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

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?

• Looks like $\sum 1/n$ to me. – David Mitra Sep 26 '14 at 16:21

Use the limit comparison test with $\sum_{n=1}^\infty n^{-1}$.
$$\frac{1}{n\pi +\frac{\pi}{2}} \geq \frac{1}{n\pi +n\pi}\geq \frac{1}{8n} =\frac{1}{8}\cdot\frac{1}{n}$$
$$\dfrac{1}{n \pi + \frac{\pi }{2}} = \dfrac{2 }{ 2n \pi + \pi } = \dfrac{2}{\pi } \cdot \dfrac{1}{2n + 1} \ge \dfrac{2}{\pi } \cdot \dfrac{1}{2n + n} = \dfrac{2}{3 \pi } \cdot \dfrac{1}{n }$$
Now use the fact that the Harmonic Series $\sum \dfrac 1 n$ diverges.
$\displaystyle\frac{1}{n\cdot \pi+\frac{\pi}{2}}\ge\frac{1}{n\cdot\pi+\pi} =\frac{1}{\pi}\cdot\frac{1}{n+1}$, and since $\displaystyle\sum\frac{1}{n+1}$ diverges thus $$\sum\displaystyle\frac{1}{n\cdot \pi+\frac{\pi}{2}}$$ diverges!