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?

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

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

  • $\begingroup$ 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

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


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