Urgent! sequence and series

2010-10-26 4:41 am

1. determine whether the sequence converges or diverges. if it converges, find the limit:

a sub n= (-3)^n / n!

2. show that the sequence defined by

a sub 1= 2
a sub (n+1) = 1/ (3- a sub n)

satifisfies 0 less than (a sub n) less than or equal to 2 and is decreasing. deduce that the sequence is convergent and find its limit.

3. determine whether the geometric series is convergent or divergent. if it converges, find its sum.

sum (n=1) to inifinty (e^n / 3^(n-1) )

please explain in details!!!
thank you very much!!

回答 (1)

Lotus_本願山彌陀講堂

2010-10-26 7:52 am

✔ 最佳答案

Q1.
a(n)= (-3)^n/ n!
| a(n+1)/a(n) | = 3/(n+1) -> 0
so, lim(n->∞) a(n) =0

Q2.
Claim1: 0< a(n) <= 2
pf. (by MI.)
n=1, a(1)=2 , holds
Suppose that 0< a(k) <= 2, then
a(k+1) = 1/[3-a(k)] > 1/3 >0
and
a(k+1) -2 = 1/[3-a(k)] -2= [2a(k) -5]/[3-a(k)] < (4-5)/[3-a(k)] <0
so that , 0<a(n)<=2 for all n=1,2,3,4,...

Claim2: a(n+1)< a(n)
pf (by MI.)
a(2)= 1/(3-2)= 1 < a(1) holds
Suppose that a(k)<a(k-1), then
a(k+1)-a(k)= 1/[3-a(k)]- 1/[3-a(k-1)]
= [a(k)-a(k-1)]/{[3-a(k)][3-a(k-1)]}
<0 (by MI and 3-a(k), 3-a(k-1)>0 )
so that a(n+1)-a(n) <0 (ie. the sequenct {a(n)} is decreasing )

Claim3: lim(n->∞) a(n)= (3-√5) /2
{ a(n) } is decreasing and a(n)>0, so that lim(n->∞) a(n) exists.
Let lim(n->∞) a(n) = x, then
lim(n->∞) a(n+1) = lim(n->∞) 1/[3-a(n)]
x= 1/(3-x), x^2- 3x+1=0, x= (3 +/- √5)/2
x= (3+√5)/2 is inconsistant, because a(n)<= 2
so, lim(n->∞) a(n) = (3-√5)/2

Q3.
Σ[n=1~∞] e^n/3^(n-1) is a geometric series with r= e/3, |r|<1
so, Σ[n=1~∞] e^n/3^(n-1) converges to a/(1-r) = e/(1- e/3)= 3e/(3-e)

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