Evaluate lim(n->inf)$\displaystyle \sum^n_{k=1}\frac{(n+2k)^2}{n^3}$ and prove your result.

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- Feb 10th 2007, 10:54 PMtttcomraderEvaluate a Riemann sum
Evaluate lim(n->inf)$\displaystyle \sum^n_{k=1}\frac{(n+2k)^2}{n^3}$ and prove your result.

- Feb 11th 2007, 03:48 AMtopsquark
Your summand is:

$\displaystyle \frac{n^2 + 4kn + 4k^2}{n^3} = \frac{1}{n} + \frac{4k}{n^2} + \frac{4k^2}{n^3}$

Now, of all of these the 1/n sum is going to be the most important because the sum of the other two as n approaches infinity converges, whereas the 1/n series does not. So your series does not converge.

There are innumerable ways to prove that the 1/n series does not converge and that the other two do so I leave this part of the proof to you.

-Dan - Feb 11th 2007, 04:41 AMCaptainBlack
Evaluate:

Lim(n -> infty) [Sum(k=1..n) (n+2k)^2 / n^3]

First look at:

S(n)= Sum(k=1..n) (n+2k)^2 / n^3 = Sum(k=1..n) (n^2+4kn+4k^2) / n^3

.....=[Sum(k=1..n)1/n] + (4/n^2)[Sum(k=1..n)k] + (4/n^3)[Sum(k=1..n)(k^2)]

.....=1 + (4/n^2) n(n+1)/2 + (4/n^3) n(n+1)(2n+1)/6

.....= 1 + (2+2/n) + (4/3 +2/n +(2/3)/n^2)

Hence:

Lim(n -> infty) [Sum(k=1..n) (n+2k)^2 / n^3]=Lim(n -> infty) S(n) = 13/3

(But I would recomend that you check my working!)

RonL - Feb 11th 2007, 04:43 AMCaptainBlack
- Feb 11th 2007, 03:00 PMtopsquark