While solving a problem based on integration, I arrived at the following
$$\sum\limits_{x = 1}^{38} \ln\left(\frac{x}{x+1}\right)$$
I'm supposed to prove that this is less than $\ln(99)$ in order to do that, I tried evaluating the summation on wolfram(as I don't know how to compute this by myself) , but in the general form, wolfram uses the gamma function, which I don't know.
This is a telescoping series: $$ \sum_{x=1}^{38}\ln\frac{x}{x+1} = \sum_{x=1}^{38}\ln x -\ln(x+1) \\ = (\ln 1-\ln 2)+(\ln 2-\ln 3)+\cdots+(\ln 38-\ln 39) \\ = \ln 1 - \ln 39 \\ = -\ln 39. $$