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

First note that the following

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

Now we have the inequalities

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

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

Then, by the root criteria (if $kneq0$) the serie converge iff $frac{1}{pk}<1$ iff $1<pkleq p$.

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

$(frac q p)^{n} to 0$ 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 $p>1$.

You could better compute $k$, but it's not the best method. Note also that the simple fact that $0le a_nle1$ doesn't guarantee the existence of $lim_{ntoinfty}a_n$, so you don't really have $k$.

First method.

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

Second method.

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