estimation can suffice to select the correct option (which should be the “a”).
– gimusi
Nov 13 at 7:34
After the editing for the multiple choices the correct aswer should be “d”.
– gimusi
Nov 13 at 8:49
|
show 2 more comments
up vote
2
down vote
up vote
2
down vote
We are looking for
$$1+sum_{k=1}^infty left(frac1{8k+1}-frac1{8k-1}right)=1-sum_{k=1}^infty frac{2}{64k^2-1}approx 1-frac1{32}zeta(2)approx frac{pi}{8}(sqrt2+1)$$
answered Nov 13 at 7:11
gimusi
86k74292
We are looking for
$$1+sum_{k=1}^infty left(frac1{8k+1}-frac1{8k-1}right)=1-sum_{k=1}^infty frac{2}{64k^2-1}approx 1-frac1{32}zeta(2)approx frac{pi}{8}(sqrt2+1)$$
answered Nov 13 at 7:11
gimusi
86k74292
edited Nov 13 at 8:48
answered Nov 13 at 7:11
gimusi
86k74292
answered Nov 13 at 7:11
gimusi
86k74292
answered Nov 13 at 7:11
gimusi
86k74292
86k74292
We need to sum...I add one more step
– gimusi
Nov 13 at 7:13
Sir i have also reached there, but after that i don't know how to go ahead, plz help me by solving this, i shall be very thankful to u
– user532616
Nov 13 at 7:14
What about $$sum_{k=1}^infty frac{2}{64k^2-1}approx (1/32) sum_{k=1}^inftyfrac1{k^2}$$
– gimusi
Nov 13 at 7:19
Since you are given a multiple choice I think that an estimation can suffice to select the correct option (which should be the “a”).
– gimusi
Nov 13 at 7:34
After the editing for the multiple choices the correct aswer should be “d”.
– gimusi
Nov 13 at 8:49
|
show 2 more comments
We need to sum...I add one more step
– gimusi
Nov 13 at 7:13
Sir i have also reached there, but after that i don't know how to go ahead, plz help me by solving this, i shall be very thankful to u
– user532616
Nov 13 at 7:14
What about $$sum_{k=1}^infty frac{2}{64k^2-1}approx (1/32) sum_{k=1}^inftyfrac1{k^2}$$
– gimusi
Nov 13 at 7:19
Since you are given a multiple choice I think that an estimation can suffice to select the correct option (which should be the “a”).
– gimusi
Nov 13 at 7:34
After the editing for the multiple choices the correct aswer should be “d”.
– gimusi
Nov 13 at 8:49
We need to sum...I add one more step
– gimusi
Nov 13 at 7:13
We need to sum...I add one more step
– gimusi
Nov 13 at 7:13
Sir i have also reached there, but after that i don't know how to go ahead, plz help me by solving this, i shall be very thankful to u
– user532616
Nov 13 at 7:14
Sir i have also reached there, but after that i don't know how to go ahead, plz help me by solving this, i shall be very thankful to u
– user532616
Nov 13 at 7:14
What about $$sum_{k=1}^infty frac{2}{64k^2-1}approx (1/32) sum_{k=1}^inftyfrac1{k^2}$$
– gimusi
Nov 13 at 7:19
What about $$sum_{k=1}^infty frac{2}{64k^2-1}approx (1/32) sum_{k=1}^inftyfrac1{k^2}$$
– gimusi
Nov 13 at 7:19
Since you are given a multiple choice I think that an estimation can suffice to select the correct option (which should be the “a”).
– gimusi
Nov 13 at 7:34
Since you are given a multiple choice I think that an estimation can suffice to select the correct option (which should be the “a”).
– gimusi
Nov 13 at 7:34
After the editing for the multiple choices the correct aswer should be “d”.
– gimusi
Nov 13 at 8:49
After the editing for the multiple choices the correct aswer should be “d”.
– gimusi
Nov 13 at 8:49
|
show 2 more comments
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How did you narrow it down to those four possibilities?
– Lord Shark the Unknown
Nov 13 at 7:02
1
Maybe you can consider it a sum of two different infinite sums? What would the two expressions be?
– Matti P.
Nov 13 at 7:03
The sum equals $$int_0^1frac{1-x^6}{1-x^8},dx.$$
– Lord Shark the Unknown
Nov 13 at 7:03
But how this sum is equivalent to this integral.. Plz explain clearly
– user532616
Nov 13 at 7:06
expand the integrand as a power series and integrate term-by-term gives the original series
– user10354138
Nov 13 at 7:46