Question

which follow these converge? Also calculate

1. F0 = F1 = 1, and Fn + 1 = Fn + Fn-1
2. a1 = 2 and a n + 1 = (to) ^ (1/2)
3. A0 = 1 and an + 1 = an + m ^ n for every m∈R

Answer 1

the first is Fibonacci in recurrence formula

Answer 2

haha go sleep
at any rate how do you decide convergence?

Answer 3

yes diverge since the ratio goes to psi which greater than 1

Answer 4

psi=golden mean

Answer 5

hmm ya you can say increasing without bound

Answer 6

limit is another way to go at it, that's i asked what does the asker know about convergence

Answer 7

haha ya, i never memorized

Answer 8

hmm well good that you mentioned boundness

Answer 9

okay!

Answer 10

hmm lets the same way loser did the first
a1=2
a2=root2
a3=root (root2)
so is this bounded or not

Answer 11

what language is that problem was written?

Answer 12

Deutsch

Answer 13

hmm i see. what level of mathematics are you doing?

Answer 14

see that our terms are getting closer and closer to zero yes
2>root2>root(root(2))>...........

Answer 15

so a_n+1<=a_n and bounded
therefore converges

Answer 16

sorry , i do not speaking well english ..
I am in my first year at the college

Answer 17

oh i see, no problem
but you understand the math yes

Answer 18

\[a_1=2\\a_2=\sqrt{2}\\a_3=\sqrt{\sqrt{2}}\\.\\.\\.\\a_{n+1}=\sqrt{a_{n}}\]
note that \[a_{n+1}\leq a_n\] and \[a_n \] is bounded
so it must converges

Answer 19

hmm there is little problem @perl
the sequence is not non increasing yes?

Answer 20

not non increasing means decreasing?

Answer 21

not really isn't non increasing assume that terms might equal

Answer 22

we have a sequence
$$ \LARGE 2, 2^{\frac 12}, 2^{\frac14},2^{\frac 18},.. 2^{\frac 1{2^{n}}},...

$$

Answer 23

that's my question this sequence is strictly decreasing

Answer 24

right

Answer 25

it is bounded above by 2, bounded below by 0 (or 1), and it is monotonic
therefore there exists a greatest lower bound

Answer 26

and that is the limit of the sequence

Answer 27

it is bounded below by zero since all the terms in the sequence are positive

Answer 28

and bounded above by 2 since it is decreasing, and 2 is the first term

Answer 29

we have a sequence
$$ \LARGE 2 \geq 2^{\frac 12} \geq 2^{\frac14} \geq 2^{\frac 18},.. \geq2^{\frac 1{2^{n}}},...

$$

Answer 30

yeah i wanted to use boundness theorem
but i thought it has to be nonincreasing

Answer 31

In the mathematical field of real analysis, the monotone convergence theorem is any of a number of related theorems proving the convergence of monotonic sequences (sequences that are increasing or decreasing) that are also bounded. Informally, the theorems state that if a sequence is increasing and bounded above by a supremum, then the sequence will converge to the supremum; in the same way, if a sequence is decreasing and is bounded below by an infimum, it will converge to the infimum
http://en.wikipedia.org/wiki/Monotone_convergence_theorem

Answer 32

\(\color{blue}{\text{Originally Posted by}}\) @xapproachesinfinity
see that our terms are getting closer and closer to zero yes
2>root2>root(root(2))>...........
\(\color{blue}{\text{End of Quote}}\)
that's what i was doing but had doubt

Answer 33

the monotonic sequences convergence theorem does not have to be strictly decreasing or increasing. monotonicity is enough

Answer 34

oh ok

Answer 35

non decreasing (sometimes called increasing)
non increasing (sometimes called decreasing)

http://en.wikipedia.org/wiki/Monotonic_function

Answer 36

love your pic by the way , i forgot who you were

Answer 37

okay i see

Answer 38

poincare is awesome haha

Answer 39

watched some videos about math history, this guy is quite something

Answer 40

check nj wilberger math history
one of the videos talk about poincare and other mathematicians

Answer 41

so in a nutshell the monotonic sequence theorem says

1. If a sequence of real numbers is strictly increasing (or not decreasing) and bounded above, then its supremum is the limit.

2. If a sequence of real numbers is strictly decreasing (or not increasing) and bounded below, then its infimum is the limit.

In our case we use part 2.
but we have to show that an+1 < an

Answer 42

show that \( \LARGE 2^{\frac {1}{n+1} } < 2^\frac 1n \)

Answer 43

i see thanks :)

Answer 44

you can show it either by induction
or how about raise both sides to the power of n+1

Answer 45

hmm we show that f(x)=1/2^x is decreasing
so 2^(1/n+1)<2(1/n)

Answer 46

hmm i went too far
induction is good actually