Question

lim

Answer 1

\[\lim_{n \rightarrow \infty} \sum_{i=1}^{n}\frac{ 3 }{ n }\left[ \left( \frac{ i }{ n } \right)^{2} +1 \right]\]

Answer 2

turn it into a riemann sum noobahz

Answer 3

why is the limit even necessary?

Answer 4

the limit is there so that the summation is proper

Answer 5

can't put the infinity on top of the sigma and call it a night?

Answer 6

that also works; the limit just makes it more clear how the expression is evaluated (you're not actually summing up infinite terms)

Answer 7

I know it works, but the limit seems to serve a very important purpose

Answer 8

the limit ensures that all the n variables in the expression are equal

Answer 9

you mean when doing partitions?

Answer 10

i mean replacing all instances of n in the expression with infinity would not work since the different infinities cannot be considered equal

Answer 11

thanks

Answer 12

Take the interval [0,1], partition it into n equal sub-intervals and evaluate the Riemann sum at right end of each sub-interval for the function 3 x^2 +1

Answer 13

The sum would converge to
\[
\int_0^( 3( x^2 +1)dx
\]
I forget the parenthesis in my previous post

Answer 14

now it makes more sense

Answer 15

so the limit imposes a very important purpose after all

Answer 16

it suggests integration

Answer 17

\[
\int_0^1 3( x^2 +1)dx

\]

Answer 18

the summation can be evaluated explicitly without integration - it is essentially the sum of squares. although, integration is easiest in this specific case

Answer 19

x^3 + x
2 - 0 = 2

Answer 20

The final answer is 4

Answer 21

is it?

Answer 22

my my

Answer 23

oh that's right, it's not 3x^2 + 1
lmaoooo

Answer 24

thank you both

Answer 25

@eliassaab
so whenever I see a limit and then sigma notation, I can rewrite it into antiderivative?

Answer 26

No not always. It has to be of the form
\[
\sum_i^n (x_{i+1}-x_i) f( c_i)
\]
Where
\[
x_0=a, x_1, x_2, \cdots x_n=b
\]
is a partition of the interval [a,b] and
\[c_i\in[x_i,x_i]
\]
and f(x) is a continuous function on [a,b]
in this case, the above Riemann sum converge to
\[
\int_a^b f(x) dx
\]

Answer 27

okay, because I was wondering how did you figure out the interval to be evaluated

Answer 28

You will have to find it from the form of the sum

Answer 29

Of course experience counts

Answer 30

If you take the 3 out of your sum, you will find

Answer 31

that 1/n is the length of the sub-intervals. If you take n sub-intervals, n x1/n =1
So the length of the interval is 1 and we are starting from 0, so the interval is [0,1)

Answer 32

It is to see that f(x) = 3( x^2 +1)

Answer 33

i can see that now

Answer 34

thank you once again