Is this a counter example for a comparison test for sequences? I’ve recently started learning about sequences and convergence and divergence, and I came across the comparison test for sequences. What I have  is that:

What if $a_n$ is defined as a periodic function, such as $a_n = 1 + sin (x)$
If $b_n$ is any function that is greater than $a_n$ that does converge, shouldn't $a_n$ converge?
As an example, let $b_n = (3x+2)/(x+1)$
This seems to satisfy the conditions, yet $a_n$ (the sinusoidal function) does not appear convergent...

 A: The Comparison Test says that, if $0 \le a_n \le b_n$, then


*

*If $\sum\limits_{n=1}^\infty b_n$ converges, then $\sum\limits_{n=1}^\infty a_n$ also converges.

*If $\sum\limits_{n=1}^\infty a_n$ diverges, then $\sum\limits_{n=1}^\infty b_n$ also diverges.
A: The result you're trying to prove is false. Let $a_n = (-1)^n + 2$, and $b_n = 5$. Clearly $b_n \to 5$, but $a_n$ oscillates between $3$ and $1$ without converging. You've basically mimicked that example, but with some functions (strictly, you should evaluate your functions at each natural number in order to get a true sequence).
There is a related result, an instance of the squeeze theorem, which states that if $0 \leq a_n \leq b_n$ and $b_n \to 0$, then $a_n \to 0$.
A: Hint. As Carl Heckman noticed, you have probably mixed up sequences and series.
For series $\sum a_n$ and $\sum b_n$, such that

$$0\leq a_n\leq b_n,$$

the following holds true:

If the series $\sum a_n$ diverges then the series $\sum b_n$ diverges
If the series $\sum b_n$ converges then the series $\sum a_n$ converges

