How do you find a formula for the sum n terms #sum_(i=1)^n (16i)/n^3# and then find the limit as #n->oo#?

Answer 1

# sum_(i=1)^n (16i)/n^3 = 18/n^2(n+1) #

# lim_(n rarr oo)sum_(i=1)^n (16i)/n^3 = 0 #

Let # S_n = sum_(i=1)^n (16i)/n^3 # # :. S_n = 16/n^3 sum_(i=1)^n i #
And using the standard results: # sum_(r=1)^n r = 1/2n(n+1) #

We have;

# S_n = 16/n^3 { 1/2n(n+1) } # # :. S_n = 18/n^2(n+1) #
Now we examine the behaviour of # S_n # as # n rarr oo #. We have;
# S_n = 18/n^2(n+1) # # :. S_n = 18/n+18/n^2 # # :. lim_(n rarr oo)S_n = lim_(n rarr oo) { 18/n+18/n^2 } # # :. lim_(n rarr oo)S_n = 18lim_(n rarr oo) (1/n) + 18lim_(n rarr oo)(1/n^2) #
And as both #1/n rarr 0# and #1/n^2 rarr 0# as #n rarr oo#, we have;
# :. lim_(n rarr oo)S_n = 0 + 0 # # :. lim_(n rarr oo)S_n = 0 #
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Answer 2

To find a formula for the sum of ( n ) terms of the series ( \sum_{i=1}^n \frac{16i}{n^3} ), we use the formula for the sum of the first ( n ) positive integers, which is ( \frac{n(n + 1)}{2} ).

The given series can be written as ( \frac{16}{n^3} \sum_{i=1}^n i ). Substituting the sum of the first ( n ) positive integers, we have ( \frac{16}{n^3} \times \frac{n(n + 1)}{2} ).

This simplifies to ( \frac{8(n + 1)}{n^2} ).

To find the limit as ( n ) approaches infinity, we divide both the numerator and the denominator by ( n^2 ), resulting in ( \frac{8}{n} + \frac{8}{n^2} ). As ( n ) approaches infinity, both terms approach zero.

Therefore, the limit of the sum as ( n ) approaches infinity is ( 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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