What is the value of $frac11+frac13-frac15-frac17+frac19+frac111-dots$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Find the sum of $sum 1/(k^2 - a^2)$ when $0<a<1$Find the limit of $sum frac1log^n(n)$Calculating the infinite series $1-frac13+frac15-frac17+frac19-frac111cdots$Another SummationNumerical convergence depending on summation orderSign flip for every TWO terms in a sum? (Instead of one)Show that $sum_n=0^infty |a_n|<infty.$A series whose terms are the products of terms of a geometric and a power seriesCheck if the series $sum_n=1^inftyfrac(-1)^fracn(n-1)2sqrt n$ convergesConvergence of $1+frac13-frac12+frac15+frac17-frac14+frac19+frac111-frac16+ldots$What is the sign and exact value of this sum: $sum_n=0^+infty(-1)^n tan(frac1n!)$?
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What is the value of $frac11+frac13-frac15-frac17+frac19+frac111-dots$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Find the sum of $sum 1/(k^2 - a^2)$ when $0<a<1$Find the limit of $sum frac1log^n(n)$Calculating the infinite series $1-frac13+frac15-frac17+frac19-frac111cdots$Another SummationNumerical convergence depending on summation orderSign flip for every TWO terms in a sum? (Instead of one)Show that $sum_n=0^infty |a_n|<infty.$A series whose terms are the products of terms of a geometric and a power seriesCheck if the series $sum_n=1^inftyfrac(-1)^fracn(n-1)2sqrt n$ convergesConvergence of $1+frac13-frac12+frac15+frac17-frac14+frac19+frac111-frac16+ldots$What is the sign and exact value of this sum: $sum_n=0^+infty(-1)^n tan(frac1n!)$?
$begingroup$
The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.
The series $sum_k=1^infty frac(-1)^frack^2+k+222k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.
What is the convergence value, explicitly, of the second series?
The first summation is noted above, because it might be a useful information to evaluate the second summation.
sequences-and-series summation integers pi
edited Apr 17 at 15:09
user21820
40.4k544163
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
3587
$endgroup$
add a comment |
$begingroup$
The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.
The series $sum_k=1^infty frac(-1)^frack^2+k+222k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.
What is the convergence value, explicitly, of the second series?
The first summation is noted above, because it might be a useful information to evaluate the second summation.
sequences-and-series summation integers pi
edited Apr 17 at 15:09
user21820
40.4k544163
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
3587
$endgroup$
1
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40
add a comment |
$begingroup$
The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.
The series $sum_k=1^infty frac(-1)^frack^2+k+222k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.
What is the convergence value, explicitly, of the second series?
The first summation is noted above, because it might be a useful information to evaluate the second summation.
sequences-and-series summation integers pi
edited Apr 17 at 15:09
user21820
40.4k544163
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
3587
$endgroup$
The series $sum_k=1^infty frac(-1)^k+12k-1=frac11-frac13+frac15-frac17+dots$ converges to $fracpi4$. Here, the sign alternates every term.
The series $sum_k=1^infty frac(-1)^frack^2+k+222k-1=frac11+frac13-frac15-frac17+dots$ also converges. Here, the sign alternates every two terms.
What is the convergence value, explicitly, of the second series?
The first summation is noted above, because it might be a useful information to evaluate the second summation.
sequences-and-series summation integers pi
sequences-and-series summation integers pi
edited Apr 17 at 15:09
user21820
40.4k544163
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
3587
edited Apr 17 at 15:09
user21820
40.4k544163
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
3587
edited Apr 17 at 15:09
user21820
40.4k544163
edited Apr 17 at 15:09
user21820
40.4k544163
edited Apr 17 at 15:09
user21820
40.4k544163
40.4k544163
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
3587
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
3587
asked Apr 17 at 10:08
Hussain-AlqatariHussain-Alqatari
3587
3587
1
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40
add a comment |
1
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40
1
1
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $$beginalign
S_1&=1-frac15+frac19-1over13+dots\
S_2&=frac13-frac17+1over11-1over15+dots
endalign$$ so that the sum we seek is $S_1+S_2.$
T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$
The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.
