# How to use direct comparison test to prove convergence of series

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?

• In fact, the series $\sum \frac{1}{n}$ diverges. Aug 3, 2019 at 22:02

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.

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

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 $$\sum_n\dfrac 1{n^p}$$.