Limit of harmonic series

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If $lim_ntoinfty nleft(b-sum_r=1^nfrac1n+rright)=a,$ find $a$ and $b.$



My progress:



As $lim_ntoinftysum_r=1^nfrac1n+r=ln 2implies b=ln 2$



Now what to do for $a$






limits







share|cite|improve this question





















  • Why does $b=ln 2?$ You've shown that $a =lim_ntoinfty n(b-ln 2)$, but nothing more...
    – Rhys Hughes
    3 hours ago











  • @RhysHughes: divide your equation by $n$ and you get $b=log 2$.
    – Paramanand Singh
    3 hours ago










  • @RhysHughes: if we admit that the existence of $a,b$ is implied in the question, then $Sigma$ tends to $b$.
    – Yves Daoust
    2 hours ago











  • Nice question and yet a downvote! +1 already given to compensate.
    – Paramanand Singh
    2 hours ago














up vote
2
down vote

favorite
1












If $lim_ntoinfty nleft(b-sum_r=1^nfrac1n+rright)=a,$ find $a$ and $b.$



My progress:



As $lim_ntoinftysum_r=1^nfrac1n+r=ln 2implies b=ln 2$



Now what to do for $a$






limits







share|cite|improve this question





















  • Why does $b=ln 2?$ You've shown that $a =lim_ntoinfty n(b-ln 2)$, but nothing more...
    – Rhys Hughes
    3 hours ago











  • @RhysHughes: divide your equation by $n$ and you get $b=log 2$.
    – Paramanand Singh
    3 hours ago










  • @RhysHughes: if we admit that the existence of $a,b$ is implied in the question, then $Sigma$ tends to $b$.
    – Yves Daoust
    2 hours ago











  • Nice question and yet a downvote! +1 already given to compensate.
    – Paramanand Singh
    2 hours ago












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





If $lim_ntoinfty nleft(b-sum_r=1^nfrac1n+rright)=a,$ find $a$ and $b.$



My progress:



As $lim_ntoinftysum_r=1^nfrac1n+r=ln 2implies b=ln 2$



Now what to do for $a$






limits







share|cite|improve this question













If $lim_ntoinfty nleft(b-sum_r=1^nfrac1n+rright)=a,$ find $a$ and $b.$



My progress:



As $lim_ntoinftysum_r=1^nfrac1n+r=ln 2implies b=ln 2$



Now what to do for $a$






limits




limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









Makar

593116




593116











  • Why does $b=ln 2?$ You've shown that $a =lim_ntoinfty n(b-ln 2)$, but nothing more...
    – Rhys Hughes
    3 hours ago











  • @RhysHughes: divide your equation by $n$ and you get $b=log 2$.
    – Paramanand Singh
    3 hours ago










  • @RhysHughes: if we admit that the existence of $a,b$ is implied in the question, then $Sigma$ tends to $b$.
    – Yves Daoust
    2 hours ago











  • Nice question and yet a downvote! +1 already given to compensate.
    – Paramanand Singh
    2 hours ago
















  • Why does $b=ln 2?$ You've shown that $a =lim_ntoinfty n(b-ln 2)$, but nothing more...
    – Rhys Hughes
    3 hours ago











  • @RhysHughes: divide your equation by $n$ and you get $b=log 2$.
    – Paramanand Singh
    3 hours ago










  • @RhysHughes: if we admit that the existence of $a,b$ is implied in the question, then $Sigma$ tends to $b$.
    – Yves Daoust
    2 hours ago











  • Nice question and yet a downvote! +1 already given to compensate.
    – Paramanand Singh
    2 hours ago















Why does $b=ln 2?$ You've shown that $a =lim_ntoinfty n(b-ln 2)$, but nothing more...
– Rhys Hughes
3 hours ago





Why does $b=ln 2?$ You've shown that $a =lim_ntoinfty n(b-ln 2)$, but nothing more...
– Rhys Hughes
3 hours ago













@RhysHughes: divide your equation by $n$ and you get $b=log 2$.
– Paramanand Singh
3 hours ago




@RhysHughes: divide your equation by $n$ and you get $b=log 2$.
– Paramanand Singh
3 hours ago












@RhysHughes: if we admit that the existence of $a,b$ is implied in the question, then $Sigma$ tends to $b$.
– Yves Daoust
2 hours ago





@RhysHughes: if we admit that the existence of $a,b$ is implied in the question, then $Sigma$ tends to $b$.
– Yves Daoust
2 hours ago













Nice question and yet a downvote! +1 already given to compensate.
– Paramanand Singh
2 hours ago




Nice question and yet a downvote! +1 already given to compensate.
– Paramanand Singh
2 hours ago










2 Answers
2






active

oldest

votes

















up vote
5
down vote



accepted










Getting the value of $a$ is tricky / non-obvious. Use the following lemma




Lemma: If $fin C^1[0,1] $ then $$lim_ntoinfty sum_k=1^nfleft(fracknright)-nint_0^1f(x),dx=fracf(1)-f(0)2$$
(proof available here)




In the above lemma it is sufficient to assume the Riemann integrability of $f'$.



