# sequences and series

• Apr 10th 2007, 02:48 AM
mathshelpneeded
sequences and series
Please i don't know how to even BEGIN this question.

A sequence defined as Sn(x) 1 + 2x + 3x^2.....nx^(n-1)

(I assume that the it is a geometric sequence and a is n and r is x? Am i correct?)

By considering (1-x)Sn(x) show that

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

Obtain the value of

lim (n tends to infinity) {2/3 + 3/2^2 + 4/3^3 + ..... + n/3^(n-1) + 3/2*n/3^n }

• Apr 10th 2007, 03:26 AM
topsquark
Quote:

Originally Posted by mathshelpneeded
A sequence defined as Sn(x) 1 + 2x + 3x^2.....nx^(n-1)

(I assume that the it is a geometric sequence and a is n and r is x? Am i correct?)

No, because 2x/1 = 2x and 3x^2/(2x) = (3/2)x. The terms are not related by a constant factor.

Quote:

Originally Posted by mathshelpneeded
By considering (1-x)Sn(x) show that

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

Your written answer is incorrect. (And you really need to use parenthesis!)

Sn(x) = 1 + 2x + 3x^2 + ... + nx^{n - 1}

Now
1*Sn(x) = 1 + 2x + 3x^2 + ... + nx^{n - 1}

x*Sn(x) = x + 2x^2 + 3x^3 + ... + nx^n

So
(1 - x)*Sn(x) = 1 + x + x^2 + ... + x^{n - 1} - nx^n

Now 1 + x + x^2 + ... + x^{n - 1} is a geometric series with n terms, so the sum of this is:
(1 - x^n)/(1 - x)

Thus:
(1 - x)*Sn(x) = (1 - x^n)/(1 - x) - nx^n

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

-Dan
• Apr 10th 2007, 03:35 AM
topsquark
Quote:

Originally Posted by mathshelpneeded
Obtain the value of

lim (n tends to infinity) {2/3 + 3/2^2 + 4/3^3 + ..... + n/3^(n-1) + 3/2*n/3^n }

In this case, by "limit" they mean they want the exact value of the series if the series has an infinite number of terms.

Note that
2/3 + 3/3^2 + 4/3^3 + ...
is just your Sn(x) series for x = 1/3 with 1 subtracted, or:
2/3 + 3/3^2 + 4/3^3 + ... = Sn(1/3) - 1 (for n -> infinity)

So
Sn(1/3) = [1 - (1/3)^n]/[1 - (1/3)]^2 - n*(1/3)^n/[1 - (1/3)]

= (9/4)*[1 - 1/3^n] - (3/2)*n/3^n

Now, for very large n, what happens to the first term? I'll let you work it on a calculator and simply say that as n grows very large, 3^n grows very large, and thus 1/3^n goes to 0. So the first term goes to a value of 9/4 as n goes to infinity.

For very large n, what happens to the second term? Using a calculator you can again verify that this term also goes to 0.

Thus as n goes to infinity:
Sn(x) -> 9/4.

Finally:
2/3 + 3/3^2 + 4/3^3 + ... = Sn(1/3) - 1 (for n -> infinity)

= 9/4 - 1 = 5/4.

-Dan
• Apr 10th 2007, 07:55 AM
mathshelpneeded
Quote:

Originally Posted by topsquark

Note that
2/3 + 3/3^2 + 4/3^3 + ...
is just your Sn(x) series for x = 1/3 with 1 subtracted, or:

Thanks a lot i never noticed that.