Question

How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given `ln(1/2)+ln(2/3)+ln(3/4)+...+ln(n/(n+1))+...`?

Answer 1

We know that `ln(a/b)=ln(a)-ln(b)`.

So, we can rewrite the sum
`ln(1/2)+ln(2/3)+ln(3/4)+...+ln((n-1)/n)+ln(n/(n+1))`
to
`=(ln(1)-ln(2))+(ln(2)-ln(3))+(ln(3)-ln(4))+…+(ln(n-1)-ln(n))+(ln(n)-ln(n+1))`.

Slightly rearrange the terms:
`=ln(1)+(ln(2)-ln(2))+(ln(3)-ln(3))+…+(ln(n-1)-ln(n-1))-ln(n+1)`

See how almost all of the terms cancel out? This is called a telescoping sum. Everything simplifies to
`=ln(1)-ln(n+1)`
`=ln(1/(n+1))`

Now, to find the sum of the infinite series, set `n->oo`:
`lim_(n->oo)ln(1/(n+1))`
`=-oo`

So, the sum of the infinite series diverges.