2
$\begingroup$

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.

$\endgroup$
2
  • 5
    $\begingroup$ Hint: $ln \frac ab=ln(a)-ln(b)$. $\endgroup$ – lulu Jan 28 '16 at 1:39
  • 1
    $\begingroup$ If you can prove that the sum diverges, then it suffices to say that it cannot converge to any real number, specifically $-1$. $\endgroup$ – Decaf-Math Jan 28 '16 at 1:41
1
$\begingroup$

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 $f(+\infty)\neq -1$.

$\endgroup$
0
$\begingroup$

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.

Here: Why does the series $\sum_{n=1}^∞ \ln ({n \over n+1})$ diverges? And general tips about series and the logarithm

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.