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Show that $\sum _{n=1 } ^{\infty } (n \pi + \pi/2)^{-1 } $ diverges.

Both the root test and the ratio test is inconclusive. Can you suggest a series for the series comparison test?

Thanks in advance!

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    $\begingroup$ Looks like $\sum 1/n$ to me. $\endgroup$ – David Mitra Sep 26 '14 at 16:21
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Use the limit comparison test with $\sum_{n=1}^\infty n^{-1}$.

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$$\frac{1}{n\pi +\frac{\pi}{2}} \geq \frac{1}{n\pi +n\pi}\geq \frac{1}{8n} =\frac{1}{8}\cdot\frac{1}{n}$$

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It is never too pretty to go for the Root and Ratio Tests on the onset. A much more elegant method would be to notice that:

$$ \dfrac{1}{n \pi + \frac{\pi }{2}} = \dfrac{2 }{ 2n \pi + \pi } = \dfrac{2}{\pi } \cdot \dfrac{1}{2n + 1} \ge \dfrac{2}{\pi } \cdot \dfrac{1}{2n + n} = \dfrac{2}{3 \pi } \cdot \dfrac{1}{n } $$

Now use the fact that the Harmonic Series $ \sum \dfrac 1 n $ diverges.

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$\displaystyle\frac{1}{n\cdot \pi+\frac{\pi}{2}}\ge\frac{1}{n\cdot\pi+\pi} =\frac{1}{\pi}\cdot\frac{1}{n+1}$, and since $\displaystyle\sum\frac{1}{n+1}$ diverges thus $$\sum\displaystyle\frac{1}{n\cdot \pi+\frac{\pi}{2}}$$ diverges!

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