If $\sum a_n$ converges, does $\sum a_n / 2^n$ converge as well?

Question

If $\sum a_n$ converges, does $\sum \dfrac{a_n}{2^n}$ converge as well?

I can't use differential or integral calculus for this.

Answer 1

The series $\sum a_n$ converges so

$$\exists n_0\in\Bbb N\;|\; \forall n\ge n_0,\quad |a_n|\le 1$$ so $$\exists n_0\in\Bbb N\;|\; \forall n\ge n_0,\quad \frac{|a_n|}{2^n}\le \frac1{2^n}$$ and we conclude the convergence of the desired series by comparison with the geometric convergent series $\sum\frac1{2^n}$.

Answer 2

Since I don't have enough reputation to add comments I will write it as an answer. Couldn't we just use Abel's Test for convergences? (see here: http://en.wikipedia.org/wiki/Abel%27s_test)

In terms of this test:

$\sum{\frac{1}{2^n}}$ is $b_n$. It is true that $b_1\geq b_2 \geq b_3 \geq... \geq 0$ Also, it is upper bounded by $1$.

And $\sum{a_n}$ is convergents. That is given.

So by this test also $\sum {a_n*b_n}$ is also convergents.