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a) $\frac{ \sin(n) + e^n}{\cos(n) + \pi^n}$ I expanded and got $\sin (n) + e^n/(cos(n) + \pi^n) x \cos (n+1) + \pi^{n+1}/\sin(n+1) + e^{n+1}$?

and also how to simplify $(n+1)^e/(n^e)e$

Edit: I totally got confused with the ratio test formula.Sorry my bad.But thanks for helping!

Wouter
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user9535888
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2 Answers2

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I can't really make out your work, but it should go something like this: $$a_n = \frac{e^n+\sin(n)}{\pi^n +\cos(n)}$$

$$\frac{a_{n+1}}{a_n} = \frac{ \frac{e^{n+1}+\sin(n+1)}{\pi^{n+1} +\cos(n+1)} }{ \frac{e^n+\sin(n)}{\pi^n +\cos(n)} } = \frac{e^{n+1}+\sin(n+1)}{e^n+\sin(n)} \cdot \frac{\pi^n+\cos(n)}{\pi^{n+1}+\cos(n+1)}$$ Using trivial bounds on $\sin$ and $\cos$,

$$\underbrace{\frac{e^{n+1} - 1}{e^n + 1}}_{\to e} \cdot \underbrace{\frac{\pi^n - 1}{\pi^{n+1}+1}}_{\to 1/\pi} \leq \frac{a_{n+1}}{a_n} \leq \underbrace{\frac{e^{n+1} + 1}{e^n - 1}}_{\to e} \cdot \underbrace{\frac{\pi^n + 1}{\pi^{n+1}-1}}_{\to 1/\pi}$$ by the squeeze theorem $\lim_{n\to\infty}\tfrac{a_{n+1}}{a_n} = \tfrac{e}{\pi} < 1$. So we may apply the ratio test.

adfriedman
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$$a_n = \frac{e^n+\sin(n)}{\pi^n +\cos(n)} = \left( \frac{e}{\pi} \right)^n \cdot \frac{1+\frac{\sin{n}}{e^n}}{1+\frac{\cos{n}}{\pi^n}} \Rightarrow$$ $$\frac{a_{n+1}}{a_n}=\frac{e}{\pi}\cdot \frac{1+\frac{\sin{(n+1)}}{e^{n+1}}}{1+\frac{\cos{(n+1)}}{\pi^{n+1}}}\cdot \frac{1+\frac{\cos{n}}{\pi^n}}{1+\frac{\sin{n}}{e^n}} \stackrel{n\rightarrow\infty}{\longrightarrow}\frac{e}{\pi}\cdot \frac{1+0}{1+0}\cdot \frac{1+0}{1+0}=\frac{e}{\pi}$$

trancelocation
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