prove that the sequence is bounded for all n>=2 , n in N
#Proof sequence is bounded
slender mason
magic kelpBOT
prove that the sequence is bounded for all n>=2 , n in N
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slender mason
i think that maybe using the definition of a bounded sequence is wrong (like $\left| w_{n} \right|\le S, S\in \mathbb{R}_{\ge 0}$ ) or maybe what I'm doing is right except if i do the triangular inequality
past cairnBOT
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slender mason
a.k.a $\left| a+b \right|\le \left| a \right|+\left| b \right| ; a,b \in \mathbb{R}$
past cairnBOT
ᚾᛁᚷᚷᛖᚱ
rustic timber
what is the limit of (5/2)^n as n goes to infinity ?
slender mason
infinity?
rustic timber
yes, so what is the limit of (5/2)^n-2 as n goes to infinity ?
slender mason
infinity
rustic timber
yes, so from this, use the definition of lim = infinity to bound 1/((5/2)^n- 2)
slender mason
what if I don't want to use the definition of a limit
rustic timber
well you can find some n0 so that (5/2)^n-2 is larger than 10-2 for all n > n0
slender mason
so what is the n0 for the expression to be larger than you're referring to
rustic timber
you get it automatically using the definition of limit, otherwise you can get an explicit one using the logarithm
slender mason
oh i see, so can you explain how to use the definition of a limit for this case
to bound w_n
rustic timber
there exists an n0 such that for alr n > n0, (5/2)^n- 2 > 8
hence, for all n>n0, wn < 1 + 3/8
thus, wn < max (w0, w1,..., wn0, 1+3/8)
if you really want to do it by hand, then you can also show that the sequence wn is decreasing for n > 3, so that wn <= max(w0, w1, w2, w3)
but seeing the limit is useful for intuition, and then later refine your proof to a simpler one
slender mason
oh i see so for a n>n0 to satisfy the upper bound we take the maximum of the finite set w0,w1,...,wn0, therefore wn<max(w0, w1,..., wn0, 1+3/8)
rustic timber
yes
slender mason
if you really want to do it by hand, then you can also show that the sequence wn...
oh that's another one but it's complicated but i prefer seeing the limit, thus making the proof simpler
well how about for the left side (lower bound of w_n)
rustic timber
you can show that wn is positive for n => 3
slender mason
yea that's it?
rustic timber
hence the lower bound is min(w0,w1,w2,0)
slender mason
oh i see
rustic timber
(actually I just reread it, and wn is only defined for n>=2, so the lower bound is really min(w2, 0)
slender mason
by the way n>=2
slender mason
(actually I just reread it, and wn is only defined for n>=2, so the lower bound ...
yea
so this makes the lower bound min(w0,w1,w2,...,1)
rustic timber
(you can also use the limit argument, really |(5/2)^n-2| goes to infinity)
rustic timber
so this makes the lower bound min(w0,w1,w2,...,1)
yes, also
slender mason
(you can also use the limit argument, really |(5/2)^n-2| goes to infinity)
so as n goes to infinity the denominator I(5/2)^n-2I goes to infinity and thus w_n is 1 + 0 = 1
rustic timber
thus lim wn = 1
slender mason
yep
rustic timber
but yeah, at some point n0, 1/2 <= wn <= 1+1/2
and then the bound is this + the bound on the first terms
slender mason
so the key idea is its limit wn = 1
with the limit wn lies on the interval [1/2,3/2]
rustic timber
yes, that wn converges
slender mason
that makes sense for any convergent series
rustic timber
merci beacoup
de rien
cinder treeBOT
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cinder treeBOT
@slender mason
slender mason
+close