What is sum_(k=0)^(4n+2)1/(k+1)+1/(k+3)4n+2∑k=01k+1+1k+3 ?
1 Answer
Explanation:
We seek a way to evaluate the sum
S=sum_(k=0)^m(1/(k+1)+1/(k+3))S=m∑k=0(1k+1+1k+3)
Note:
- The substitution
color(red)(m=4n+2m=4n+2 , is made for simplicity
Definition:
The
color(blue)(H_n=sum_(k=1)^(n)1/kHn=n∑k=11k
Our goal will be to express the original sum in terms of a harmonic number.
It may seems strange to express the original sum, by another sum. But, you may think of the harmonic numbers, similar to the factorial, in the way evaluate them.
Express the sum in terms of a harmonic number:
Using some basic summation identities
S=sum_(k=0)^m1/(k+1)+sum_(k=0)^m1/(k+3)S=m∑k=01k+1+m∑k=01k+3
color(white)(S)=sum_(k=1)^(m+1)1/k+sum_(k=3)^(m+3)1/kS=m+1∑k=11k+m+3∑k=31k
color(white)(S)=sum_(k=1)^(m+1)1/k+sum_(k=1)^(m+3)1/k-3/2S=m+1∑k=11k+m+3∑k=11k−32
color(white)(S)=2sum_(k=1)^(m+1)1/k+1/(m+2)+1/(m+3)-3/2S=2m+1∑k=11k+1m+2+1m+3−32
Or in terms of a harmonic number
S=2H_(m+1)+1/(m+2)+1/(m+3)-3/2S=2Hm+1+1m+2+1m+3−32
For your problem you may substitute back
S=2H_(4n+3)+1/(4n+4)+1/(4n+5)-3/2S=2H4n+3+14n+4+14n+5−32
Bonus info
A fairly reasonable approximation of a harmonic number is
H_n~~ln(n)-gammaHn≈ln(n)−γ
The drawing at the upper right hand corner illustrates this quite well
