Question

What is the general form of telescoping series?

Answer 1

there isn't one

Answer 2

actually u can find it in wiki :)

Answer 3

so would you recognise that an expression is a telescoping series?

Answer 4

Like how do i know that the thing i posted is a telescopic series?

Answer 5

ok and your job is to add this right?

Answer 6

it's pretty obvious from observation

Answer 7

yes

Answer 8

you will see it if you start replacing \(r\) by integers starting at 6 to see what you get. don't subtract, just substitute
all but the first term or two will be killed off
try it

Answer 9

so must the expression have at least two terms?

Answer 10

\[\frac{1}{13}-\frac{1}{11}+\frac{1}{15}-\frac{1}{13}+\frac{1}{17}-\frac{1}{15}...\]

Answer 11

this is also telescopic?

Answer 12

just a direct substitution. see what is left
if you were adding up from 6 to \(\infty\) as in
\[\sum_6^{\infty}\frac{1}{2r+1}-\frac{1}{2r-1}\] it looks like you would get \(-\frac{1}{11}\) but in your case you are adding up to \(n\) so you should get an expression involving \(n\)

Answer 13

as for the second one, see what you get when you make \(r=1, r=2, r=3,...\)

Answer 14

oh ok, so the step to do this, is to first see the sum to infinity and then see the sum to n?

Answer 15

no it is the other way around

Answer 16

first sum up to n, then get a formula involving n, and then take the limit as n goes to infinity

Answer 17

for example in your second question you have a sum that looks like
\(-1+1-1+1...\)

Answer 18

oh ok, so how would i use ur method to approach this one?

Answer 19

for the -1+1-1...sum to infinity should give 0?

Answer 20

you should compute the partial sums
\(S_1=-1, S_2=0, S_3=-1,S_4=0, \) etc so in general \(S_n=-1\) if \(n\) is odd and \(S_n=0\) if \(n\) is even
this sequence has no limit so not, the answer is not zero

Answer 21

the sequence of partial sums is \[\{-1,0,-1,0,-1,0,...\}\] and this sequence has no limit for sure

Answer 22

if i use wolfram, how did they produce a 1/2 in the answer?

Answer 23

is it clear what a "partial sum" is?
\[S_n=\sum_{i=m}^na_i\]

Answer 24

yes

Answer 25

i don't know it is a very clever machine that wolfram, but the sum is not 1/2

Answer 26

I guess
\[ \sum_{r=1}^{n}\frac 1 {r+2} + \frac 1 r - \frac 2 {r +2 } = \sum_{r=1}^{n}\frac 1 {r+2} - \frac 1 {r +2 } + \sum_{r=1}^{n}\frac 1 r - \frac 1 {r +2 } \text { back to original method :D}\]

Answer 27

in order to understand this you need to know how to compute a limit if a sequence, and what a partial sum is. for your very first problem see if you can find a general from involving \(n\) for
\[S_n=\sum_{r=6}^n\frac{1}{2r+1}-\frac{1}{2r-1}\]

Answer 28

shouldn't it be
\[ - \frac{1 - (-1)^n}{2}\]

Answer 29

do it by example
in other words find
\[S_1, S_2, S_3, ...\] and then \(S_n\) will be obvious

Answer 30

hmmmm ok~~~

Answer 31

so Find Sn first and then see what the limit to infinity implies?