Answer to Question #297344 in Real Analysis for Tokyo

Question

{F} test the series : infinity sigma n=1 (-1) ^n-1 sin nx/n√n for absolute and conditional convergence

Answer 1

Solution:

$\Sigma_{n=1}^{\infty} \dfrac{(-1)^{n-1}\sin nx}{n\sqrt n} \\=\Sigma_{n=1}^{\infty} \dfrac{(-1)^{n-1}\sin nx}{n^{3/2}}$

Let $a_n=\dfrac{(-1)^{n-1}\sin nx}{n^{3/2}}$

$\Rightarrow a_{n+1}=\dfrac{(-1)^{n}\sin (n+1)x}{(n+1)^{3/2}}$

Now, $p=\lim_{n\rightarrow \infty}|\dfrac{a_{n+1}}{a_n}|=\lim_{n\rightarrow \infty}|\dfrac{\dfrac{(-1)^{n}\sin (n+1)x}{(n+1)^{3/2}}}{\dfrac{(-1)^{n-1}\sin nx}{n^{3/2}}}|$

$=\lim_{n\rightarrow \infty}|{\dfrac{(-1)\sin (n+1)x.n^{3/2}}{(n+1)^{3/2}\sin nx}}| \\=\lim_{n\rightarrow \infty}|{\dfrac{\sin (n+1)x}{(1+\frac 1n)^{3/2}\sin nx}}| \\=\lim_{n\rightarrow \infty}|{\dfrac{1}{(1+\frac 1n)^{3/2}}}|\times \lim_{n\rightarrow \infty}|{\dfrac{\sin (n+1)x}{\sin nx}}| \\=1\times \infty \\= \infty$

So, $p>1$

Thus, the series is divergent by the ratio test.