# True of false: The sum of this infinite series. [duplicate]

I'm fairly sure it is false, but I'm not quite sure about which test I should use to prove it.

$$\sum_{n=2}^\infty \ln\left(\frac{n-1}{n}\right) = -1$$ I think using the integral test should work, but it may get kind of messy, so I'm looking for som advice. All I have to do is prove that it diverges, right? Thanks in advance.

• Hint: $ln \frac ab=ln(a)-ln(b)$. – lulu Jan 28 '16 at 1:39
• If you can prove that the sum diverges, then it suffices to say that it cannot converge to any real number, specifically $-1$. – Decaf-Math Jan 28 '16 at 1:41

It is actually very easy. See that

$$f(N)=\sum_{n=2}^N\ln \left(\frac{n-1}{n}\right)= \sum_{n=2}^N\ln \left(1-\frac{1}{n}\right)$$

Is decreasing.

Now $f(3)\simeq-1.09861>f(+\infty)$. So $f(+\infty)\neq -1$.

If the series above did converged to -1, then

$$1=(-1)(-1)=-1\left(\sum^\infty_{n=2}ln\left(\frac{n-1}{n}\right)\right)=\sum^\infty_{n=2}-ln\left(\frac{n-1}{n}\right)=\sum^\infty_{n=2}ln\left(\frac{n}{n-1}\right)$$

would also be true. The series on the right was shown to be divergent elsewhere in this forum.