Question

Hi everyone...Can anyone help me try and figure out which of 5 series are telescoping? I have a picture of the problem...I don't just want the answer but I would like to discuss if possible! Thanks! :o)
Please just view 17.bmp

Answer 1

Am I correct in assuming I just need to start writing out each series to see if stuff starts to cancel out? And if so, isn't there a better way to solve this problem?

Answer 2

a) looks like a P-Series if you turn the denominator into x^(1/2)...so I am leaning towards a) not being a telescoping series...

Answer 3

for b), the first 3 terms become 13/6, 13/40, and 13/126...this looks like it is reducing and positive, so this seems like it would be a geometric series...hmmm

Answer 4

for c) the order goes...7/4 +7/5 -7/6 +7/6 -7/7 +7/7+...+7/(k-4) -7/(k-2)
...that would make the 7/6's cancel as well as the 7/7's etc all the way up until leaving the 7/(k-4) -7/(k-2)...
I believe this series written out would then be 7/4 +7/5+7/(k-4) -7/(k-2)...
Can anyone confirm if this is correct and if this is a telescoping series? Thanks!

Answer 5

for d) I get the series of 4/30, 4/42, then 4/56 which is a reducing, positive series so this one looks like a geometric series and not a telescoping

Answer 6

and for f), it seems like alot of work to write out and calculate but i think you should get an increasingly large, positive, continuing series which smells like a geometric to me...

Answer 7

So am I correct when I say that thje telescoping series here is choice c) ?

Answer 8

a) isn't a telescoping series

Answer 9

b) is neither and it's not geometric, but it's still not a telescoping series

Answer 10

c) definitely looks like a telescoping series due to the cancellations of terms like that

Answer 11

that's hatI thought too but I needed confirmation as I am a little shaky on these

Answer 12

not sure what kind of series d) is, but it's not telescopic

Answer 13

f) is a geometric series, so that's not telescopic either

Answer 14

both b) and d) look like they are positive, reducing, and continuous...i don't know what that is called but aren't those the 3 criteria for utilizing the integral test for converge or diverge? just as a side note anyway?

Answer 15

that sounds about right, basically the nth term is a very small term (for some very large n), so you're adding a smaller and smaller bit to the sum....which is some finite number that isn't diverging

Answer 16

so I got it right then?

Answer 17

yes you did

Answer 18

I think c) is the only telescoping series

Answer 19

yes!

Answer 20

you don't need to find the sum since they just want to know which ones are telescoping series or not

Answer 21

ok...but you do have to write them out so you can tell if they cancel or not right?

Answer 22

yes that's one of the best ways of seeing if it is a telescoping series or not

Answer 23

can you also tell me if this is correct and what the official name is for finding this?
7/4 +7/5+7/(k-4) -7/(k-2)

Answer 24

is this called tfinding the k^th term or something?

Answer 25

well I think the 7/(k-4) and -7/(k-2) terms will cancel with terms somewhere nearby

Answer 26

so you'll be left with just 7/4 +7/5

Answer 27

7/4 +7/5

35/20 +28/20

63/20

Which is equal to the sum of the series in c)

Answer 28

confirmation

http://www.wolframalpha.com/input/?i=sum%287%2F%28n-4%29-7%2F%28n-2%29%2C+n+%3D+8+..+infinity%29+

Answer 29

ok...well what's bugging me is that i had a question before that asked to find the k^th term or something like that or n^th term or something and the answer had k's or n's on the back end...any idea what i'm talking about? lol

Answer 30

you mean the kth term? or the kth partial sum?

Answer 31

yeah, it was like the kth partial sum i think but i didn't actually have to find the answer, only the statement with the k-1 and k at the end or something...in that respect, I was just wondering if I was asked tht in this problem, would the answer 7/4 +7/5+7/(k-4) -7/(k-2) be correct?

Answer 32

hmm let me check

Answer 33

thanks!

Answer 34

i think i have the terms, just not adding them right, one sec

Answer 35

Just a thought...if Sn is called a partial sum, would what I'm asking you possible be called finding the formula for the Nth partial sum?

Answer 36

yes that's exactly what I'm trying to do lol

want me to show you what I have so far?

Answer 37

please, yes lol

Answer 38

oh wait, i misplaced a sign lol, i found it

Answer 39

First term: 7/(8-4) - 7/(8-2) ---> 7/4 - 7/6


Second term: 7/(9-4) - 7/(9-2) ---> 7/5 - 7/7


Third term: 7/(10-4) - 7/(10-2) ---> 7/6 - 7/8 (+7/6 cancels with -7/6 back in first term)


Fourth term: 7/(11-4) - 7/(11-2) ---> 7/7 - 7/9 (+7/7 cancels with -7/7 back in second term)


Fifth term: 7/(12-4) - 7/(12-2) ---> 7/8 - 7/10 (+7/8 cancels with -7/8 back in third term)


Sixth term: 7/(13-4) - 7/(13-2) ---> 7/9 - 7/11 (+7/9 cancels with -7/9 back in fourth term)


Seventh term: 7/(14-4) - 7/(14-2) ---> 7/10 - 7/12 (+7/10 cancels with -7/10 back in fifth term)


Eighth term: 7/(15-4) - 7/(15-2) ---> 7/11 - 7/13 (+7/11 cancels with -7/11 back in sixth term)



General Rule: For any term, the first piece cancels out with the second piece 2 terms back

...
...
...

