Question

omg, how do you evaluate an infinite series? lim
To infinity, Sigma sign
n=1
1
-----
n2 + n

Answer 1

i think you have to compute partial sums yes?

Answer 2

hmm, not sure

Answer 3

I think I have to figure out what type of series it is first or some pellet

Answer 4

i just did this in my summer class today let me get you a reference sheet my instructor gave to everyone it should help alot one moment

Answer 5

wow holy pellet I love the internet!

Answer 6

this site is unbelievable

Answer 7

partial fractions i think gets it. let me see

Answer 8

good work

going to take a look now

Answer 9

this converges by integral test, and in fact the sum is 1. but we need partial fractions i think to get the partial sums. i have to compute for a minute or several

Answer 10

OH I get it now, partial fractions right

Answer 11

for the part on e where it says "converges is |r| > 1" its suppose to say converges if |p| > 1

Answer 12

how do you figure out if it is an "alternating series"

Answer 13

riddle me this

Answer 14

negative then to the power?

Answer 15

ok yes if you use partial fractions you get it

Answer 16

\[\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\] gives a telescoping sum

Answer 17

keep going, and also, see my question above about 'alternating series'

Answer 18

in fact you can instantly see the series converges to 1

Answer 19

you take the limit of each term?

Answer 20

I can't instantly see anything like that

Answer 21

\[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...=1\]

Answer 22

it telescopes

Answer 23

oh so it's sort of like limit pellet, you just need to recognize those patterns

Answer 24

a, c d, e on the study guide are the only helpful things on it and i just started today on it so i hope satellite can clear the rest of it up XD

Answer 25

yeah helps if you know the tricks

Answer 26

even though it's 'converging to 1', doesn't it go to 0 since it starts at 1 and you are subtracting terms?

Answer 27

bu you can see it yes? and if you want to be rigorous i guess using the telescoping nature you can actually write out what the partial sums look like.

Answer 28

what are the "tricks"

Answer 29

so some in here - Q4d, e, 5, 6

Answer 30

the meaning of the infinite sum is the limit of the partial sums. the individual terms of course go to zero, otherwise you can forget it. but the question was not "does this converge" but rather "what is the limit?"

Answer 31

http://www.maths.uq.edu.au/~bmm/MATH1050/Handouts/problems4.pdf

and here are the worked solutions

http://www.maths.uq.edu.au/~bmm/MATH1050/Handouts/problems4sols.pdf

Answer 32

thanks broseph

Answer 33

you can actually write out the partial sums and see what they are. i mean write out
\[\sum_{k=1}^n\frac{1}{k^2+k}\]

Answer 34

ok hang on

Answer 35

i believe if you do this, using the partial fraction decomposition, you will see that you get


\[\sum_{k=1}^n\frac{1}{k^2+k}=\frac{n}{n+1}\] and then if you take
\[\lim_{n\rightarrow \infty}\frac{n}{n+1}\] you get 1 by inspection

Answer 36

forget that noise it is even easier

Answer 37

ok so it's 1/k^2 + k = A/K + B/K+1 what do I do from here??

Answer 38

\sum_{k=1}^n\frac{1}{k^2+k}=\frac{n}{n+1} where did you get that from?

Answer 39

put
\[S_n=\sum_{k=1}^n \frac{1}{k^2+k}=\sum_{k=1}^n\frac{1}{k}-\frac{1}{k+1}\]

Answer 40

here we will see it. but actually easier.

Answer 41

ok

Answer 42

so what do we have?
\[S_1=1-\frac{1}{2}\]
\[S_2=1-\frac{1}{3}\]
\[S_3=1-\frac{1}{4}\] and in general by the telescope
\[S_n=1-\frac{1}{n+1}\] which happens to be
\[S_n=\frac{n}{n+1}\] but we don't really need that form

Answer 43

the definition of the infinite sum is the limit as n goes to infinity of the partial sums

Answer 44

and that limit is clear as day 1

Answer 45

the whole gimmick was to use partial fractions and see that you ended up with the telescoping sum, making the partial sums easy to compute, and therefor the limit easy as well

Answer 46

ok what are the tricks? which are the common ones to look for there are like 4

Answer 47

n/n+1 is one of them for sure

Answer 48

not sure what exactly you mean. sometimes the sums are impossible. sometimes you get
\[\frac{\pi^2}{6}\] and some genius figured it out 200 years ago. but this partial sum one is a good one.

Answer 49

i mean "partial fractions" one

Answer 50

ok first step see if limit is equal to 0 that could mean it converges. lets see this baby

Answer 51

you mean the limit of the terms, not the limit of the partial sums. the limit of the terms must be zero but you need lots more than that.

Answer 52

right
what else do you need?

Answer 53

oh is that the one I'm supposed to be trying?

Answer 54

i cant believe no one gave you a medal Satellite!