求和的一些估计式
计算limn→∞(2√n−n∑k=11√k).
Use √n=∑nk=1(√k−√k−1), then
2√n−n∑k=11√k=n∑k=1(2√k−2√k−1−1√k)=n∑k=11√k(√k−√k−1)2=n∑k=11√k((√k−√k−1)(√k+√k−1)(√k+√k−1))2=n∑k=11√k(√k+√k−1)2.
This shows the limit does exist and limn→∞(2√n−∑nk=11√k)=∑∞k=11√k(√k+√k−1)2.
The value of this sums equals −ζ(12)≈1.4603545. This value is found by other means, though:
2√n−n∑k=11√k=2√n−(ζ(12)−ζ(12,n+1))∼−ζ(12)−12√n+o(1n).
取整函数的求和问题
证明n−1∑k=1sin((2⌊√kn⌋+1)π2n)=cot(π2n)cos(π2n),
where ⌊⋅⌋ is greatest integer function.
Lemma Summation by Pasts (1)http://en.wikipedia.org/wiki/Summation_by_parts
b∑k=afkΔgk=fkgk|b+1k=a−b∑k=agk+1Δfk=fb+1gb+1−faga−b∑k=agk+1Δfk
with difference operator Δ:Δfk=fk+1−fk.
My solution
Get j=⌊√kn⌋⇒j≤√kn<j+1⇒j2n≤k<(j+1)2n
Therefore if ⌊j2n⌋+1≤k≤⌊(j+1)2n⌋ then √kn=j
Note:
j=0⇒⌊j2n⌋+1=1
j=n−2⇒⌊(j+1)2n⌋=n−2
j=n−1⇒⌊(j+1)2n⌋=n>n−1
So that sum became
S=n−1∑k=1sin((2⌊√kn⌋+1)π2n)=sin((2⌊√(n−1)n⌋+1)π2n)+n−2∑k=1sin((2⌊√kn⌋+1)π2n)=sin((2n−1)π2n)+n−2∑j=0(⌊(j+1)2n⌋−⌊j2n⌋)sin((2j+1)π2n)=sin(π2n)+n−2∑j=0Δ(⌊j2n⌋)sin((2j+1)π2n)
Using the [1], we get
S=sin(π2n)+[sin((2j+1)π2n)⌊j2n⌋]n−1j=0−n−2∑j=0⌊(j+1)2n⌋(sin((2j+3)π2n)−sin((2j+1)π2n))=(n−1)sin(π2n)−2sin(π2n)n−2∑j=0⌊(j+1)2n⌋cos((j+1)πn)=(n−1)sin(π2n)−2sin(π2n)n−1∑j=1⌊j2n⌋cos(jπn)⏟=A
With sum A, using reverse summand property we get
A=n−1∑j=1⌊j2n⌋cos(jπn)=n−1∑j=1⌊(n−j)2n⌋cos((n−j)πn)=−n−1∑j=1⌊n−2j+j2n⌋cos(jπn)=−A+n−1∑j=1(2j−n)cos(jπn)⇒A=12n−1∑j=1(2j−n)cos(jπn)
We get
cos(jπn)=12sin(π2n)[sin((2j+1)π2n)−sin((2j−1)π2n)]=12sin(π2n)Δ[sin((2j−1)π2n)]
continue using [1] :)
A=(2j−n)4sin(π2n)sin((2j−1)π2n)|nj=1−14sin(π2n)n−1∑j=12sin((2j+1)π2n)
A=n−12+14sin2(π2n)n−1∑j=1Δ[cos(jπn)]=n−12+14sin2(π2n)⋅cos(jπn)|nj=1=n−12−1+cos(πn)4sin2(π2n)=n−12−2cos2(π2n)4sin2(π2n)
Therefore:
S=(n−1)sin(π2n)−2sin(π2n)A=(n−1)sin(π2n)−(n−1)sin(π2n)+cos2(π2n)sin(π2n)=cot(π2n)cos(π2n)