Logarithms

Question

If u_n= (1-1/n)² , where n is an integer greater than 1, find the value of lgu₂ + lg u₃+ lg u₄ +...+lg u_(10^k), where k is a positive integer and lg=log₁₀ .

Answer 1

u(n) = (1 - 1/n)^2, n integer, n > 1

log(u(n)) = 2 log(1 - 1/n)

log(u(n)) = 2 log((n - 1)/n)

log(u(n)) = 2 log(n - 1) - 2 log(n)

log(u(2)) = 2 log(2 - 1) - 2 log(2)
= 2 log(1) - 2 log(2)                          b/c log(1) = 0
= - 2 log(2)

log(u(3)) = 2 log(3 - 1) - 2 log(3)
= 2 log(2) - 2 log(3)

log(u(2)) + log(u(3)) = - 2 log(2) + 2 log(2) - 2 log(3)
= - 2 log(3)

log(u(4)) = 2 log(4 - 1) - 2 log(4)
= 2 log(3) - 2 log(4)

log(u(2)) + log(u(3)) + log(u(4)) = - 2 log(3) + 2 log(3) - 2 log(4)
= - 2 log(4)

Following the pattern - 2 log(2), - 2 log(3), - 2 log(4):

sum{n=2,10^k}(log(u(n))) = - 2 log(10^k)
= - 2k log(10)
= - 2k                          b/c log(10) = 1

Answer 2

u_n  = (1-1/n)^2, so lg(u_n) = 2lg(1-1/n)  = 2(lg(n-1) - lg(n)).  So lg(u₂) = 2(lg(1) -lg(2)), and  lg(u₃)  = 2(lg(2) - lg(3)), and lg(u₂) + lg(u₃) =  2(lg(1) - lg(3))

 

So ∑_n=2,10^klg(u_n) = 2(lg(1) - lg(10^k))   = 2(0-k) = -2k