Thread: Evaluate a Riemann sum

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

Originally Posted by tttcomrader

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

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

Originally Posted by tttcomrader

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

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

Originally Posted by topsquark

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

But still it does

(did it really take me most of an hour to type this? Judging by the time stamps on my post and
topsquarks it must have taken one of us fifty odd mins.)

RonL

Originally Posted by CaptainBlack

But still it does

(did it really take me most of an hour to type this? Judging by the time stamps on my post and
topsquarks it must have taken one of us fifty odd mins.)

RonL

Ah, I've run into this SNAFU before. I forgot we are considering the sum, not just the individual terms.

-Dan