2
$\begingroup$

I would like to solve the following series: $$\sum_{n=3}^\infty \frac{\log(n-1)-\log(n)}{\log(n-1)\log(n)}$$

Here's what I did: $$\sum_{n=3}^\infty \frac{\log(n-1)-\log(n)}{\log(n-1)\log(n)}=\sum_{n=3}^\infty \frac{\log\bigl(\frac{n-1}{n}\bigr)}{\log(n-1)\log(n)}=\sum_{n=3}^\infty\frac{\log\bigl(1-\frac{1}{n}\bigr)}{\log(n-1)\log(n)}$$

Apparently I can solve this by using the comparison test, but I don't know how this would be implemented in this exercise. Also, if it is possible, how can I find the result of the series?

$\endgroup$
  • 3
    $\begingroup$ Write it as $\frac{1}{\log(n)} - \frac{1}{\log(n-1)}$. Telescoping. $\endgroup$ – Winther Jul 5 '19 at 15:10
4
$\begingroup$

$$\begin{align} \sum_{n=3}^{\infty} \frac{\log(n-1)-\log(n)}{\log(n-1) \log(n)} &=\sum_{n=3}^{\infty}\left( \frac{\log(n-1)}{\log(n-1) \log(n)}-\frac{\log(n)}{\log(n-1) \log(n)}\right) \\ &=\sum_{n=3}^{\infty}\left( \frac{1}{\log(n)}-\frac{1}{\log(n-1)}\right) \\ &=\lim_{N\to \infty}\sum_{n=3}^N\left( \frac{1}{\log(n)}-\frac{1}{\log(n-1)}\right) \\ &=\lim_{N\to \infty}\left(\frac{1}{\log N}-\frac{1}{\log 2}\right) \\ &=-\frac{1}{\log 2}. \end{align}$$

$\endgroup$
  • $\begingroup$ Thanks, how did you get $\frac{1}{log2}$ from the third to the fourth passage? $\endgroup$ – Jonathan S. Jul 5 '19 at 15:39
  • 1
    $\begingroup$ It's a telescoping series. In the partial sum note the following: $$\sum_{n=3}^N \frac{1}{\log (n)}-\sum_{n=3}^N \frac{1}{\log(n-1)}=\sum_{n=3}^N \frac{1}{\log (n)}-\sum_{n=2}^{N-1} \frac{1}{\log(n)}=\left(\sum_{n=3}^{N-1} \frac{1}{\log (n)}+\frac{1}{\log N}\right)-\left(\frac{1}{\log 2}+\sum_{n=3}^{N-1} \frac{1}{\log (n)}\right)=\frac{1}{\log N}-\frac{1}{\log 2}$$ $\endgroup$ – Azif00 Jul 5 '19 at 15:47
1
$\begingroup$

Hint: It is $$\sum_{n=3}^{\infty}\frac{1}{\log(n)}-\frac{1}{\log(n-1)}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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