1
$\begingroup$

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?

  • 1
    Kindly refer here (http://meta.math.stackexchange.com/questions/107/) on how to typeset (i.e. how to write equations etc) on this website. Typesetting makes it easier to read.2012-05-21
  • 0
    Also, did you mean $\sum_{n=1}^{\infty} \log(n+1)$ or $\sum_{n=1}^{\infty} \log((n+1)/n)$?2012-05-21
  • 0
    @Henry T. Horton: Great work!2012-05-21
  • 3
    Remember that if the terms of a series don't go to zero, the series can't converge. The integral test also comes with conditions which you haven't checked.2012-05-21
  • 0
    You could also notice that $\sum_{n=1}^{\infty} \log(n+1) = \log((n+1)!)$.2012-05-21
  • 0
    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

8

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.