For your question $f(x) = 1/(1+x)$ and the limit $a$ is $$lim_ntoinfty nint_0^1f(x),dx-sum_k=1^nfleft(frack n right)$$ which by above lemma equals $(f(0)-f(1))/2=1/4$.






share|cite|improve this answer




















  • I need to study that! Very nice, thanks!
    – gimusi
    2 hours ago










  • @gimusi: in that case see an even more powerful application of this lemma at math.stackexchange.com/a/2847768/72031
    – Paramanand Singh
    2 hours ago

















up vote
2
down vote













Recall that by the rate of convergence of the Harmonic series



$$sum_r=1^nfrac1n+r=sum_r=1^2nfrac1r-sum_r=1^nfrac1r= ln 2n+gamma+frac14n-ln n-gamma-frac12n+oleft(frac1nright)=$$$$=ln 2-frac14n+oleft(frac1nright)$$



then we have



$$lim_ntoinfty nleft(b-sum_r=1^nfrac1n+rright)=lim_ntoinfty nleft(b-ln 2+frac14n+oleft(frac1nright)right)=a$$



therefore $b=ln 2$ and $a =frac14$.






share|cite|improve this answer






















  • +1 already there, but you may replace $sim$ with $=$ and add $o(1/n)$.
    – Paramanand Singh
    2 hours ago










  • @ParamanandSingh Yes that's better as you suggested! Thanks
    – gimusi
    2 hours ago










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










Getting the value of $a$ is tricky / non-obvious. Use the following lemma




Lemma: If $fin C^1[0,1] $ then $$lim_ntoinfty sum_k=1^nfleft(fracknright)-nint_0^1f(x),dx=fracf(1)-f(0)2$$
(proof available here)




In the above lemma it is sufficient to assume the Riemann integrability of $f'$.



For your question $f(x) = 1/(1+x)$ and the limit $a$ is $$lim_ntoinfty nint_0^1f(x),dx-sum_k=1^nfleft(frack n right)$$ which by above lemma equals $(f(0)-f(1))/2=1/4$.






share|cite|improve this answer




















  • I need to study that! Very nice, thanks!
    – gimusi
    2 hours ago










  • @gimusi: in that case see an even more powerful application of this lemma at math.stackexchange.com/a/2847768/72031
    – Paramanand Singh
    2 hours ago














up vote
5
down vote



accepted










Getting the value of $a$ is tricky / non-obvious. Use the following lemma




Lemma: If $fin C^1[0,1] $ then $$lim_ntoinfty sum_k=1^nfleft(fracknright)-nint_0^1f(x),dx=fracf(1)-f(0)2$$
(proof available here)




In the above lemma it is sufficient to assume the Riemann integrability of $f'$.



For your question $f(x) = 1/(1+x)$ and the limit $a$ is $$lim_ntoinfty nint_0^1f(x),dx-sum_k=1^nfleft(frack n right)$$ which by above lemma equals $(f(0)-f(1))/2=1/4$.






share|cite|improve this answer




















  • I need to study that! Very nice, thanks!
    – gimusi
    2 hours ago










  • @gimusi: in that case see an even more powerful application of this lemma at math.stackexchange.com/a/2847768/72031
    – Paramanand Singh
    2 hours ago












up vote
5
down vote



accepted







up vote
5
down vote



accepted






Getting the value of $a$ is tricky / non-obvious. Use the following lemma




Lemma: If $fin C^1[0,1] $ then $$lim_ntoinfty sum_k=1^nfleft(fracknright)-nint_0^1f(x),dx=fracf(1)-f(0)2$$
(proof available here)




In the above lemma it is sufficient to assume the Riemann integrability of $f'$.



For your question $f(x) = 1/(1+x)$ and the limit $a$ is $$lim_ntoinfty nint_0^1f(x),dx-sum_k=1^nfleft(frack n right)$$ which by above lemma equals $(f(0)-f(1))/2=1/4$.






share|cite|improve this answer












Getting the value of $a$ is tricky / non-obvious. Use the following lemma




Lemma: If $fin C^1[0,1] $ then $$lim_ntoinfty sum_k=1^nfleft(fracknright)-nint_0^1f(x),dx=fracf(1)-f(0)2$$
(proof available here)




In the above lemma it is sufficient to assume the Riemann integrability of $f'$.