answered Apr 17 at 11:03
saulspatzsaulspatz
17.8k31536
$endgroup$
2
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
add a comment |
$begingroup$
Hint:
$$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$
$$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$
$$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$
$$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
$$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$
then use
$$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$
answered Apr 17 at 12:30
E.H.EE.H.E
17k11969
$endgroup$
add a comment |
$begingroup$
$$
beginalign
&sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
&=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
&=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
&=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
&=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
&=fracpisqrt24tag5
endalign
$$
Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $mathbbZ$
$(3)$: factor $frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$
answered Apr 17 at 17:33
robjohn♦robjohn
271k27316643
$endgroup$
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $$beginalign
S_1&=1-frac15+frac19-1over13+dots\
S_2&=frac13-frac17+1over11-1over15+dots
endalign$$ so that the sum we seek is $S_1+S_2.$
T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$
The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.
answered Apr 17 at 11:03
saulspatzsaulspatz
17.8k31536
$endgroup$
2
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
add a comment |
$begingroup$
Let $$beginalign
S_1&=1-frac15+frac19-1over13+dots\
S_2&=frac13-frac17+1over11-1over15+dots
endalign$$ so that the sum we seek is $S_1+S_2.$
T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$
The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.
answered Apr 17 at 11:03
saulspatzsaulspatz
17.8k31536
$endgroup$
2
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
add a comment |
$begingroup$
Let $$beginalign
S_1&=1-frac15+frac19-1over13+dots\
S_2&=frac13-frac17+1over11-1over15+dots
endalign$$ so that the sum we seek is $S_1+S_2.$
T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$
The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.
answered Apr 17 at 11:03
saulspatzsaulspatz
17.8k31536
$endgroup$
Let $$beginalign
S_1&=1-frac15+frac19-1over13+dots\
S_2&=frac13-frac17+1over11-1over15+dots
endalign$$ so that the sum we seek is $S_1+S_2.$
T0 compute $S_1,$ consider $$f(x) = 1-x^5over5+x^9over9-x^13over13+dots$$ so that $$f'(x)=-x^4+x^8-x^12+dots=-x^4over1+x^4, |x|<1$$
and $$f(x)=int_0^x-t^4over1+t^4mathrmdt+f(0), |x|<1$$
By Abel's limit theorem, $$S_1=lim_xto1-int_0^x-t^4over1+t^4mathrmdt+f(0)=1-int_0^1t^4over1+t^4mathrmdt$$ and we can do a similar calculation for $S_2$ to get $$S_2=int_0^1t^2over1+t^4mathrmdt$$
The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.
answered Apr 17 at 11:03
saulspatzsaulspatz
17.8k31536
answered Apr 17 at 11:03
saulspatzsaulspatz
17.8k31536
answered Apr 17 at 11:03
saulspatzsaulspatz
17.8k31536
answered Apr 17 at 11:03
saulspatzsaulspatz
17.8k31536
17.8k31536
2
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
add a comment |
2
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
2
2
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms.
$endgroup$
– Hussain-Alqatari
Apr 17 at 12:56
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
$begingroup$
Note that $S_1 = int_0^1 mathrmdt -int_0^1t^4over1+t^4mathrmdt = int_0^11over1+t^4mathrmdt$, which might be marginally easier to integrate.
