How do you find a formula for the sum n terms #Sigma1/n^3(i-1)^2# and then find the limit as #n->oo#?

Answer 1

# sum_(i=1)^n 1/n^3(i-1)^2 = 1/(6n^2)(2n-1)(n-1) #

# lim_(n rarr oo)sum_(i=1)^n 1/n^3(i-1)^2 = 1/3 #

Let # S_n = sum_(i=1)^n 1/n^3(i-1)^2 # # :. S_n = 1/n^3 sum_(i=1)^n (i^2-2i+1) # # :. S_n = 1/n^3 { sum_(i=1)^n i^2 - 2sum_(i=1)^ni + sum_(i=1)^n1 } #
And using the standard results: # sum_(r=1)^n r = 1/2n(n+1) # and # sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) #

We have;

# S_n = 1/n^3 { 1/6n(n+1)(2n+1) -(2)1/2n(n+1) + n } # # :. S_n = n/(6n^3) { (n+1)(2n+1) - 6(n+1) + 6 } # # :. S_n = 1/(6n^2) { 2n^2+3n+1-6n-6+6 } # # :. S_n = 1/(6n^2) { 2n^2-3n+1 } # # :. S_n = 1/(6n^2)(2n-1)(n-1) #
Now we examine the behaviour of # S_n # as # n rarr oo #. We have;
# S_n = 1/(6n^2) { 2n^2-3n+1 } # # :. S_n = 1/3 -1/(2n)+1/(6n^2) # # :. lim_(n rarr oo)S_n = lim_(n rarr oo) { 1/3 -1/(2n)+1/(6n^2) } # # :. lim_(n rarr oo)S_n = lim_(n rarr oo) (1/3) - lim_(n rarr oo)(1/(2n)) + lim_(n rarr oo)(1/(6n^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 = 1/3 - 0 + 0 # # :. lim_(n rarr oo)S_n = 1/3 #
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Answer 2

To find a formula for the sum of n terms of the series ( \sum_{i=1}^{n} \frac{1}{n^3(i-1)^2} ), we can first rewrite the series in a telescoping form. Factoring out ( \frac{1}{n^3} ) from each term yields:

[ \frac{1}{n^3} \sum_{i=1}^{n} \frac{1}{(i-1)^2} ]

Next, we can use the formula for the sum of squares of consecutive integers:

[ \sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6} ]

By substituting ( i - 1 ) for ( i ), we get:

[ \sum_{i=1}^{n} (i-1)^2 = \frac{n(n - 1)(2n - 1)}{6} ]

So, the sum of the series becomes:

[ \frac{1}{n^3} \cdot \frac{n(n - 1)(2n - 1)}{6} ]

[ = \frac{(n - 1)(2n - 1)}{6n^2} ]

To find the limit of this expression as ( n ) approaches infinity, we can apply the limit properties. After simplifying the expression, we get:

[ \lim_{n \to \infty} \frac{(n - 1)(2n - 1)}{6n^2} = \lim_{n \to \infty} \frac{2n^2 - 3n + 1}{6n^2} ]

[ = \lim_{n \to \infty} \left( \frac{2}{6} - \frac{3}{6n} + \frac{1}{6n^2} \right) ]

[ = \frac{1}{3} ]

Therefore, the limit of the series ( \sum_{i=1}^{n} \frac{1}{n^3(i-1)^2} ) as ( n ) approaches infinity is ( \frac{1}{3} ).

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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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