How to check if the sum of infinite series is convergent?

Question

I have this exercise where I need to find if the sum of infinite series is convergent:

$\sum_{n=1}^ \infty \frac{(\sin^2(x) - \sin (x) +1)^n}{\ln(1+n)} $ for x $ \in (\pi/2,\pi) $

Now I decided to do a ratio test for $ \frac{|a_n+1|}{a_n} $ but I am currently stuck on simplifying the result and proceeding with the solution.

$ \frac{(\sin^2(x) - \sin (x) +1)^{n+1}}{\ln(2+n)} \cdot \frac{\ln(1+n)}{(\sin^2(x) - \sin (x) +1)^{n}}$

I am not really sure how to proceed from here, any further help would be appreciated, thanks!

Answer 1

hint

$$\ln(n+2)=\ln\Bigl((n+1)(1+\frac{1}{n+1})\Bigr)$$

The limit of the ratio is

$$L=\sin^2(x)-\sin(x)+1$$

but

$$-1<\sin(x)\Bigl(\sin(x)-1\Bigr)<0$$

thus

$$0<L<1$$ the series is convergent.

Answer 2

As an alternative by root test

$$\sqrt[n]{\frac{(\sin^2(x) - \sin (x) +1)^n}{\ln(1+n)}}\to\sin^2(x) - \sin (x) +1$$

and on the interval

$$0<\sin^2(x) - \sin (x) +1<1$$

Answer 3

PARTIAL PROOF

Find the radius of convergence of the series $\frac{|a_{N+1}|}{|a_N|}$=$\left|\frac{\left(\ln\ \left(n+1\right)\right)}{\ln\left(n+2\right)}\left(\sin x^2-\sin x+1\right)\right|$

As n tends to ∞, Using L'Hospital's Rule,

$\frac{|a_{N+1}|}{|a_N|}$=$\left|\frac{\left(\left(n+2\right)\right)}{n+1}\left(\sin x^2-\sin x+1\right)\right|$

=$\left|\left(\sin x^2-\sin x+1\right)\right|$ (As n tends to ∞)

This must be <1 for the series to converge

Therefore,

$\left|\left(\sin x^2-\sin x+1\right)\right|<1$ for the series to converge

Prove that the interval lies in these values.

Answer 4

Show that if $0<u<1,$ then $0<u^2-u+1<1.$ Now fix $x\in (\pi/2,\pi).$ Then $0<\sin x <1.$ From the above, $0<\sin^2 x - \sin x + 1 <1.$ Hence

$$\sum_{n=1}^{\infty} (\sin^2 x - \sin x + 1)^n$$

is a positive convergent series. Dividing by $\ln (n+1)$ only helps. Thus your series converges for all $x\in (\pi/2,\pi).$