How can one evaluate the following series:

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$$sum_k=1^infty (-1)^kfrac(lnk)^2k spacespacespace ?$$



From $sum_ngeq 1frac(-1)^n ln nn$
it has been answered that



$$ sum_k=1^infty (-1)^k fraclnkk = space gamma cdot ln2 space - space frac(ln2)^22 space space space , $$



where $gamma$ is the Euler-Mascheroni constant.



$ \ $




sequences-and-series eulers-constant




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  • 2




    In the answer given by Olivier, maybe try to differentiate (3) one more time?
    – user223391
    Jul 29 at 18:44







  • 1




    Differentiate twice the Eta function and let $zto 1^+$,
    – Mark Viola
    Jul 29 at 18:46














up vote
0
down vote

favorite












$$sum_k=1^infty (-1)^kfrac(lnk)^2k spacespacespace ?$$



From $sum_ngeq 1frac(-1)^n ln nn$
it has been answered that



$$ sum_k=1^infty (-1)^k fraclnkk = space gamma cdot ln2 space - space frac(ln2)^22 space space space , $$



where $gamma$ is the Euler-Mascheroni constant.



$ \ $




sequences-and-series eulers-constant




share|cite|improve this question

















  • 2




    In the answer given by Olivier, maybe try to differentiate (3) one more time?
    – user223391
    Jul 29 at 18:44







  • 1




    Differentiate twice the Eta function and let $zto 1^+$,
    – Mark Viola
    Jul 29 at 18:46












up vote
0
down vote

favorite









up vote
0
down vote

favorite











$$sum_k=1^infty (-1)^kfrac(lnk)^2k spacespacespace ?$$



From $sum_ngeq 1frac(-1)^n ln nn$
it has been answered that



$$ sum_k=1^infty (-1)^k fraclnkk = space gamma cdot ln2 space - space frac(ln2)^22 space space space , $$



where $gamma$ is the Euler-Mascheroni constant.



$ \ $




sequences-and-series eulers-constant




share|cite|improve this question













$$sum_k=1^infty (-1)^kfrac(lnk)^2k spacespacespace ?$$



From $sum_ngeq 1frac(-1)^n ln nn$
it has been answered that



$$ sum_k=1^infty (-1)^k fraclnkk = space gamma cdot ln2 space - space frac(ln2)^22 space space space , $$



where $gamma$ is the Euler-Mascheroni constant.



$ \ $





sequences-and-series eulers-constant





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 29 at 18:49









Mark Viola

126k1172167




126k1172167









asked Jul 29 at 18:41









Leonardo Bohac

648




648







  • 2




    In the answer given by Olivier, maybe try to differentiate (3) one more time?
    – user223391
    Jul 29 at 18:44







  • 1




    Differentiate twice the Eta function and let $zto 1^+$,
    – Mark Viola
    Jul 29 at 18:46












  • 2




    In the answer given by Olivier, maybe try to differentiate (3) one more time?
    – user223391
    Jul 29 at 18:44







  • 1




    Differentiate twice the Eta function and let $zto 1^+$,
    – Mark Viola
    Jul 29 at 18:46







2




2




In the answer given by Olivier, maybe try to differentiate (3) one more time?
– user223391
Jul 29 at 18:44





In the answer given by Olivier, maybe try to differentiate (3) one more time?
– user223391
Jul 29 at 18:44





1




1




Differentiate twice the Eta function and let $zto 1^+$,
– Mark Viola
Jul 29 at 18:46




Differentiate twice the Eta function and let $zto 1^+$,
– Mark Viola
Jul 29 at 18:46










1 Answer
1






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up vote
2
down vote



accepted










Hint. From the standard relation
$$sum_ngeq 1frac(-1)^n n^s=left(frac12^s-1-1right)zeta(s)$$ one may differentiate twice and set $s to 1^+$ using the Laurent series of the Riemann zeta function.