For your question $f(x) = 1/(1+x)$ and the limit $a$ is $$lim_ntoinfty nint_0^1f(x),dx-sum_k=1^nfleft(frack n right)$$ which by above lemma equals $(f(0)-f(1))/2=1/4$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









Paramanand Singh

46.5k555147




46.5k555147











  • I need to study that! Very nice, thanks!
    – gimusi
    2 hours ago










  • @gimusi: in that case see an even more powerful application of this lemma at math.stackexchange.com/a/2847768/72031
    – Paramanand Singh
    2 hours ago
















  • I need to study that! Very nice, thanks!
    – gimusi
    2 hours ago










  • @gimusi: in that case see an even more powerful application of this lemma at math.stackexchange.com/a/2847768/72031
    – Paramanand Singh
    2 hours ago















I need to study that! Very nice, thanks!
– gimusi
2 hours ago




I need to study that! Very nice, thanks!
– gimusi
2 hours ago












@gimusi: in that case see an even more powerful application of this lemma at math.stackexchange.com/a/2847768/72031
– Paramanand Singh
2 hours ago




@gimusi: in that case see an even more powerful application of this lemma at math.stackexchange.com/a/2847768/72031
– Paramanand Singh
2 hours ago










up vote
2
down vote













Recall that by the rate of convergence of the Harmonic series



$$sum_r=1^nfrac1n+r=sum_r=1^2nfrac1r-sum_r=1^nfrac1r= ln 2n+gamma+frac14n-ln n-gamma-frac12n+oleft(frac1nright)=$$$$=ln 2-frac14n+oleft(frac1nright)$$



then we have



$$lim_ntoinfty nleft(b-sum_r=1^nfrac1n+rright)=lim_ntoinfty nleft(b-ln 2+frac14n+oleft(frac1nright)right)=a$$



therefore $b=ln 2$ and $a =frac14$.






share|cite|improve this answer






















  • +1 already there, but you may replace $sim$ with $=$ and add $o(1/n)$.
    – Paramanand Singh
    2 hours ago










  • @ParamanandSingh Yes that's better as you suggested! Thanks
    – gimusi
    2 hours ago














up vote
2
down vote













Recall that by the rate of convergence of the Harmonic series



$$sum_r=1^nfrac1n+r=sum_r=1^2nfrac1r-sum_r=1^nfrac1r= ln 2n+gamma+frac14n-ln n-gamma-frac12n+oleft(frac1nright)=$$$$=ln 2-frac14n+oleft(frac1nright)$$



then we have



$$lim_ntoinfty nleft(b-sum_r=1^nfrac1n+rright)=lim_ntoinfty nleft(b-ln 2+frac14n+oleft(frac1nright)right)=a$$



therefore $b=ln 2$ and $a =frac14$.






share|cite|improve this answer






















  • +1 already there, but you may replace $sim$ with $=$ and add $o(1/n)$.
    – Paramanand Singh
    2 hours ago










  • @ParamanandSingh Yes that's better as you suggested! Thanks
    – gimusi
    2 hours ago












up vote
2
down vote










up vote
2
down vote









Recall that by the rate of convergence of the Harmonic series



$$sum_r=1^nfrac1n+r=sum_r=1^2nfrac1r-sum_r=1^nfrac1r= ln 2n+gamma+frac14n-ln n-gamma-frac12n+oleft(frac1nright)=$$$$=ln 2-frac14n+oleft(frac1nright)$$



then we have



$$lim_ntoinfty nleft(b-sum_r=1^nfrac1n+rright)=lim_ntoinfty nleft(b-ln 2+frac14n+oleft(frac1nright)right)=a$$



therefore $b=ln 2$ and $a =frac14$.






share|cite|improve this answer














Recall that by the rate of convergence of the Harmonic series



$$sum_r=1^nfrac1n+r=sum_r=1^2nfrac1r-sum_r=1^nfrac1r= ln 2n+gamma+frac14n-ln n-gamma-frac12n+oleft(frac1nright)=$$$$=ln 2-frac14n+oleft(frac1nright)$$



then we have



$$lim_ntoinfty nleft(b-sum_r=1^nfrac1n+rright)=lim_ntoinfty nleft(b-ln 2+frac14n+oleft(frac1nright)right)=a$$



therefore $b=ln 2$ and $a =frac14$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 3 hours ago









gimusi

78.4k73890




78.4k73890











  • +1 already there, but you may replace $sim$ with $=$ and add $o(1/n)$.
    – Paramanand Singh
    2 hours ago










  • @ParamanandSingh Yes that's better as you suggested! Thanks
    – gimusi
    2 hours ago
















  • +1 already there, but you may replace $sim$ with $=$ and add $o(1/n)$.
    – Paramanand Singh
    2 hours ago










  • @ParamanandSingh Yes that's better as you suggested! Thanks
    – gimusi
    2 hours ago















+1 already there, but you may replace $sim$ with $=$ and add $o(1/n)$.
– Paramanand Singh
2 hours ago




+1 already there, but you may replace $sim$ with $=$ and add $o(1/n)$.
– Paramanand Singh
2 hours ago












@ParamanandSingh Yes that's better as you suggested! Thanks
– gimusi
2 hours ago




@ParamanandSingh Yes that's better as you suggested! Thanks
– gimusi
2 hours ago

















 

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