1
$\begingroup$

I'm having trouble identifying which test to use since the terms in the series oscillate between positive and negative values.

$\endgroup$
  • 2
    $\begingroup$ Note that $$\frac{|\sin 6n|}{1+2^n}\leq\frac{1}{1+2^n}<\frac{1}{2^n}$$ and $\sum_{n=1}^\infty\frac{1}{2^n}$ converges. $\endgroup$ – DiegoMath Apr 17 at 22:39
  • $\begingroup$ Thus, the series converges absolutely. Now, by comparison test... $\endgroup$ – DiegoMath Apr 17 at 22:39
  • $\begingroup$ @DonAntonio Indeed, thanks! I fixed the comment $\endgroup$ – DiegoMath Apr 17 at 22:44
  • $\begingroup$ So by taking the absolute value of the original expression, does this mean that the series is absolutely convergent? $\endgroup$ – Jeremiah Apr 17 at 22:49
  • 1
    $\begingroup$ @Jeremiah Indeed so, and thus it also converges. $\endgroup$ – DonAntonio Apr 17 at 23:59
0
$\begingroup$

We know that

$$\int_0^{\infty} \dfrac{sin(x)}{x} \ dx$$ and therefor $$\sum_{n=0}^{\infty} \dfrac{sin(n)}{n}$$

converges.

There are many detailed proofs on the web so I will not focus on proving that.

But you can see that:

$$|\dfrac{sin(n)}{2^n}| \leq |\dfrac{sin(n)}{n}|$$

And therefor it must be convergent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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