Is $\sum_{n=1}^\infty\frac{\lfloor\sin(\frac{4n}{\pi})+3\rfloor}{4^n}$ convergent? I have to determine whether the following series is convergent or not, and if it converges I have to find its sum. The series is:
$$\sum_{n=1}^\infty\frac{\lfloor\sin(\frac{4n}{\pi})+3\rfloor}{4^n}$$
The problem is that I don´t have any idea how to proceed with this excersice. I have been thinking about using the comparison test but the serie changes the value of its numerator. For example when $n=8$ and $n=9$ the numerator has a value of $2$ in other values for n the numerator is $3$ so I am confused about it. Could anyone help me?
 A: HINT 
$$|\sin{(\frac{4n}{\pi})}| \leq 1$$ 
Use comparison test and prove the absolute convergence of the series which will give you the general convergence
A: We know when $\sin t=\pm1$ iff $t=k\pi$ here $\dfrac{4n}{\pi}=k\pi$ shows $n=\dfrac{k\pi^2}{4}$ that's never happen. Then $|\sin\dfrac{4n}{\pi}|<1$ and 
$|\lfloor\sin(\dfrac{4n}{\pi})+3\rfloor|\leq3$ for all $n$. Then
$$\sum_{n=1}^\infty\frac{\lfloor\sin(\frac{4n}{\pi})+3\rfloor}{4^n}\leq3\sum_{n=1}^\infty\frac{1}{4^n}=1$$
A: We are going to compare the given series to a convergent geometric series.  Recall that if $|r| < 1$, then
$$ \sum_{n=0}^{\infty} r^n < \infty. $$
(Indeed, we can actually do slightly better in this case, but it won't really help us to solve the given problem, so lets not worry about what this series actually converges to).
First, note that $\lfloor x \rfloor \le x$ for all real $x$, from which we obtain
$$ \lfloor \sin(\theta) + 3 \rfloor \le \sin(\theta) + 3$$
for all real $\theta$.  Next, observe that $-1 \le \sin(\theta) \le 1$ for any $\theta\in\mathbb{R}$ and so
$$ \sin(\theta) + 3 \le 1 + 3 = 4 \qquad\forall\theta\in\mathbb{R}.$$
Combining the inequalities, and applying them to the series, we conclude that
$$ \sum_{n=1}^{\infty} \frac{\left\lfloor \sin\left( \frac{4n}{\pi} \right) + 3 \right\rfloor}{4^n}
\le  \sum_{n=1}^{\infty} \frac{\sin\left( \frac{4n}{\pi} \right) + 3}{4^n}
\le \sum_{n=1}^{\infty} \frac{4}{4^n}
= \sum_{n=0}^{\infty} \left(\frac{1}{4} \right)^n
< \infty.$$
We have shown that the original series converges, and we can even get a reasonable bound on its limit (if we work out what the geometric series that we are bounding by converges to), but we have not actually computed the value of the series.  That requires more advanced techniques (either numerical, or analytic).
A: We have $|\sin(x)+3|\leq 1+3=4$ for all $x\in\mathbb R$. The floor-function won't change this so you can adapt that and use it for your series.
A: Hint:
This is a series with positive terms. You can use asymptotic analysis:
$$\frac{\lfloor\sin(\frac{4n}{\pi})+3\rfloor}{4^n}=O\Bigl(\frac 1{4^n}\Bigr),$$
and the latter is a convergent geometric series.
A: HINT:
We can see this serie as $\sum_{i=1}^\infty(\frac{1}{4})^n\lfloor sin(\frac{4n}{\pi})+3\rfloor$, also $\lfloor sin(\frac{4n}{\pi})+3\rfloor \le 1 + 3$
