# series questionWatch

#1
S(x) = 1 + 2x +3x^2 +.....+nx^(n-1)

consider (1-x)S(x) and show

S(x) = (((1-x)^n)/((1-x)^2)) - ((nx^n)/(1-x)) x=/=1

and also obtain the limit as n tends to infinity of

[2/3 + 3/3^2 + 4/3^3 +.....+n/3^(n-1) + (3/2)(n/3^n)]

for last bit i think the answer is 5/4 but dont understand

thanking you
0
12 years ago
#2
Suppose x≠1. Then:
(1 - x)S(x) = S(x) - xS(x)
= 1 + x + x^2 + ... + x^(n-1) + nx^n
= (1 - x^n)/(1 - x) + nx^n, since the n terms on the left form a geometric series.

=> S(x) = (1 - x^n)/(1 - x)^2 + (nx^n)/(1 - x), as required.

Now:
2/3 + 3/3^2 + 4/3^4 + ... + n/3^(n-1) + (3/2) n/3^n = S(1/3) - 1 + (3/2) n/3^n
= (1 - (1/3)^n)/(2/3)^2 + (n (1/3)^n)/(2/3) - 1 + (3/2) n/3^n
-> 1/(2/3)^2 + 0 - 1 + 0
-> 5/4
0
#3
thanks for the first bit but u totally lost me on part 2 could u go over it in more steps and give me a written explanation of what youre doing i dont really see where youre first line comes from!

i.e.
2/3 + 3/3^2 + 4/3^4 + ... + n/3^(n-1) + (3/2) n/3^n = S(1/3) - 1 + (3/2) n/3^n
0
#4
looking at it again i get this bit apart from why u make

2/3 + 3/3^2 + 4/3^4 + ... + n/3^(n-1)=S(1/3)

surely this isnt in the same format as the original S(x) series
0
#5
i appreciate that my argument isnt strengthened by the fact your answer is correct but could u explain it to me please - i am a bit dunse!
0
12 years ago
#6
You're right: it isn't in the same format as S(x) - it looks like S(x) - 1!

S(x) - 1 = 2x + 3x^2 +.....+ nx^(n-1)

S(1/3) - 1 = 2(1/3) + 3(1/3)^2 + 4(1/3)^3 + ... n(1/3)^(n-1)
= 2/3 + 3/3^2 + 4/3^3 + ... + n/3^(n-1)

Is it clear now?
0
#7
yeah cheers
0
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