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Finding Radius of convergence of series $$\sum^{\infty}_{r=1}x^{r}\cdot \cos^2(r)$$ is

What i try

Let $a_{n}=\cos^2(n)\cdot x^{n}.$ Then $a_{n+1}=\cos^2(n+1)\cdot x^{n+1}$

Now $$\lim_{n\rightarrow \infty}\bigg|\frac{\cos^2(n+1)\cdot x^{n+1}}{\cos^2(n)\cdot x^{n}}\bigg|<1$$

$$\lim_{n\rightarrow \infty}|x|\bigg|\frac{\cos^2(n+1)}{\cos^2(n)}\bigg|<1$$

How do i solve it After that . Help me please

Thanks

jacky
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    It appears you are trying to use the Ratio Test, but I don't believe that will be conclusive here. The ratio of $\cos(r+1)$ to $\cos(r)$ will be unbounded. – 2'5 9'2 May 11 '20 at 04:50
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    Also, some places you are using "$r$" and other places "$n$". But it seems like you mean those to be the same thing. – 2'5 9'2 May 11 '20 at 04:52
  • Thanks i have edited it. – jacky May 11 '20 at 04:55

3 Answers3

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Or you can notice that since $\cos^2 r = \frac{1}{2} + \dfrac{ \cos(2r) }{2}$, then your series is

$$ \frac{1}{2} \sum x^r + \dfrac{1}{2} \sum \cos (2r) x^r $$

Now the first series have radious of convergence $R=1$ and notice that for the second sum $|\cos 2r| < 1$

James
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  • Thanks James i have understand first part. (Geometric sum which is defined when $|x|<1$). But did not understand second part. Can you explain little more). Thanks – jacky May 11 '20 at 04:57
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    Just like $|\cos2r|<1$, also $|\cos^2(r)|<1$. So the new series has the same feature as the original. – 2'5 9'2 May 11 '20 at 05:06
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If $|x|>1$, your terms do not converge to $0$. So the radius is at most $1$.

If $|x|<1$, then your $n$th term has absolute value at most $|x|^n$. So then the series absolutely converges, and therefore the radius is at least $1$.

It follows that the radius is $1$.

2'5 9'2
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  • Thanks alex. When i substitute $|x|>1$ in $\displaystyle \lim_{n\rightarrow \infty}\bigg|\frac{\cos^2(n+1)\cdot x^{n+1}}{\cos^2(n)\cdot x^{n}}\bigg|<1$. Then How can i find value of $\cos$ ratio. Plesse explain me. – jacky May 11 '20 at 04:58
  • Can we Directly say $$\sum^{\infty}{r=1}\cos^2(r)\cdot x^{r}\leq \sum^{\infty}{r=1}x^{r}=\frac{x}{1-x}.$$ which is converge when $|x|<1.$ So our original series is converge $|x|<1.$ – jacky May 11 '20 at 05:51
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$\displaystyle\lim_{n\rightarrow \infty}|x|\bigg|\dfrac{\cos^2(n+1)}{\cos^2(n)}\bigg|\lt 1$

$\displaystyle\implies |x| \lim_{n\rightarrow \infty}\bigg|\dfrac{\cos^2(n+1)}{\cos^2(n)}\bigg|\lt 1$

$\begin{align}\implies |x| &\lt\dfrac{1}{\displaystyle\lim_{n\rightarrow \infty}\bigg|\frac{\cos^2(n+1)}{\cos^2(n)}\bigg|}\\&=\displaystyle\lim_{n\rightarrow \infty}\dfrac{1}{\bigg|\frac{\cos n\cos 1-\sin n\sin 1} {\cos n}\bigg|^2}\\&=\displaystyle\lim_{n\rightarrow \infty}\dfrac{1}{\bigg|\cos 1-\tan n\sin 1\bigg|^2}\\&= \dfrac{1}{\bigg|\cos 1-\dfrac{\pi}{2}\sin 1\bigg|^2}\end{align}$

Jimmy
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  • Can we say directly $$\sum^{\infty}{r=1}\cos^2(r)x^r\leq \sum^{\infty}{r=1}x^r=\frac{x}{1-x}$$ which is converge when $|x|<1$ . So our original series is converge for $|x|<1.$ – jacky May 11 '20 at 05:53
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    Sure, why not! I've just continued from where you got stuck. In that case you can say better : $x\lt 1$. But here you see $|x|\lt\dfrac{1}{\cos1-\frac{\pi}{2}\sin1}=1.0283\ldots$. Certainly $x\lt 1$ is better approximation than $x\lt 1.0283\ldots$ for the radius of convergence. And as @alex proves the radius of convergence is indeed $1$. – Jimmy May 11 '20 at 06:05