Question

Sn=Suumation from i=2 to inf = 3/4 -1/2n -1/2(n+1) who it happen?

Answer 1

i am a bit lost on this one

Answer 2

sorry typing mistake * how it happen?

Answer 3

because i can't really read it but it looks like you have a telescoping sum
write out the first few terms and see if that is true

Answer 4

how we found the formula of this partial sum?

Answer 5

can you try in equation editor? i can only make a guess as to what this says

Answer 6

looks like maybe
\[\frac{3}{4}-\frac{1}{2n}-\frac{1}{2(n+1)}\]

Answer 7

\[Sn = \sum_{i=2}^{\infty} 1/i^2 -1 = 3/4 - 1/2n -1/2(n+1)\]

Answer 8

1/(i^2-1)

Answer 9

satelite you are right

Answer 10

oh ok you are looking for
\[\sum_i \frac{1}{i^2-1}\]

Answer 11

how did we find this formula

Answer 12

i bet we can use partial fractions for this one

Answer 13

YES

Answer 14

how did 3/4 -1/2n -1/2(n+1) come?

Answer 15

is it a general formula for this series?

Answer 16

typo there
\[\frac{1}{i^2-1}=\frac{1}{2(i-1)}-\frac{1}{2(i+1)}\]

Answer 17

ok then?

Answer 18

now imagine what happens if you start replacing i by 2, 3, 4, ...

Answer 19

don't simplify, just replace

Answer 20

how can we break it like this 1/(2(i-1) -1/2(i+1) will give us the denominator 2(i^2-1) whereas the original denominator is i^2-1

Answer 21

ok that is a separate question
lets take the first question first, that is, assume for a moment that we know
\[\frac{1}{i^2-1}=\frac{1}{2(i-1)}-\frac{1}{2(i+1)}\]

Answer 22

ok

Answer 23

and start replacing i by 2, 3, 4, 5, ... and see what happens
you get
\[\frac{1}{2}-\frac{1}{6}+\frac{1}{4}-\frac{1}{8}+\frac{1}{6}-\frac{1}{10}+\frac{1}{8}-\frac{1}{12}...\]

Answer 24

with a little practice you will see this right away. the sum "telescopes" that is, you are adding and then subtracting the same terms, so they will end up canceling

Answer 25

i.e. add up to zero

Answer 26

okkkkk gotcha thanks bro

Answer 27

why does the sum go to infinity and have an answer in terms of n ?
me thinks we have a few typos here...

Answer 28

that is, where you have
\[\frac{1}{6}\] there will be a
\[-\frac{1}{6}\] and then
\[\frac{1}{8}\] and a
\[-\frac{1}{8}\] and so on
so what is left?
the only numbers that do not have a "match" that is, the only numbers that remain are
\[\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\] which explains why the first term is
\[\frac{3}{4}\]

Answer 29

now to be pricise, we need to write out what
\[\sum_{i=2}^n\frac{1}{i^2-1}\] is in terms of n

Answer 30

the limit is zero so the sum won't be infinite

Answer 31

isn't? @ satelite

Answer 32

and from the telescoping nature of the sum, we see that what you will have is
\[\frac{3}{4}-\frac{1}{2n}-\frac{1}{n(n+1)}\]

Answer 33

the last two terms didn't get canceled yet, because we only went up to n
now take the limit as n goes to infinity and you get
\[\sum_{i=1}^{\infty}\frac{1}{i^2-1}=\lim_{n\to \infty}\sum_{i=2}^n\frac{1}{i^2-1}=\frac{3}{4}\]

Answer 34

now as to how you get
\[\frac{i}{i^2-1}=\frac{1}{2(i-1)}-\frac{1}{2(i+1)}\] do you know "partial fraction decomposition"? because that is what you need for this one

Answer 35

yes i do

Answer 36

but have you read what I said just above?

Answer 37

then it should be good from here yes?

Answer 38

this line
the limit is zero so the sum won't be infinite

Answer 39

the limit of the second two terms is zero, so the limit of the whole thing is 3/4

Answer 40

yes exactlly

Answer 41

because that is the precise definition of the infinite sum, namely
\[\sum_{i=2}^{\infty} a_i=\lim_{n\to \infty}\sum_{i=2}^na_i\] so you have to find the partial sums, and then compute the limit of the partial sums

i would like to add that most people would say "the sum telescopes" and the only thing that survives is 3/4 and leave it at that, but it is good to be precise

Answer 42

ok got it now I have lil' seperate question but related to series

Answer 43

do me a favor and repost
it is hard to keep scrolling down

Answer 44

ok