How to use direct comparison test to prove convergence of series

Question

Use direct comparison test to prove if the following series converge or not.

A) $\sum_{n=0}^\infty (\frac{8}{3^n +2})$

B) $\sum_{n=0}^\infty(\frac{1}{2^n +3^n})$

Well, I don't understand very well the direct comparison test. I know what it says but I don't know how to apply it. I was told that when using direct comparison test you should use p-series, but is it true? I mean, I could say that $3^n+2>n$ so $\frac{8}{3^n +2}<\frac{8}{n}$. But is $\frac{8}{n}$ convergent or divergent? Because, $\frac{1}{n}$ diverges, but does $\frac{8}{n}$ diverge too?

Answer 1

The series $\sum_{n=1}^\infty\frac8n$ diverges, but $\frac8{3^n+2}<\frac8{3^n}$ and $\sum_{n=1}^\infty\frac8{3^n}$ converges (apply the ratio test). And $\frac1{2^n+3^n}<\frac1{2^n}$ and the series $\sum_{n=1}^\infty\frac1{2^n}$ converges.

Answer 2

$3^n +2 > 3^n$ implies that $$\frac{8}{3^n+2} < \frac{8}{3^n}$$ and the series $$\sum_{n=0}^\infty \frac{1}{3^n}$$ clearly converges.

Answer 3

Note the harmonic series $\sum_n\dfrac 1n$ diverges. But $\sum_n\dfrac 8n=8\sum_n\dfrac1n$, thus it also diverges.

The geometric series $\sum_n x^n$ converges (to $\dfrac 1{1-x}$) iff $\vert x\vert\lt1$.

Thus $\sum_n( \dfrac 13)^n$ converges.

Now by the comparison test, $\sum_n\dfrac8{3^n+2}\lt\sum_n\dfrac8{3^n}=8\sum_n(\dfrac13)^n \lt\infty $.

Similarly $\sum_n \dfrac 1{2^n+3^n}\lt\sum_n\dfrac 1{3^n}\lt\infty $.

(Or you could do $\sum_n\dfrac 1{2^n+3^n}\lt\sum_n\dfrac 1{2^n}\lt\infty $.)

I don't see any way to use p-series here, since those are sums of the form $\sum_n\dfrac 1{n^p}$.