Test for convergence - Log operation

Question

∑ Log (n / n+ 1)

I solved the above problem in two different ways.

1st Method

= Log n - Log n+1
= Log n - Log n . Log 1
= Log n - Log n . 0
= Log n
= Log ∞
= ∞ (Diverges)

Method 2

= Log ( n / n + 1 )
= Log ( n / n . (1 + 1 /n))
= Log ( 1 / 1 + (1/n))
Substituting Limit, n=> ∞
= Log ( 1 / 1 + 1 / ∞))
= Log ( 1 / 1 + 0 )
= Log ( 1 / 1 )
= Log 1
= 0 (Converges)

Method 3

u1 = -log2
u2 = log2 - log3
u3 = log3 - log4
un = logn - log(n+1)

sn = u1 + u2 + ... un
sn = - log(n+1)

sub method 1
Substituting Limit, n=> ∞

sn = - log (∞)
sn = - ∞ (diverges)

sub method 2

sn = - log(n+1)
sn = - (log n . log 1)
sn = - (log n . 0)
sn = - (∞ . 0)
sn = indeterminate form

Where am i making the mistakes. Which one is the right way to solve this problem?

Answer 1

hint: $\log\left(\dfrac{n}{n+1}\right)=\log n - \log(n+1)\Rightarrow S_n =-\log(n+1)$