Apply the Integral Test: $\int_1^\infty \ln(x + 1) ~dx.$ Let $t=x+1$, $dt = dx$: $\int_1^\infty \ln(x+1) ~dx = \int_2^\infty \ln t ~dt.$ Integrate by parts: $dv = dt; \quad v=t; $ $u = \ln t; \quad du= 1/t ~dt.$ \begin{align*} \int u ~dv & = uv - \int v ~du \\ \int \ln t ~dt & = t \ln t - \int t \cdot \frac{1}{t} ~dt \\ & = t \ln t - t \\ & = (x+1) \ln(x+1) - (x+1). \end{align*} At the upper limit infinity, the integral diverges to infinity.
Therefore, the series diverges as well.
Is my procedure correct?
Is there an easier way to do it?