A = (n+ 0.5)ln(1 + 1/n) - 1
Use a series expansion to show that if n is large enough for 1/n^3 and higher powers of 1/n to be neglected, A approximately equals to 1/12n^2
Thanks in advance for your help
Hi,
Use the fact that ln(1 + x) = x - x^2/2 + x^3/3 - ... from the formulae list.
Thanks!
Cheers,
Wen Shih
yeah i tried that and got -1/4n^2 as the ans
(n + 0.5) [ 1/n - 1/2n^2] -1
did i make some careless mistake somewhere?
Originally posted by quailmaster:yeah i tried that and got -1/4n^2 as the ans
(n + 0.5) [ 1/n - 1/2n^2] -1
did i make some careless mistake somewhere?
Hi quailmaster,
When you expand out In(1+(1/n)), you will have to include 1/(3n^3) as well. It is only after simplifying the entire term, then should you get rid of terms including 1/n^3.
Thus, you should get (1/3n^3)(n) - (1/2)(1/2n^2) after the other terms cancel each other out. This will simplify to 1/12n^2.
Cheers.