How to check if the sum of infinite series is convergent?
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!
$\endgroup$ 34 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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 $x\in (\pi/2,\pi).$
$\endgroup$ 5 $\begingroup$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
$$0<\sin^2(x) - \sin (x) +1<1$$
$\endgroup$ $\begingroup$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.
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