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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?

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    The upper limit of my summation should be $N$, of course. And the $n$ in the factorial should be $N$. Oops.2012-05-21

1 Answers 1

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I see no problems assuming you kept track of your bounds of integration properly. However, you do not need to do the integral test to see that $\sum\limits_{n=1}^\infty \ln(n+1)$ diverges. If it were to converge, then $\lim\limits_{n\to\infty} \ln(n+1)=0$ but this is clearly not the case.