Question

Calc 2,

Can you please tell me why is this considered a 'geometric Serie" instead of arithmetic ?

Answer 1

@AllTehMaffs

Answer 2

could you post the series please?

Answer 3

It's number 7 , I am sorry that I forgot to mention this .

Answer 4

number 7 ?
I'm afraid that nobody here is reading your particluar book at the moment, so you'll have to actually write out the series ;)

Answer 5

How silly of me D: https://www.dropbox.com/s/08x8kcrjisdzdqx/Screenshot%202013-12-14%2017.26.13.jpg

Answer 6

here its this eerie

Answer 7

serie*

Answer 8

this is a series?

Answer 9

you can break \(\dfrac1{(k+2)(k+3)}\) into \(\dfrac1{k+3}-\dfrac1{k+2}\) and expand the summation

Answer 10

you'll find that the terms cancel out each other

Answer 11

isn't this a sum/sequence ? Or am I wrong ? It's my first day in this chapter so parton me if I make some horrible mistakes , I wish its ok :/

Answer 12

oh wait sorry my fault this is a series

Answer 13

I already did that , but what is next

Answer 14

after expanding it *

Answer 15

and i would say that this is neither an arithmetic series nor a geometric series

Answer 16

(P.S. and the singular form of series is, you guessed it, still series)

Answer 17

and why are you saying its neither of those? then what is it ? can you please tell me how did you find out this?

Answer 18

so the terms will cancel out each other and you'll be left with \(\dfrac1{\infty+2}-\dfrac13\) which is actually just \(\dfrac13\)

Answer 19

an arithmetic sequence has to have a common difference while a geometric sequence has to have a common ratio

Answer 20

i mean \(-\dfrac13\) sorry

Answer 21

do you need the steps or you can actually do it myself or you want me to guide you instead of just giving you all the steps

Answer 22

I prefer to guide me of its possible bro :D

Answer 23

But I am really confused

Answer 24

I can understand that a sequence can have a same difference between its term

Answer 25

the series sums to 1/2, but I thought the question was just why it's geometric, i.e. what's the common ratio?

Answer 26

but what do you mean by different ratio ? basically what exactly is ratio ?

Answer 27

lemme write down what we have now, we have:
\[\Large\sum_{k=1}^\infty\frac1{(k+2)(k+3)}=\sum_{k=1}^\infty\left(\dfrac1{k+3}-\dfrac1{k+2}\right)\]

Answer 28

I have the solution and it says that it sums to ⅓ , but I don't know how tha process took place

Answer 29

ratio of a and b is \(\dfrac ba\)

Answer 30

aaah

Answer 31

ok so far I have come across, I even expanded the s eirie ,but after that , I have to admit that I am completely clueless

Answer 32

sorry i made a mistake there lemme write it again

Answer 33

\[\frac{1}{(k+2)(k+3)}=\frac{1}{k+2}-\frac{1}{k+3}\]

Answer 34

we have:
\[\large\sum_{k=1}^\infty\frac1{(k+2)(k+3)}=\sum_{k=1}^\infty\left(\frac1{k+2}-\frac1{k+3}\right)\]

Answer 35

yes

Answer 36

and the answer should be \(\dfrac13\) instead of 0.5

Answer 37

yes but how, the processs s what I really need to know and be able to reproduce

Answer 38

sorry, it is 1/3

Answer 39

This is called a telescoping series and the methods described above are the right way to do it. The answer is 1/3

Answer 40

expand that and we have:
\[\frac1{1+2}-\frac1{1+3}+\frac1{1+3}-\frac1{1+4}+\cdots+\frac1{\infty+2}-\frac1{\infty+3}\]

Answer 41

cancel them and we got \(\dfrac1{1+2}\) which is \(\dfrac13\)

Answer 42

what's the "k" here ? http://screencast.com/t/ojwnBozWwZ

Answer 43

1

Answer 44

The formal way of doing it is to compute the partial sum
\[
s_n=\sum _{k=1}^n \frac{1}{(k+2) (k+3)}=\frac{1}{3}-\frac{1}{n+3}
\]
And let n go to infinity

Answer 45

How did you find out this part in one step http://screencast.com/t/6K4sooTrfU5 @eliassaab

Answer 46

I cheated, all the remaining terms cancel as described above/

Answer 47

To determine whether it's an arithmetic sequence or a geometric sequence, we consider the first three terms:
Difference of first two terms\(=\dfrac1{3\times4}-\dfrac1{4\times5}=\dfrac1{30}\)
Difference of second two terms\(=\dfrac1{4\times5}-\dfrac1{5\times6}=\dfrac1{60}\)
Since they are not equal, it is not an arithmetic sequence.
Ratio of the first two terms\(=\dfrac1{4\times5}\div\dfrac1{3\times4}=\dfrac35\)
Ratio of the second two terms\(=\dfrac1{5\times6}\div\dfrac1{4\times5}=\dfrac23\)
Since they are not equal, it is not a geometric sequence.

Answer 48

I do not want to repeat what the others have helped with so far.

Answer 49

so if its not a geographic neither an arithmetic the its a telescopic ?

Answer 50

then*

Answer 51

no

Answer 52

it can be neither of the three

Answer 53

ok I wil practice this and I will learned the technique, thanks so much kc_kennylau for your patience and help

Answer 54

no problem :) give credits to others too :)

Answer 55

ofc !, thank you so much @TuringTest @eliassaab @Zarkon

Answer 56

Here all the steps to compute the partial sum
\[
s_n=\sum _{k=1}^n \frac{1}{(k+2) (k+3)}=\sum _{k=1}^n
\left(\frac{1}{k+2}-\frac{1}{k+3}\right)=\\\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+ \cdots -\frac{1}{n+2}+\frac{1}{n+2}-\frac{1}{n+3}=\frac{1}{3}-\frac{1}{n+3}

\]

Answer 57

YW