Question

last question about series.

Answer 1

\[\sum_{n=1}^{\infty}\left(\frac{1}{n^2+3n+2}\right)\]

Answer 2

\[\lim_{n \rightarrow \infty}A_n=0\] so I know it is going to converge for the least part (same I would obtain by integration - that is converges to something, BUT integration is longer)

Answer 3

you just want to knw whether it converges or not right

Answer 4

i am given a hint that it is a telescoping series.
I need to know converges to waht

Answer 5

I got the converges part (hopefully)

Answer 6

start by writing it as partial fractions

Answer 7

hint given to me is: it is a telescoping series

Answer 8

Well, would it be\[\frac{A}{(n+2)}+\frac{B}{(n)}=\frac{A}{n(n+2)}\]
\[A(n)+B(n+2)=1\]\[A=-1/2\]\[B=1/2\]\[\frac{1}{2n}-\frac{1}{2n+4}\]

Answer 9

\[\frac{-1/2}{n+2}+\frac{1/2}{n}\]this is how I got the last

Answer 10

\[\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n+2}\]??

Answer 11

don't say it I think I got it

Answer 12

hey wait

\[n^2+3n+2 = (n+1)(n+2)\]

right

Answer 13

I still got it, but tnx

Answer 14

\[\frac{1}{2}\left(\color{blue}{\sum_{n=1}^{\infty}\frac{1}{n+1}-\sum_{n=1}^{\infty}\frac{1}{n+2}}\right)\]

Answer 15

that 1/2 should not be there

Answer 16

the A and B

Answer 17

http://www.wolframalpha.com/input/?i=partial+fractions+%5Cfrac%7B1%7D%7Bn%5E2%2B3n%2B2%7D

Answer 18

oh I will look over that later then

Answer 19

but I think I grasped the more importnat concept right now

Answer 20

\[\color{blue}{\sum_{n=1}^{\infty}\frac{1}{n+1}-\sum_{n=1}^{\infty}\frac{1}{n+2}}\]\[\color{blue}{\sum_{n=2}^{\infty}\frac{1}{n}-\sum_{n=3}^{\infty}\frac{1}{n}}\]so we will get just the second term, and the rest will be canceled out.

my answer is: Converges to 1/2

Answer 21

(if I did it great, if I did something crazy then I apologize for rushing)

Answer 22

that doesnt make sense

Answer 23

I got the answer correctly though, acc to wolfram

Answer 24

and what makes no sense, the entire thing or part(s) ?

Answer 25

telescoping works like this :

\[\sum\limits_{n=1}^{\infty} f(n) - f(n+1) = f(1)\]

provided \(\lim\limits_{n\to \infty} f(n) = 0\)

Answer 26

the provided is true, and what I wrote was very different but I think I conceptualized it differently.

1/[n+1] MINUS 1/[(n+1) + 1 ]

Answer 27

and f(1) is 1/2 in this case

Answer 28

we have
\[\color{blue}{\sum_{n=1}^{\infty}\frac{1}{n+1}-\frac{1}{n+2}}\]

first define \(f(n) = \dfrac{1}{n+1}\), then above series is of form

\[\color{blue}{\sum_{n=1}^{\infty}f(n) - f(n+1)}\]

Answer 29

well, but the reason it works it because f(1) is the only thing that remains from subtraction.

that is (in this case)

[1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ....]
minus
[1/3 + 1/4 + 1/5 + 1/6 + ....]
=
1/2

Answer 30

I was trying to say this, but kind of failed to do so

Answer 31

I see you have the correct concept

Answer 32

\[\color{blue}{\sum_{n=1}^{\infty}\frac{1}{n+1}-\sum_{n=1}^{\infty}\frac{1}{n+2}}\]\[\color{blue}{\sum_{n=2}^{\infty}\frac{1}{n}-\sum_{n=3}^{\infty}\frac{1}{n}}\]\[\color{blue}{\sum_{n=2}^{1}\frac{1}{n}+}\color{blue}{\sum_{n=3}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\frac{1}{n}}\]
=1/2

this is how I was trying t put it

Answer 33

but you are saying, it is just like saying

(Undefined ) - (Undefined) ?

Answer 34

yes

Answer 35

yes, that is a good line.

Answer 36

just should not have separated those sigmas.... tnx once again !!

Answer 37

hey no, actually it is okay to separate them like that.. sry i was confusing with something else

Answer 38

so the way I wrote it at first is valid (as far as notations and defined/undefined) goes ?

Answer 39

very cool:)
well... prohibiting something that is permitted is better than Visa versa.

Answer 40

\[\begin{align}\color{blue}{\sum_{n=1}^{\infty}\frac{1}{n+1}-\frac{1}{n+2}} &= \color{blue}{\sum_{n=1}^{\infty}\frac{1}{n+1}-\sum_{n=1}^{\infty}\frac{1}{n+2}} \\~\\&= \color{blue}{\sum_{n=0}^{\infty}\frac{1}{n+2}-\sum_{n=1}^{\infty}\frac{1}{n+2}} \\~\\&= \color{blue}{\dfrac{1}{0+2}+\sum_{n=1}^{\infty}\frac{1}{n+2}-\sum_{n=1}^{\infty}\frac{1}{n+2}} \\~\\&= \color{blue}{\frac{1}{2} + 0} \end{align}\]

I don't see anything wrong in writing like this..

Answer 41

in this case though the answer is exactly 1/2.
I mean (even) without "approaches 1/2"

Answer 42

yes, that is what I was doing ... tnx for verifying !!

Answer 43

I know why got the decompositon incorrrectly

Answer 44

A/(n+1) + B/(n+2) = 1/(n+1)(n+2)
A(n+2)+B(n+1)=1
B=-1
A=1

1/(n+1) - 1/(n+2)

Answer 45

yes, made mistake in factoring hehe....