I mainly ask this because I was thinking about doing Direct Comparison Test for Convergence testing on Series.

I am looking at

$$\sum_{n=2}^{\infty} \frac{(\arctan n)(\ln n)}{n^2}$$

which I am going to compare with

$$\sum_{n=2}^{\infty} \frac{\frac{\pi}{2}(\ln n)}{n^2}$$

can I say that

$$\sum_{n=2}^{\infty} \frac{\frac{\pi}{2}(\ln n)}{n^2} < \sum_{n=2}^{\infty} \frac{n^{0.5}}{n^2} =\sum_{n=2}^{\infty} \frac{1}{n^{1.5}}$$

The final series converges by P series and because that's greater than the original series, the original series also converges.

Did I make any incorrect assumptions?

  • 1
    $\begingroup$ Looks like a good approach. $\endgroup$
    – abiessu
    Commented Nov 14, 2017 at 18:06
  • $\begingroup$ Is the chain of inequalities in the title correct? $\endgroup$
    – Smit Shah
    Commented Nov 14, 2017 at 18:11
  • 3
    $\begingroup$ $\arctan(n)<\frac{\pi}{2}, \log(n)<\sqrt{n}$ hence the given series is absolutely convergent to some number $\leq \frac{\pi}{2}\left[\zeta(\frac{3}{2})-1\right]$, correct. (You dropped the $\pi/2$ constant but that is not really relevant for proving convergence). $\endgroup$ Commented Nov 14, 2017 at 18:22


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