share|cite|improve this answer





















  • Hi Oliver, thanks for replying. From your answer to the other post, when you considered the expansion for 2^(1-s) , isn't the signal preceding the third term positive instead of negative? I'm trying to go through what you've done there together with one additional differentiation.
    – Leonardo Bohac
    Jul 29 at 19:27










  • @LeonardoBohac You are right, typo has been corrected. Thank you.
    – Olivier Oloa
    Jul 30 at 8:25










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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Hint. From the standard relation
$$sum_ngeq 1frac(-1)^n n^s=left(frac12^s-1-1right)zeta(s)$$ one may differentiate twice and set $s to 1^+$ using the Laurent series of the Riemann zeta function.






share|cite|improve this answer





















  • Hi Oliver, thanks for replying. From your answer to the other post, when you considered the expansion for 2^(1-s) , isn't the signal preceding the third term positive instead of negative? I'm trying to go through what you've done there together with one additional differentiation.
    – Leonardo Bohac
    Jul 29 at 19:27










  • @LeonardoBohac You are right, typo has been corrected. Thank you.
    – Olivier Oloa
    Jul 30 at 8:25














up vote
2
down vote



accepted










Hint. From the standard relation
$$sum_ngeq 1frac(-1)^n n^s=left(frac12^s-1-1right)zeta(s)$$ one may differentiate twice and set $s to 1^+$ using the Laurent series of the Riemann zeta function.






share|cite|improve this answer





















  • Hi Oliver, thanks for replying. From your answer to the other post, when you considered the expansion for 2^(1-s) , isn't the signal preceding the third term positive instead of negative? I'm trying to go through what you've done there together with one additional differentiation.
    – Leonardo Bohac
    Jul 29 at 19:27










  • @LeonardoBohac You are right, typo has been corrected. Thank you.
    – Olivier Oloa
    Jul 30 at 8:25












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Hint. From the standard relation
$$sum_ngeq 1frac(-1)^n n^s=left(frac12^s-1-1right)zeta(s)$$ one may differentiate twice and set $s to 1^+$ using the Laurent series of the Riemann zeta function.






share|cite|improve this answer













Hint. From the standard relation
$$sum_ngeq 1frac(-1)^n n^s=left(frac12^s-1-1right)zeta(s)$$ one may differentiate twice and set $s to 1^+$ using the Laurent series of the Riemann zeta function.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 29 at 18:51









Olivier Oloa

106k17173292




106k17173292











  • Hi Oliver, thanks for replying. From your answer to the other post, when you considered the expansion for 2^(1-s) , isn't the signal preceding the third term positive instead of negative? I'm trying to go through what you've done there together with one additional differentiation.
    – Leonardo Bohac
    Jul 29 at 19:27










  • @LeonardoBohac You are right, typo has been corrected. Thank you.
    – Olivier Oloa
    Jul 30 at 8:25
















  • Hi Oliver, thanks for replying. From your answer to the other post, when you considered the expansion for 2^(1-s) , isn't the signal preceding the third term positive instead of negative? I'm trying to go through what you've done there together with one additional differentiation.
    – Leonardo Bohac
    Jul 29 at 19:27










  • @LeonardoBohac You are right, typo has been corrected. Thank you.
    – Olivier Oloa
    Jul 30 at 8:25















Hi Oliver, thanks for replying. From your answer to the other post, when you considered the expansion for 2^(1-s) , isn't the signal preceding the third term positive instead of negative? I'm trying to go through what you've done there together with one additional differentiation.
– Leonardo Bohac
Jul 29 at 19:27




Hi Oliver, thanks for replying. From your answer to the other post, when you considered the expansion for 2^(1-s) , isn't the signal preceding the third term positive instead of negative? I'm trying to go through what you've done there together with one additional differentiation.
– Leonardo Bohac
Jul 29 at 19:27












@LeonardoBohac You are right, typo has been corrected. Thank you.
– Olivier Oloa
Jul 30 at 8:25




@LeonardoBohac You are right, typo has been corrected. Thank you.
– Olivier Oloa
Jul 30 at 8:25












 

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