Question

If sequence diverges will series also automatically diverge?

Answer 1

For instance,
\[2,~~~-4,~~~2,~~-4,~~~2,~-4....\]

Answer 2

So try and think of a counter example here

Answer 3

I know that lim n-> INFINITY of An= diverges (it does not approach ONE value).
and series will get to neg. infinity as we add more terms.

Answer 4

maybe check out the harmonic

Answer 5

but, although I know this series will diverge, are there any possibilities for having sequence diverge BUT yet so series converge ?

Answer 6

What I am thinking is that this is NOT possible.

Answer 7

and that it must be a "law" that is sequence diverges then series divreges.

Answer 8

wait, oh yes... it is not bound so it must diverge

Answer 9

yes, yes.... I went kind of silly on this one.
it is not bound (if sequence diverges) and therefore series will diverge.

Answer 10

Got it:)

Answer 11

uhm, pretty sure the alternating harmonic converges

Answer 12

let me double check

Answer 13

harmonic series diverges, sequence converges

Answer 14

but I am not saying the converse (i.e. that if it is bound it will converge)
I am saying if NOT bound - will diverge.

Answer 15

nah, the alternating sequence doesn't converge

Answer 16

yes, that is what I though;)

Answer 17

and nor does the series.

Answer 18

oh oh oh, give me a sec then. here is the http://mathworld.wolfram.com/HarmonicSeries.html

Answer 19

the series converges

Answer 20

series does not

Answer 21

but does the sequence, I need to think about

Answer 22

it does it converges to ln2

Answer 23

*Alternating harmonic series**

Answer 24

alternating harmonic series ?

Answer 25

yea, read the link

Answer 26

like
1 -1/2 1/3 -1/4 ?

Answer 27

yea

Answer 28

well if signs change then converges

Answer 29

and anyways, as for if it it is bounded it converges, the theorem for that is if it is bounded and monotone it converges on the interval

Answer 30

bounded and monotonic (with exception of harmonic series)

Answer 31

nah that is for every function that satisfies those two qualities

Answer 32

there are other functions that are not monotone that converge

Answer 33

but bound and monotonic converges - you are saying the series converges ... ?

Answer 34

yes.

Answer 35

alrighty... I guess, ty once again.

Answer 36

and your initial question should be answered by this test http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx

Answer 37

Fact 1:
so for any geometric series, if |r|<1 the series automatically converges.

Answer 38

sorry, it's been like 5 years since i've done this, I had to refresh before I could really answer. The proof of the limit test is at the bottom

Answer 39

yes

Answer 40

lamar tutorials.... one more thing on my online reference list.

Answer 41

For the series to converge, the corresponding sequence must converge to 0

Answer 42

yes, that is very important. tnx

Answer 43

he is amazing, and yes. I read it wrong originally

Answer 44

\[\lim\limits_{n\to\infty} a_n \ne 0 \implies \sum\limits_{n=1}^{\infty} a_n~\text{does not converge}\]

Answer 45

yes, sequence to 0, or else you are adding same term (even if sequence converges) infinite number of times, giving a series that approaches infinity.

Answer 46

Exactly

Answer 47

like if sequence converges to 3, then series would be
3 + 3 +3 +3 ....
= 3 times (Infinity)

Answer 48

= inf.

Answer 49

roughly

Answer 50

tnx rational. Indeed very rational !

Answer 51

the converse is not true however
\[\lim\limits_{n\to\infty} a_n = 0~~~ ~ \nRightarrow ~~~~\sum\limits_{n=1}^{\infty} a_n~\text{ converges}\]

This is bit hard to make sense of when you think of it for the first time

Answer 52

well, the harmonic series proves that the converse is false, just as you said.

Answer 53

Yes to me divergence of harmonic series was hard to digest in sequences and series class

Answer 54

yes, my teacher showed it too as, and he made it very easy.
\[1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \]\[1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8}+ \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \]

Answer 55

so you are essentially adding 1/2 infinite number of times.

Answer 56

it would take million of terms to get to 1000, but with many many many terms you can reach any positive large number. (slowly, but still)

Answer 57

that was my favorite thing in the course, probably.

Answer 58

my favorite is harmonic series too and there are like more than hundred proofs for its divergence.. let me see if i still have that pdf..

Answer 59

I guess if those wouldn't be too hard for me. I am not very smart in math.

Answer 60

if you say that enough it will become true. please don't

Answer 61

here it is

Answer 62

you're in calc, you are not bad in math

Answer 63

well not to be very smart I don't have to be stupid....
(never called myself stupid)

Answer 64

rational, did you make this file ?

Answer 65

oh, my bad... I see the authors

Answer 66

psychologists call that a self fulfilling prophecy/expectation

Answer 67

you just have not seen or practiced enough yet :)

Answer 68

self fulfilling prophesy, yes:) I heard that term don't remember where....
this file is awesome, I read (and understood) 2 proves so far... it is beautiful!!

Answer 69

also, you two may appreciate this trick, when reviewing for an exam if you wannna do a practice one google "site: .edu [subject] exam" It will pop up a bunch of practice tests

Answer 70

cool:) or I can use Galileo

Answer 71

i liked this proof very much :

We know that \(\large e^x \gt 1+x\)

\[\begin{align}e^{1+\frac{1}{2} + \frac{1}{3}+\cdots +\frac{1}{n}} &= e^{1}\cdot e^{\frac{1}{2}}\cdot e^{\frac{1}{3}}\cdots \cdot e^{\frac{1}{n}}\\~\\&\gt (1+1)\cdot\left( 1+\frac{1}{2}\right) \cdot \left( 1+\frac{1}{3}\right) \cdots \cdot \left( 1+\frac{1}{n}\right) \\~\\&= 2\cdot \dfrac{3}{2}\cdot \dfrac{4}{3}\cdots \cdot \dfrac{n+1}{n}\\~\\&=n+1\end{align}\]

That means we can make \(e^{H_n}\) as large as \(n+1\) which is arbitrary. so the harmonic series \(H_n\) diverges

Answer 72

galileo?

Answer 73

by the way as we were sitting I did another example (and got a snack).
\[\sum_{n=1}^{\infty}\frac{n^2+2}{3n^2}\]\[\lim_{n \rightarrow \infty}A_n=(1/3)\]
So therefore sequence converges to 1/3
and series diverges

Answer 74

I need help with one more, but I will open a new question for that...
yes database

Answer 75

that looks good

Answer 76

never heard of it hmm

Answer 77

don't you normally use database for research papers and such stuff.
(Haven't really been using Galileo for maths)

Answer 78

anyway... I need to get it going, sorry.

Answer 79

hahahahaha research papers. we don't do them that way

Answer 80

whenever I do research papers I need to use "authorized sources" so I have to use peer-reviewed (forgot other requirements ) for all sources.

Answer 81

we don't do them at all

Answer 82

well, you have already gone over that I guess.

Answer 83

I hope to grow out of this once too.

Answer 84

i've never written one actually

Answer 85

Lucky

Answer 86

I have 5,000 words requirement.

Answer 87

anyway.... I have one more ques. and I am done for today

Answer 88

I will open a new thred

Answer 89

alright well have funzies!

Answer 90

sure, u2