Question

Using a pattern, what is the sum of:

1/1 + 1/2 x 1/3 +1/3 x 1/4 + ....... + 1/99 x 1/100 + 1/100 x 1/101?

Answer 1

We can find a very good approximation of this sum by using the formula:

Hn≈(2n+1)[arccot(2n+1)]+(1/2)[ln(n2+n+1/2)]+(γ−1)

For n=100 , we take:

H100≈5.18737751…

The exact value of this sum is the given by the fraction:

144666362795203511602215180431041314477112788815009188499086581352357412492142272

This fraction has the estimated value of 5.18737751763962…

The integer part of it is 5 .

Answer 2

mmmhhhh?

I see a telescopic series

1/1 + 1/2 x 1/3 +1/3 x 1/4 + ....... + 1/99 x 1/100 + 1/100 x 1/101
= 1 + (1/2)(1/3) + (1/3)(1/4) + ... + (1/99)(1/100) + (1/100)(1/101)
the general term is 1/(n(n+1))
using partial fractions I can break this up as 1/n - 1/(n+1)

so we have
1 + Sigma [1/n - 1/(n+1) ] for n = 2 to 100
= 1 + (1/2-1/3) + (1/3-1/4) + (1/4-1/5) + ... + (1/99-1/100) + (1/100-1/101)
= 1 + 1/2 - 1/101 , all the other terms in between drop out
= 301/202 or appr. 1.49

( I have no idea where "helper" got his stuff from, just adding the first 4 terms
by calculator gives us 1.3 )

Answer 3

okay thx mathhelper