Question

I want to clarify something about the convergence of series and sequence.

Answer 1

So, for sequence to converge

1) Bound below and above
2) Monotonic (can be always decreasing or always increasing)

Answer 2

and for series:

3) The limit [n→ infinity] A(n) = 0

(am I correct)

Answer 3

yes, but the converse does not hold on 3)

Answer 4

yes, what other conditions are there?

Answer 5

well, yes not on those 2 - harmonic series.

Answer 6

not those 3*

Answer 7

But, what else must be true?

Answer 8

of series?

Answer 9

yes convergence of series

Answer 10

there are many many theorems

Answer 11

Oh, like Σ 1/n^p then p>1
geometric series |r|<1 (and that is where ratio test is coming from)

and others....

Answer 12

incl alternating series test.... and others

Answer 13

so it basically depends on a series in every unique cse?

Answer 14

case*

Answer 15

pretty much. there are some nice ways for some families of functions, but in general there is nothing that will deal with them all.

Answer 16

if you're comfortable with sequences, then you may think of any series as a sequence of partial sums

Answer 17

I also will assume we are talking about the real numbers with the regular topology.

Answer 18

yes, that is what i am reading on wiki and other sites ganeshie

Answer 19

yes. no imaginary sequences. I am not considering anything crazy

Answer 20

Getting good with sequences is a great idea if you are planning on doing analysis. There are definitions of continuity that uses sequences that make things much nicer than dealing with the standard \(\epsilon -\delta\) definition

Answer 21

alright

Answer 22

I was reading wiki just 3-5 minutes ago, and I saw that

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \frac{(-1)^{n+1}}{n}=\ln(2)}\)

(they don't post any link to an explanation)

Answer 23

(i can create a new Q. if that is how the policy would play this time)

Answer 24

(sorry if i am typing to much, don't be afraid to tell me to shut up:))

Answer 25

This does not easily fall without other identities.

Answer 26

which identities?

Answer 27

series expansion of ln(x)

Answer 28

criterian for convergence of an alternating series is an interesting one
i remember seeing a very nice proof from analysis

Answer 29

Maybe I am wrong, but I don't any basic ways to show this without more tools.

Answer 30

You can find the taylor expansion and plug in two...

Answer 31

But you are not proving anything....

Answer 32

wait can i start from taylor series of 1/x and integrate?

Answer 33

yeah

Answer 34

but you already know you can do that, so what is the point :)

Answer 35

unless you just feel like symbol pushing :)

Answer 36

\[\frac 1{1+x}=\sum_{n=0}^\infty (-x)^n\]

integrate both sides and plugin \(x=1\)
this is fun but yeah doesn't give you the bigger picture of beatiful alternating series

Answer 37

oh, we can get a series represenation for ln(x+1) :) nice

Answer 38

\(\large\color{black}{ \displaystyle \ln(x+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^nx^{n+1}}{n+1}}\)

Answer 39

ln(x) has no hope at x=0
but surely it behaves well for all finite positive values of x

Answer 40

oh yeah my index should have been n=0, i made that err

Answer 41

and nice to hear it works for other ln values

Answer 42

\(\large\color{black}{ \displaystyle \ln(x+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^nx^{n+1}}{n+1}}\)

\(\large\color{black}{ \displaystyle \ln(1+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^n1^{n+1}}{n+1}}\)

Answer 43

oh, sorry n=0

Answer 44

\(\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n}{n+1}}\)

Answer 45

now we're getting into radius of convergence and stuff
for what values of x does that series converge ?

Answer 46

ok, let me do ratio test:)