$$ \sum_{n\ge 0} (3+\cos n)x^n ; a_n = (3+\cos n) $$ I used d'Alembert : $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{3+\cos(n+1)}{3+\cos n} $$ Nw I'm stuck With How To get Rid Of "cosine" functions which Doesn't have limits at infinity ; a Hint Pleaze
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The limit you're after does not exist. Also, use $\to$ instead of $->$. – Ivan Neretin Jan 29 '16 at 10:23
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how so , it certainly has a Radius But how to find it – Serius Black Jan 29 '16 at 10:27
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Root test: $$\sqrt[n]{|(3+\cos n)x^n|}\to |x|$$ because $$2\le(3+\cos n)\le 4\implies \sqrt[n]{2}\le\sqrt[n]{(3+\cos n)}\le\sqrt[n]{4}.$$ So the radius is 1.
Martín-Blas Pérez Pinilla
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thank you Very Much ; So I think D'ALmebert Doesn't Work with Some trigonometrics a_n – Serius Black Jan 29 '16 at 10:34