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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 $0<r,p<1$ the series will converge?

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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 $p$).

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    $\begingroup$ OP said $r,p>0$. $\endgroup$ – Kavi Rama Murthy May 19 '19 at 0:01
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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 $r >1$.

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