Question

Show that 𝒍𝒐𝒈𝟏 + 𝒍𝒐𝒈 𝟐 + 𝒍𝒐𝒈𝟑 + ⋯ + 𝒍𝒐𝒈𝒏 = 𝜣(𝒏𝒍𝒐𝒈𝒏) for 𝑛 > 1

Answer 1

Solution:

To show: log1 + log2 + log3 + .....+ logn = (nlogn) for n > 1

Explanation:

=>Let say f(n) = log1 + log2 + log3 + .....+ logn and g(n) = nlogn

Simplifying f(n):

=>f(n) = log1 + log2 + log3 + .....+ logn

Using log property loga + logb = logab

=>f(n) = log(1*2*3*4...*n)

We know that 1*2*3...*n = n!

=>f(n) = logn(n!)

=>f(n) = nlogn asymptotically

Proving using limits:

=>Limit(n -> ){f(n)/g(n)} = Limit(n -> ){nlogn/nlogn}

=>Limit(n -> ){f(n)/g(n)} = Limit(n -> ){1}

=>Limit(n -> ){f(n)/g(n)} = 1(constant)

=>As Limit(n -> ){f(n)/g(n)} = 1(constant) so f(n) and g(n) have equal growth rate hence we can write f(n) = (g(n)) or log1 + log2 + log3 + .....+ logn = (nlogn).