$\require{cancel}$ I'm trying to find whether the series $\sum_{n=1}^{\infty} \left(\frac{1}{\ln(n+2)} - \frac{1}{\ln(n+1)}\right)$ diverges or converges. It is a telescoping series. My approach was to write down the few terms and find the patterns of which items are being removed in each term.
Here's my attempt, \begin{align*} \sum_{n=1}^{\infty} \left(\frac{1}{\ln(n+2)} - \frac{1}{\ln(n+1)}\right) = \left(\frac{1}{\ln(3)} - \frac{1}{\ln(2)}\right) + \left(\frac{1}{\ln(4)} - \frac{1}{\ln(3)}\right) + \left(\frac{1}{\ln(5)} - \frac{1}{\ln(4)}\right) + ... + \left(\frac{1}{\ln(n+2)} - \frac{1}{\ln(n+1)}\right) \end{align*}
As we can see, some of the terms are canceled. \begin{align*} \sum_{n=1}^{\infty} \left(\frac{1}{\ln(n+2)} - \frac{1}{\ln(n+1)}\right) = \left(\cancel{\frac{1}{\ln(3)}} - \frac{1}{\ln(2)}\right) \quad + \left(\cancel{\frac{1}{\ln(4)}} - \cancel{\frac{1}{\ln(3)}}\right) \quad + \left(\cancel{\frac{1}{\ln(5)}} - \cancel{\frac{1}{\ln(4)}}\right) \quad + \cdots \quad + \left(\frac{1}{\ln(n+2)} - \cancel{\frac{1}{\ln(n+1)}}\right) \end{align*}
At the end, all we have is \begin{align*} \sum_{n=1}^{\infty} \left(\frac{1}{\ln(n+2)} - \frac{1}{\ln(n+1)}\right) = \left( - \frac{1}{\ln(2)} + \frac{1}{\ln(n+2)}\right) \end{align*}
And this is where the difference between my answer and solution sheet exist. In the solution, after the expanding the series and canceling some items, the result is \begin{align*} \left( - \frac{1}{\ln(2)} - \frac{1}{\ln(n+2)}\right) \end{align*}
Since I'm a bit confused about telescoping series, I'm worried that it might be my mistake. However, I couldn't find any error in my calculations. So, I would like to request for your help to classify if there is anything wrong with my approach or the solution provided from book has maybe a typo between - and +.
Thank you.
Yours is correct. The book has a typo.
However, your attempt is not correctly written. What you (and your book) calculated is not $\sum_{n=1}^{\infty} \left(\frac{1}{\ln(n+2)} - \frac{1}{\ln(n+1)}\right)$ but $\sum_{k=1}^n\left(\frac{1}{\ln(k+2)} - \frac{1}{\ln(k+1)}\right)$.
Note that likewise, a telescoping series $\sum_{n=1}^\infty(a_{n+1}-a_n)$ converges if and only if the sequence $(a_n)$ has a finite limit $\ell$, and the sum of the series is then $\ell-a_1$, because $$\sum_{k=1}^n(a_{k+1}-a_k)= a_{n+1}-a_1.$$