Vtyjkyuk

$$lim_nto infty sum_r=1^n fracr^44r^2-1$$
I am a class 12th student and this question is part of a previously given assignment.My solution is as follows...
$$lim_nto infty sum_r=1^n fracr^44r^2-1$$
$$=frac116lim_nto infty
sum_r=1^nleft(4r^2+1+ frac14r^2-1right)$$
$$=lim_nto inftyfracn(n+1)(2n+1)24 ;+;lim_nto infty(n/16)+;;lim_nto inftysum_r=1^nfrac116(4r^2-1)$$
what should i do after that? ,
$mathbf i,think,answer,is , infty $

Just notice that $$fracn(n+1)(2n+1)24+fracn16+sum_r=1^nfrac116(4r^2-1)gefracn16 forall ninmathbbN,$$ so $$lim_ntoinftyleft[fracn(n+1)(2n+1)24+fracn16+sum_r=1^nfrac116(4r^2-1)right]gelim_nto +inftyfracn16=+infty$$

You could write
$$fracr^44 r^2-1=frac116+fracr^24+frac132left(frac12 r-1-frac12 r+1right)$$ making, as you almost did,
$$sum_r=1^n fracr^44r^2-1=fracn16+frac n (n+1) (2 n+1)24 +frac132sum_r=1^nleft(frac12 r-1-frac12 r+1right)$$ and notice a telescoping sum
$$sum_r=1^nleft(frac12 r-1-frac12 r+1right)=1-frac12 n+1$$
Adding all terms and simplifying, this would give
$$S_n=sum_r=1^n fracr^44r^2-1=fracn^4+2 n^3+2 n^2+n6 (2 n+1)$$

You can immediately tell the sum diverges, because as n goes to infinity, so does the terms of the sum. Think about it, what should happen as n goes to infinity ? If you are adding an infinite amount of terms, an infinite amount of them must go to cero, as n goes to infinity. Otherwise, your sum blows up!