$endgroup$
– Michael Seifert
Apr 17 at 14:11
add a comment |
$begingroup$
Hint:
$$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$
$$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$
$$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$
$$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
$$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$
then use
$$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$
answered Apr 17 at 12:30
E.H.EE.H.E
17k11969
$endgroup$
add a comment |
$begingroup$
Hint:
$$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$
$$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$
$$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$
$$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
$$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$
then use
$$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$
answered Apr 17 at 12:30
E.H.EE.H.E
17k11969
$endgroup$
add a comment |
$begingroup$
Hint:
$$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$
$$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$
$$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$
$$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
$$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$
then use
$$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$
answered Apr 17 at 12:30
E.H.EE.H.E
17k11969
$endgroup$
Hint:
$$frac11+frac13-frac15-frac17+dots=sum_k=1^infty frac18k-7+frac18k-5-frac18k-3-frac18k-1$$
$$=sum_k=1^infty (frac18k-7-frac18k-1)+(frac18k-5-frac18k-3)$$
$$=sum_k=1^infty (1+frac18k+1-frac18k-1)+(frac13+frac18k+3-frac18k-3)$$
$$=frac43+sum_k=1^infty (frac-264k^2-1+frac-664k^2-9)$$
$$=frac43-frac132sum_k=1^infty frac1k^2-frac164-frac332sum_k=1^infty frac1k^2-frac964$$
then use
$$frac1-pi x cot(pi x)2x^2=sum_k=1^infty frac1k^2-x^2$$
answered Apr 17 at 12:30
E.H.EE.H.E
17k11969
edited Apr 17 at 14:52
answered Apr 17 at 12:30
E.H.EE.H.E
17k11969
answered Apr 17 at 12:30
E.H.EE.H.E
17k11969
answered Apr 17 at 12:30
E.H.EE.H.E
17k11969
17k11969
add a comment |
add a comment |
$begingroup$
$$
beginalign
&sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
&=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
&=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
&=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
&=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
&=fracpisqrt24tag5
endalign
$$
Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $mathbbZ$
$(3)$: factor $frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$
answered Apr 17 at 17:33
robjohn♦robjohn
271k27316643
$endgroup$
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
add a comment |
$begingroup$
$$
beginalign
&sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
&=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
&=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
&=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
&=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
&=fracpisqrt24tag5
endalign
$$
Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $mathbbZ$
$(3)$: factor $frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$
answered Apr 17 at 17:33
robjohn♦robjohn
271k27316643
$endgroup$
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
add a comment |
$begingroup$
$$
beginalign
&sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
&=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
&=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
&=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
&=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
&=fracpisqrt24tag5
endalign
$$
Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $mathbbZ$
$(3)$: factor $frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$
answered Apr 17 at 17:33
robjohn♦robjohn
271k27316643
$endgroup$
$$
beginalign
&sum_k=0^inftyleft(frac18k+1+frac18k+3-frac18k+5-frac18k+7right)\
&=sum_k=0^inftyleft(frac18k+1-frac18k+7right)+sum_k=0^inftyleft(frac18k+3-frac18k+5right)tag1\
&=sum_kinmathbbZfrac18k+1+sum_kinmathbbZfrac18k+3tag2\
&=frac18sum_kinmathbbZfrac1k+frac18+frac18sum_kinmathbbZfrac1k+frac38tag3\
&=fracpi8left[cotleft(fracpi8right)+cotleft(frac3pi8right)right]tag4\[6pt]
&=fracpisqrt24tag5
endalign
$$
Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $mathbbZ$
$(3)$: factor $frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $cotleft(fracpi8right)=1+sqrt2$ and $cotleft(frac3pi8right)=-1+sqrt2$
answered Apr 17 at 17:33
robjohn♦robjohn
271k27316643
edited Apr 17 at 17:49
answered Apr 17 at 17:33
robjohn♦robjohn
271k27316643
answered Apr 17 at 17:33
robjohn♦robjohn
271k27316643
answered Apr 17 at 17:33
robjohn♦robjohn
271k27316643
271k27316643
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
add a comment |
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
$begingroup$
For sign change every $3$ terms, $cotleft(fracpi12right)=2+sqrt3$, $cotleft(frac3pi12right)=1$, $cotleft(frac5pi12right)=2-sqrt3$.
$endgroup$
– robjohn♦
Apr 18 at 12:05
add a comment |
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1
$begingroup$
Wolfram says the result is $fracsqrt2pi4$
$endgroup$
– Peter Foreman
Apr 17 at 10:28
$begingroup$
@PeterForeman yes, but how to approach that value?
$endgroup$
– Hussain-Alqatari
Apr 17 at 10:40