[Guest]

Prove which of the following converge or diverge. Where convergence exists, what is this limit?

1. (1/1) + (1/3) + (1/6) + (1/10) + (1/15) + .....

2. (ln2 - ln1) + (ln4 - ln3) + (ln6 - ln5) + (ln8 - ln7) + .....

3. S₂ * S₃ * S₄ * S₅ * ..... where S_n = 1/1ⁿ + 1/2ⁿ + 1/3ⁿ + ...

Hint: S₂ = pi²/6

[Guest]

I'm stumped on the pattern for #1

At first I though 1/(n+2^n) but that breaks down at 1/10.

#2 diverges (ln(n+1)-ln(n)) will always be greater than 0.

I'm stumped on #3

[unreality]

littlej:

the inverted triangular numbers

http://en.wikipedia.org/wiki/Triangular_number

it's the infinite sum of 2 over n^2 plus n

(1/1) + (1/3) + (1/6) + (1/10) + (1/15) + .....

= (2/(1+1)) + (2/(2+4)) + (2/(3+9)) + (2/(4+16)) + .....

= 2 * sigma(from 1 to infinity) of 1 / (n + n^2)

now

(1/n) - (1/(n+1)) = (n+1)/(n + n^2) - (n)/(n + n^2) = 1 / (n + n^2)

therefore using this partial fraction we can break 1/(n + n^2) into 1/n - 1/(n+1)

so we're left with

= 2 * [ sigma(from 1 to infinity) of 1/n ... minus ... sigma(from 1 to infinity) of 1/(n+1) ]

break down the first sigma to get 1/1 + 1/2 + 1/3 + 1/4 + ....

break down the 2nd sigma to get 1/2 + 1/3 + 1/4 + ...

so subtracting second from first, we get 1/1 = 1

so the answer is 2 * 1 = 2

[unreality]

2. (ln2 - ln1) + (ln4 - ln3) + (ln6 - ln5) + (ln8 - ln7) + .....

lnA - lnB = ln(A/B)

so rewriting:

ln(2) + ln(4/3) + ln(6/5) + ln(8/7) + ...

and lnA + lnB = ln(AB)

so rewriting...

ln(2 * 4 * 6 * 8 * 10 * ..etc. divided by 1 * 3 * 5 * 7 * 9 * ..etc) = L

the limit L only exists if the thing inside the natural log exists. This thing is equal to e^L

so we have

e^L = (2*4*6*8*10*12*...) / (1*3*5*7*9*11*...)

I believe the thing on the right diverges, therefore L doesn't exist.

[Guest]

1 + 1/3 + 1/6 + 1/10 + ....

= 1 + 2 ( 1/6 + 1/12 + 1/20 + ...）

= 1 + 2[(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+.....]

=2

[Guest]

The second one diverges.

As n gets sufficeitnly large, ln(n+11) - ln(n) = ln (n+1)/n is about 1/n which diverges.