(k-2)th term: 7/((k-2)-4) - 7/((k-2)-2) ----> 7/(k-6) - 7/(k-4) (the +7/(k-6) will cancel with -7/(k-6) in the (k-4)th term)


(k-1)th term: 7/((k-1)-4) - 7/((k-1)-2) ----> 7/(k-5) - 7/(k-3) (the +7/(k-5) will cancel with -7/(k-5) in the (k-3)th term)


kth term: 7/(k-4) - 7/(k-2) (+7/(k-4) cancels with -7/(k-4) back (k-2)th term)


-------------------------------------------------------


After all the cancellations occur, you are left with the terms: 7/4, 7/5, -7/(k-3) , and -7/(k-2)


Add them up and simplify


(7/4) + (7/5) - ( 7/(k-3) ) - ( 7/(k-2) )


35/20 + 28/20 - ( 7/(k-3) ) - ( 7/(k-2) )


63/20 - ( 7/(k-3) ) - ( 7/(k-2) )


63/20 - 7(k-2)/( (k-3)(k-2) ) - 7(k-3)/( (k-3)(k-2) )


63/20 + ( -7(k-2) - 7(k-3) )/( (k-3)(k-2) )


63/20 + ( -7k+14 - 7k+21 )/( (k-3)(k-2) )


63/20 + ( -14k+35 )/( (k-3)(k-2) )


(63(k-3)(k-2))/( 20(k-3)(k-2) ) + ( 20(-14k+35) )/( 20(k-3)(k-2) )


(63(k-3)(k-2))/( 20(k-3)(k-2) ) + ( 20(-14k+35) )/( 20(k-3)(k-2) )


(63(k^2 - 5k + 6))/( 20(k-3)(k-2) ) + ( -280k+700 )/( 20(k-3)(k-2) )


(63k^2 - 315k + 378)/( 20(k-3)(k-2) ) + ( -280k+700 )/( 20(k-3)(k-2) )


(63k^2 - 315k + 378 - 280k+700 )/( 20(k-3)(k-2) )


( 63k^2-595k+1078 )/( 20(k-3)(k-2) )


( 7(9k^2-85k+154) )/( 20(k-3)(k-2) )



So the kth partial sum, or the nth partial sum (however you look at it) is: Sk = ( 7(9k^2-85k+154) )/( 20(k-3)(k-2) )


As confirmation, you can see the mth partial sum found by wolfram alpha is the same thing (just using m instead of k) (look under the section "Partial sum formula")

http://www.wolframalpha.com/input/?i=sum%287%2F%28n-4%29-7%2F%28n-2%29%2C+n+%3D+8+..+infinity%29+

Answer 40

Here's how I cancelled terms (visually)

Answer 41

i'm looking...

Answer 42

while I'm looking, fast forward this video I just found to almost the end...he leave k's in the equations...

Answer 43

http://www.youtube.com/watch?v=xmyLUFv2mDM

Answer 44

yeah he's basically doing what I did above, just with ln() terms

Answer 45

woa...that will take me awhile to go through!...i'd give you another medal but it won't let me! lol :o)

Answer 46

lol it's ok

Answer 47

and you didn't need to go out as far as i did for the terms, but it helps see what's going on

Answer 48

i need to figure out via your work why i wanted to go k-1 and a k while you went After all the cancellations occur, you are left with the terms: 7/4, 7/5, -7/(k-3) , and -7/(k-2) hmmmm

Answer 49

well I noticed that any particular term had the first piece of it cancel out with the second piece of the previous 2 terms

Answer 50

ex: the first piece of term #6 cancels out with the second piece of term #4

Answer 51

ex: the first piece of term #8 cancels out with the second piece of term #6

Answer 52

etc

Answer 53

so i was on the right track, i just didn't anaylze my work well enough eh?

Answer 54

so that's why I started at the kth term and went back two terms to get the k-1 term and the k-2 terms to see what would be left after cancelling

Answer 55

yes you were, just either left a term out or added wrong

Answer 56

not sure exactly

Answer 57

well between your work and that video, I should be able to figure it out within a half hour or so...i dread my midterm on monday night! ugh! :o(

Answer 58

well just practice practice practice

Answer 59

and you'll be able to do it quicker

Answer 60

and keep in mind that you don't need to extend the terms out so far as I did

Answer 61

i'm trying! thanks to your help! u r a superstar Jim! :o)

Answer 62

lol thx, and you're doing great btw

Answer 63

thanks...that's nice to hear! :o)...btw...i am officially letting you off the hook now and free to return to your life! thanks soooo much!

Answer 64

lol alright, have a good night and good luck on the test

Answer 65

you're welcome

Answer 66

:o) ditto