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

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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

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

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