Question

Summation Notation

http://puu.sh/ecGJz/65efeab072.png

See Below!

Answer 1

\[\lim_{N \rightarrow \infty} \sum_{s=1}^{N} \frac{ 3s^2-2s-4 }{ N^3 }\]

Answer 2

Hint was to use linearity and power sums

Answer 3

do you know sum s=1 to N of s ?

Answer 4

\[ \sum_{s=1}^{N} s=? \]

Answer 5

is it infinity?

Answer 6

assume N is some fixed number.

Answer 7

if n is 5, then it should be 1+2+3+4+5

Answer 8

Here is a list of some summations
https://en.wikipedia.org/wiki/Summation#Some_summations_of_polynomial_expressions

Answer 9

are u asking for the power sum?

Answer 10

\[\frac{ N^2 }{ 2} + \frac{ N }{ 2 }\]

Answer 11

yes. but there is a "closed formula"
if the last number is N, the sum is N*(N+1)/2
so, in your example, with N=5, the sum should be (is!) 5*(5+1)/2= 5*6/2 = 5*3= 15

Answer 12

sorry i got confused, i just self-learned it in few minutes ago

Answer 13

the point is, for your original problem, we can break it into 3 separate problems

Answer 14

yes, that is the same as N*(N+1)/2 (if we distribute the N we get
(N^2 +N)/2 = N^2 / 2 + N/2

in your problem you have -2s (inside the sum)
but we know
\[ \sum_{s=1}^{N} s= \frac{N^2}{2} + \frac{N}{2} \]
so
\[\sum_{s=1}^{N}-2 s= -2\sum_{s=1}^{N} s= -2\left(\frac{N^2}{2} + \frac{N}{2} \right)\]

Answer 15

there is also the sum of -4
but that simplifies to -4N

now do the 3 s^2
which is 3 * sum(s^2)

Answer 16

\[\frac{ 1 }{ N^3 } \sum_{N=1}^{N} 3 ( \frac{ N^3 }{ 3 } + \frac{ N^2 }{ 2 } + \frac{ N }{ 6 }) - 2 (\frac{ N^2 }{ 2 }+\frac{ N }{ 2 })-4\]

Answer 17

ok except for the last term
\[ \sum_{s=1}^{N} -4 = -4 \sum_{s=1}^{N} 1 = -4 N \]
notice the sum of N -4's or -4 times the sum of N 1's is -4 N (not just -4)

Answer 18

-4N*

Answer 19

now distribute the 3 in the first part, distribute the -2 in the middle part...

Answer 20

\[1 + \frac{ 3 }{ 2N } + \frac{ 3 }{ 6N^2 } - \frac{ 1 }{ N } - \frac{ 1 }{ N^2 } - \frac{ 4 }{ N^2 }\]

Answer 21

i distributed the 1/n^3 as well

Answer 22

I would not have done the 1/N^3 (yet)
next, combine like terms

Answer 23

in that case i get \[N^3 + \frac{ 3N^2 }{ 2 } + \frac{ N }{ 2 } - N^2 - N -4N\]

Answer 24

so we have N^3
we have 3N^2/2 - 2N^2/2 which can be combined.
we have +N/2 - 2N/2 - 8N/2 , ditto

Answer 25

I got \[N^3 + \frac{ N^2 }{ 2 } - \frac{ 9N }{ 2 }\]

Answer 26

now divide each term by N^3

Answer 27

\[1 + \frac{ 1 }{ 2N } - \frac{ 9 }{ }\]

Answer 28

9 over 2N^2

Answer 29

yes. now as N gets "big", the 2 terms with N in the bottom approach ?

Answer 30

0

Answer 31

and the summation (in the limit which means "approaches" very very closely)
?

Answer 32

1

Answer 33

we could have done this problem by "taking the limit" when you wrote
\[ 1 + \frac{ 3 }{ 2N } + \frac{ 3 }{ 6N^2 } - \frac{ 1 }{ N } - \frac{ 1 }{ N^2 } - \frac{ 4 }{ N^2 } \]
but I like doing the extra work to first "clean it up"
then take the limit.

Answer 34

o

Answer 35

Would you recommend memorizing the power sums?

Answer 36

I know sum of integers, but for sum of squares I tend to look it up just to be sure.
For a test (assuming they expect you to memorize it, then I would). Otherwise, just remember there is such a thing and you can google for the formula.

Answer 37

Ok, got it. Thanks man!