-4
$\begingroup$

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?

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Haris Gusic, Gabriel Romon, Martin R, user1551, Jendrik Stelzner
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What have you tried? Show some effort, for God's sake! $\endgroup$ – El Ectric May 21 at 18:38
  • 1
    $\begingroup$ @ElEctric No need to get God involved. $\endgroup$ – Gabriel Romon May 21 at 18:39
  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax. $\endgroup$ – dantopa May 21 at 21:37
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I just don't understand how sinx/x*y/siny parts equal to 1. $\endgroup$ – Arda Erem Karagoz May 21 at 19:02
  • $\begingroup$ @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... $\endgroup$ – DonAntonio May 21 at 19:03
  • $\begingroup$ 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. $\endgroup$ – Arda Erem Karagoz May 21 at 19:10
  • $\begingroup$ @ArdaEremKaragoz Did you see what I did? Figure it out algebraically, and nothice that $\;\frac1{4^n}\xrightarrow[n\to\infty]{}0\;$ ... $\endgroup$ – DonAntonio May 21 at 19:15
0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.