What is the value of #lim_(n->oo)sum_(k=0)^n(2n+1)/(n+k+1)^2#?

Question

1918 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-05T15:29:51

#1#

We have #sum_(k=1)^n(2n+1)/(n+k+1)^2 le sum_(k=1)^n(2n+1)/((n+k+1)^2-1)# but

#1/((n+k+1)^2-1) = 1/2(1/(n+k)-1/(n+k+2))# and

#sum_(k=1)^n1/((n+k+1)^2-1) = 1/2(1/(n+1)+1/(n+2)-1/(2n+1)-1/(2n+2))# or more compactly

#sum_(k=1)^n1/((n+k+1)^2-1) =1/4(4n^2+5n)/(2 n^3 + 7 n^2+ 7 n+2)#

now we have

#lim_(n->oo)sum_(k=1)^n(2n+1)/((n+k+1)^2-1)=lim_(n->oo)1/4((2n+1)(4n^2+5n))/(2 n^3 + 7 n^2+ 7 n+2) = 1#

so

#lim_(n->oo)sum_(k=1)^n(2n+1)/(n+k+1)^2 le 1#

or

#lim_(n->oo)sum_(k=0)^n(2n+1)/(n+k+1)^2-lim_(n->oo)(2n+1)/(n+1)^2 le 1# so

#lim_(n->oo)sum_(k=0)^n(2n+1)/(n+k+1)^2 le 1#

Now #int_(xi=0)^(xi=n) (d xi)/(n+xi+1)^2 le lim_(n->oo)sum_(k=0)^n 1/(n+k+1)^2#

but
#int_(xi=0)^(xi=n) (d xi)/(n+xi+1)^2 =1/(n+1)-1/(2n+1)# and

#lim_(n->oo)(2n+1)(1/(n+1)-1/(2n+1))=1#

so

#1 le lim_(n->oo)sum_(k=0)^n(2n+1)/(n+k+1)^2 le 1# and concluding

# lim_(n->oo)sum_(k=0)^n(2n+1)/(n+k+1)^2=1#