# For what value of r and p is the series convergent $\sum _{ n=1 }^{ \infty } \frac{r^n}{n^p}$

I have been given the series

$$\sum _{ n=1 }^{ \infty } \frac{r^n}{n^p}$$, where $$r, p > 0$$

Which seems to be a combination of a geometric series and a p-series.

The summation of geometric series is finite and has the formular $$\sum _{ n=1 }^{ \infty } ar^n=\frac{a}{1-r}$$

It converges for $$r \in (-1,1)$$

whereas the p series does not have a direct formular $$\sum _{ n=1 }^{ \infty } \frac{1}{n^p}$$

Converges for $$p>1$$

Can I use ratio test if I am interested to find values for r and p so the series will converge? I do not know if the above is relevant to find the values

What I have been thinking so far by applying ratio test $$\lim_{n \rightarrow\infty}\frac{\frac{r^{n+1}}{(n+1)^p}}{\frac{r^n}{n^p}}$$

which simplifies to $$\lim_{n \rightarrow\infty}\frac{rn^p}{(n+1)p}$$

Is it then correct if $$0 the series will converge?

We can use the root test to test for absolute convergence. The $$n$$th root of the absolute value of the series is given by $$\frac{|r|}{n^{p/n}}\to|r|$$ as $$n\to\infty$$. Hence the series converges if $$|r|<1$$ (regardless of the value of $$p$$). If $$r=1$$, the series converges iff $$p>1$$ by the $$p$$ series test. If $$r=-1$$, the series converges iff $$p>0$$ by the alternating series test (if $$p\leq 0$$ the terms don't go to zero in this case). If $$|r|>1$$, the series does not converge (as the terms don't go to zero regardless of the value of $$p$$).
• OP said $r,p>0$. – Kavi Rama Murthy May 19 '19 at 0:01
You missed one point. If $$r=1$$ then the series converges iff $$p >1$$. Also if $$r<1$$ then it converges for all $$p$$. No need for $$p<1$$. Ratio test is the right approach. [Use the fact that $$(1+\frac 1 n)^{p} \to 1]$$. The series diverges for all $$p>0$$ if $$r >1$$.