Summations and limits

[bcddd214]

I hope my computation is correct and the question is clear.

lim(n??)??(i=1)^n->(1/n^3(i-1)^2)

The instructions are:
use the properties of sigma notation to find a formula for the given sum of n terms. Then use the formula to find the limit as n---) oo

I am pasting from word so it might be a little dirty but I get this far;
=1/n^3 ?_(i=1)^n?(i-1)(i-1)
=1/n^3 ?_(i-1)^n?(i^2-2i-1)
=1/n^3 [?_(i=1)^n?i^2 -2?_(i=1)^n?i-?_(i=1)^n?1]
1/n^3 ([n(n+1)(2n+1)/6]-2[n(n+1)/2]-1(n) )
1/n^3 ([n(n+1)(2n+1)/6]-2[n(n+1)/2]-1(n) )???

The last line is a dup because at this point I am supposed to be plugging in a number but when I go to look for that number, it's an 'n' and at that point, well, just think "monkey with a lightbulb".
I know I am supposed to be testing the limit at this point but the end result is a whole number but I just don't see the connect.

 

[galactus]

This is a Riemann sum.

If you expand $\displaystyle \frac{(i-1)^{2}}{n^{3}}$ you get $\displaystyle \frac{i^{2}}{n^{3}}-\frac{2i}{n^{3}}+\frac{1}{n^{3}}$

Now, the sum of the i^2 is $\displaystyle \sum_{i=1}^{n}i^{2}=\frac{n(n+1)(2n+1)}{6}$ and the sum of the i's is $\displaystyle \sum_{i=1}^{n}i=\frac{n(n+1)}{2}, \;\ \text{and} \;\ \sum_{i=1}^{n}1=n$

Sub these into $\displaystyle \sum_{i=1}^{n}\left(\frac{i^{2}}{n^{3}}-\frac{2i}{n^{3}}+\frac{1}{n^{3}}\right)=\frac{1}{n^{3}}\sum_{i=1}^{n}i^{2}-\frac{2}{n^{3}}\sum_{i=1}^{n}i+\frac{1}{n^{3}}\sum_{i=1}^{n}1$.

After subbing in the identities, It is now in terms of n alone. Take your limit.

Also, if we look close we can put this into the form of the integral it represents.

It appears to be the left endpoint Riemann sum, $\displaystyle x_{i}=a+(i-1)\Delta x$.

In this case, a=0, $\displaystyle \Delta x=\frac{1}{n}, \;\ x_{i}=\frac{i-1}{n}$

This works out to be $\displaystyle f(x_{i})\Delta x=\left(\frac{i-1}{n}\right)^{2}\cdot\frac{1}{n}$

$\displaystyle \int_{0}^{1}x^{2}dx$

You should get the same for the integration and your limit.

 

[bcddd214]

\displaystyle\sum_{i=1}^n\{[{n{n+1}{2n+1}/6}/n^3 ]-2[{n{n+1}/2}/n^3 ]-1{n}}

Is this what you mean?

 

[bcddd214]

My dirty word paste on how I got that madness
lim?(n??)???_(i=1)^n?1/n^3 (i-1)^2 ?
=1/n^3 ?_(i=1)^n??((i-1))/n^3 ((i-1))/n^3 ?
=?_(i-1)^n?((i^2-2i+1))/n^3
=?_(i-1)^n??i^2/n^3 -2i/n^3 ?+1/n^3
([(n(n+1)(2n+1)/6)/n^3 ]-2[(n(n+1)/2)/n^3 ]-1(n) )

 

[bcddd214]

My dirty word paste on how I got that madness
lim?(n??)???_(i=1)^n?1/n^3 (i-1)^2 ?
=1/n^3 ?_(i=1)^n??((i-1))/n^3 ((i-1))/n^3 ?
=?_(i-1)^n?((i^2-2i+1))/n^3
=?_(i-1)^n??i^2/n^3 -2i/n^3 ?+1/n^3
([(n(n+1)(2n+1)/6)/n^3 ]-2[(n(n+1)/2)/n^3 ]-1(n) )

Not sure which is worse, paste from word or my horrible attempt at LaTex

 

[bcddd214]

I found something on the internet that leads me to this.
Is this correct?

lim?(n??)???_(i=1)^n->1/n^3 (i-1)^2 ?
=1/n^3 ?_(i=1)^n->(i-1)(i-1)
=1/n^3 ?_(i-1)^n->?i^2-2i+1?
=1/n^3 [?_(i-1)^n->?i^2-?_(i=1)^n->2i?+?_(i=1)^n->1]
=1/n^3 ([n(n+1)(2n+1)/6]-2[n(n+1)/2]-1(n) )
lim(n??)??=(1/n^3 [n(n+1)(2n+1)/6]-2/n^3 [n(n+1)/2]-1/n^3 (n) )?
lim(n??)??=(1/6 [(2n^3….)/n^3 ]-2/2 [n(n+1)/n^3 ]-1/1 (n/n^3 ) )?
lim(n??)??=(1/6 [2]-2/2 [0]-1/1 (0) )?
lim(n??)??=2/6?

 

[galactus]

I did not peruse all that, but yes, 2/6=1/3 is the limit.

 

[bcddd214]

Can you check my work on this one?


lim(n??)??_(i=1)^n?16i/n^2
=1/n^2 ?_(i=1)^n?16i
=(16/n^2 [n(n+1)/2] )
=(16/2 [n(n+1)/n^2 ] )
=(16/2 (0) )=0

 

[galactus]

Please do not copy and paste from Word. It doe snot display correctly.

Is that $\displaystyle \lim_{n\to \infty}\sum_{i=1}^{n}\frac{16i}{n^{2}}$

If so, the sum is not 0. It is 8

You have 16/2=8. Where did that 0 come from?.

 

[mmm4444bot]

= (16/2 [n(n + 1)/n^2])

The expression in blue simplifies to 1 + 1/n.

That expression does not approach zero, as n approaches infinity.

 

[bcddd214]

Please do not copy and paste from Word. It doe snot display correctly.

Is that $\displaystyle \lim_{n\to \infty}\sum_{i=1}^{n}\frac{16i}{n^{2}}$

If so, the sum is not 0. It is 8

You have 16/2=8. Where did that 0 come from?.

Why does that blur space instead of a white space scare you so much?

 

[bcddd214]

Re:

= (16/2 [n(n + 1)/n^2])

The expression in blue simplifies to 1 + 1/n.

That expression does not approach zero, as n approaches infinity.

16/2 or 8!

 

[mmm4444bot]

or 8!

Eight factorial ? :shock:

:wink: