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Eufisky - The lost book

求和的一些估计式

计算limn(2nnk=11k).


Use n=nk=1(kk1), then
2nnk=11k=nk=1(2k2k11k)=nk=11k(kk1)2=nk=11k((kk1)(k+k1)(k+k1))2=nk=11k(k+k1)2.
 
This shows the limit does exist and limn(2nnk=11k)=k=11k(k+k1)2.
 
The value of this sums equals ζ(12)1.4603545. This value is found by other means, though:
2nnk=11k=2n(ζ(12)ζ(12,n+1))ζ(12)12n+o(1n). 
 

取整函数的求和问题

证明n1k=1sin((2kn+1)π2n)=cot(π2n)cos(π2n),

where is greatest integer function.


Lemma Summation by Pasts (1)http://en.wikipedia.org/wiki/Summation_by_parts
 
bk=afkΔgk=fkgk|b+1k=abk=agk+1Δfk=fb+1gb+1fagabk=agk+1Δfk
with difference operator Δ:Δfk=fk+1fk.
 
My solution
 
Get j=knjkn<j+1j2nk<(j+1)2n
 
Therefore if j2n+1k(j+1)2n then kn=j
 
Note:
 
j=0j2n+1=1
 
j=n2(j+1)2n=n2
 
j=n1(j+1)2n=n>n1
 
So that sum became 
 
S=n1k=1sin((2kn+1)π2n)=sin((2(n1)n+1)π2n)+n2k=1sin((2kn+1)π2n)=sin((2n1)π2n)+n2j=0((j+1)2nj2n)sin((2j+1)π2n)=sin(π2n)+n2j=0Δ(j2n)sin((2j+1)π2n)
Using the [1], we get
S=sin(π2n)+[sin((2j+1)π2n)j2n]n1j=0n2j=0(j+1)2n(sin((2j+3)π2n)sin((2j+1)π2n))=(n1)sin(π2n)2sin(π2n)n2j=0(j+1)2ncos((j+1)πn)=(n1)sin(π2n)2sin(π2n)n1j=1j2ncos(jπn)=A
 
With sum A, using reverse summand property we get
A=n1j=1j2ncos(jπn)=n1j=1(nj)2ncos((nj)πn)=n1j=1n2j+j2ncos(jπn)=A+n1j=1(2jn)cos(jπn)A=12n1j=1(2jn)cos(jπn)
 
We get
 
cos(jπn)=12sin(π2n)[sin((2j+1)π2n)sin((2j1)π2n)]=12sin(π2n)Δ[sin((2j1)π2n)]
 
continue using [1] :)
 
A=(2jn)4sin(π2n)sin((2j1)π2n)|nj=114sin(π2n)n1j=12sin((2j+1)π2n)
 
A=n12+14sin2(π2n)n1j=1Δ[cos(jπn)]=n12+14sin2(π2n)cos(jπn)|nj=1=n121+cos(πn)4sin2(π2n)=n122cos2(π2n)4sin2(π2n)
 
Therefore:
 
S=(n1)sin(π2n)2sin(π2n)A=(n1)sin(π2n)(n1)sin(π2n)+cos2(π2n)sin(π2n)=cot(π2n)cos(π2n)
 
来源:http://math.stackexchange.com/questions/404573/sum-sum-sin-left-frac2-lfloor-sqrtkn-rfloor-1-pi2n-right?rq=1