Sequences and series question (1 Viewer)

[sinophile]

A) Find A and B such that 1/(x-1)x = A/(x-1) + B/x

B) Let Sn= 1/(1x2) + 1/(2x3) + 1/(3x4) +...+ 1/(n-1)n.
Show that Sn=1-1/n

C) Hence or otherwise evaluate
infinity
(SIGMA) 1/(n-1)n
n=2

 

[lyounamu]

A) Find A and B such that 1/(x-1)x = A/(x-1) + B/x

B) Let Sn= 1/(1x2) + 1/(2x3) + 1/(3x4) +...+ 1/(n-1)n.
Show that Sn=1-1/n

C) Hence or otherwise evaluate
infinity
(SIGMA) 1/(n-1)n
n=2

a)

1/((x-1)x = Ax/((x-1)x) + B(x-1)/((x-1)x)
1 = Ax + B(x-1)
1 = Ax + Bx - B
therefore, A+B = 0 and -B = 1
B = -1 and A = 1

b)

 

[sinophile]

a)

1/((x-1)x = Ax/((x-1)x) + B(x-1)/((x-1)x)
1 = Ax + B(x-1)
1 = Ax + Bx - B
therefore, A+B = 0 and -B = 1
B = -1 and A = 1

b)

b?

 

[lyounamu]

b?

doing it now

hm..I cannot find any special relationship...

c)

Sn = 1-1/n
n -> infinity

Sn = 1.

 

[Trebla]

A) Find A and B such that 1/(x-1)x = A/(x-1) + B/x

B) Let Sn= 1/(1x2) + 1/(2x3) + 1/(3x4) +...+ 1/(n-1)n.
Show that Sn=1-1/n

C) Hence or otherwise evaluate
infinity
(SIGMA) 1/(n-1)n
n=2

b)

 

[lyounamu]

b)

That question is stupid then.

You cannot really use 2 unit method to get it out.

I thought of partial fraction, but using that isn't really right because this is 2 unit question.

 

[Trebla]

Well the partial fraction bit came from part a), so it's just a matter of using part a) to apply on part b)...

 

[lyounamu]

Well the partial fraction bit came from part a), so it's just a matter of using part a) to apply on part b)...

Ah, okay. I didn't see that they were in part A, B and C...