How to evaluate the limit $lim_(n -> oo) sum_(j=1)^n(5j)/n^2$ ?

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Answer 1 · 2015-08-29T18:49:36

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$lim_(nrarroo) sum_(j=1)^n (5j)/n^2$

$lim_(nrarroo) underbrace(sum_(j=1)^n (5j)/n^2)_(n" is a constant here")$

$= lim_(nrarroo) 5/n^2 underbrace(sum_(j=1)^n j)_((n(n+1))/2$

$= lim_(nrarroo) 5/n^2 (n(n+1))/2$

$= 5/2 lim_(nrarroo) (n(n+1))/n^2$

To evaluate $lim_(nrarroo) (n(n+1))/n^2$

Either use:

$lim_(nrarroo)(n(n+1))/n^2 = lim_(nrarroo)(n^2+n)/n^2 = 1$

Or use

$lim_(nrarroo)(n(n+1))/n^2 = lim_(nrarroo)(n/n (n+1)/n)$

$= lim_(nrarroo)(1 * (1+1/n)) = 1*1 =1$

So we finish:

$lim_(nrarroo) sum_(j=1)^n (5j)/n^2 = 5/2 lim_(nrarroo) (n(n+1))/n^2$

$= 5/2*1 = 5/2$

How to evaluate the limit lim_(n -> oo) sum_(j=1)^n(5j)/n^2lim_(n -> oo) sum_(j=1)^n(5j)/n^2?