It is my calculations about the convergence of a series are correct?

Question

Consider $0<q<p$ real numbers. I wana find for which values of $p$ the series $\sum\frac{1}{p^n-q^n}$ is convergent.

First note that the following

$$ \sum\frac{1}{p^n-q^n}=\sum\frac{1}{p^n(1-(\frac{q}{p})^n)} $$

Now we have the inequalities

$$ 0\leq\sqrt[n]{1-\left(\frac{q}{p}\right)^n}\leq 1 $$

So we have that $\lim_{n\rightarrow\infty}\sqrt[n]{1-\left(\frac{q}{p}\right)^n}=k$ where $0\leq k\leq 1$

Then, by the root criteria (if $k\neq0$) the serie converge iff $\frac{1}{pk}<1$ iff $1<pk\leq p$.

My question: As you see, my process depend of $k$ because I assume that $k\neq 0$ but I can't prove it. Can someone give me a hint?

Answer 1

$(\frac q p)^{n} \to 0$ because $q<p$. Hence $1-(\frac q p)^{n} \to 1$. Comparing the given series with $\sum \frac 1 {p^{n}}$ we see that the series converges iff $p>1$.

Answer 2

You could better compute $k$, but it's not the best method. Note also that the simple fact that $0\le a_n\le1$ doesn't guarantee the existence of $\lim_{n\to\infty}a_n$, so you don't really have $k$.

First method.

Note that $$ \lim_{n\to\infty}\frac{\dfrac{1}{p^n}}{\dfrac{1}{p^n-q^n}}= \lim_{n\to\infty}\left(1-\left(\dfrac{q}{p}\right)^{\!n}\right)=1 $$ so the two series $\sum_n\frac{1}{p^n-q^n}$ and $\sum_{n}\frac{1}{p^n}$ are either both convergent or both divergent.

Second method.

The ratio test, setting $r=q/p<1$, $$ \lim_{n\to\infty}\frac{\dfrac{1}{p^{n+1}-q^{n+1}}}{\dfrac{1}{p^n-q^n}}= \lim_{n\to\infty}\frac{p^{n}-q^n}{p^{n+1}-q^{n+1}}= \lim_{n\to\infty}\frac{1-r^n}{p-qr^n}=\frac{1}{p} $$ would only leave the case $p=1$ to be determined. Can you do the case $p=1$?