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#