# Determine the series whether convergence or divergence with using ratio rest. [closed]

This is the problem:

$$\sum_{n=0}^\infty 3^n\sin((\frac{1}{4})^n)$$

I can't prove the convergence of this series, how can we solve it?

## closed as off-topic by Haris Gusic, Gabriel Romon, Martin R, user1551, Jendrik StelznerMay 22 at 0:55

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• What have you tried? Show some effort, for God's sake! – El Ectric May 21 at 18:38
• @ElEctric No need to get God involved. – Gabriel Romon May 21 at 18:39
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With the ratio test:

$$\frac{a_{n+1}}{a_n}=3\frac{\sin\frac1{4^{n+1}}}{\sin\frac1{4^n}}=3\cdot\frac{\sin\frac1{4^{n+1}}}{\frac1{4^{n+1}}}\cdot\frac{\frac1{4^n}}{\sin\frac1{4^n}}\cdot\frac{4^n}{4^{n+1}}\xrightarrow[n\to\infty]{}3\cdot1\cdot1\cdot\frac14=\frac34<1$$

and thus the series (a positive one, indeed) converges.

• I just don't understand how sinx/x*y/siny parts equal to 1. – Arda Erem Karagoz May 21 at 19:02
• @ArdaEremKaragoz Don't you know the basic limit $$\lim_{x\to0}\frac{\sin x}x=1\;?$$ You must have studied this if you're studying now infinite series... – DonAntonio May 21 at 19:03
• Of course i know this but it works when limit goes to 0. In the ratio test we take the limit goes to infinity, right? If your answer is yes then I will again asking to how it is equal to 1. – Arda Erem Karagoz May 21 at 19:10
• @ArdaEremKaragoz Did you see what I did? Figure it out algebraically, and nothice that $\;\frac1{4^n}\xrightarrow[n\to\infty]{}0\;$ ... – DonAntonio May 21 at 19:15

As this is a series with positive terms, using equivalence, makes it very simple:

Near $$0$$, $$\;\sin u\sim u$$, so $$\;\sin\dfrac 1{4^n}\sim_{n\to \infty} \dfrac 1{4^n}$$, whence $$3^n\sin\dfrac 1{4^n}\sim_{n\to \infty} \dfrac {3^n}{4^n}=\Bigl(\frac34\Bigr)^n,$$ a convergent geometric series.