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I need some real help with this question. I have tried this question already and I am so stressed out from it. So what I did was that I used the Ratio Test but I couldn't get the result.

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Rabia K.
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2 Answers2

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Fix an $N$ such that $\frac{q^N}{p^N} < \frac{1}{2}$. Then for $n \ge N$, you have:

$$p^n > p^n - q^n = \left(1 - \frac{q^n}{p^n}\right)p^n > \frac{p^n}{2}$$

Now use the comparison test (ignoring the finite parts of the sums up to the $N$-th term) to conclude that your series converges iff $\sum_{n=1}^\infty \frac{1}{p^n}$ converges (which happens iff $p > 1$).

Rob Arthan
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$$p^n-q^n=p^n\Bigl(1-(\frac qp)^n\Bigr)$$

$$\implies p^n-q^n\sim p^n$$

your series has the same nature as the geometric series $$\sum \frac{1}{p^n}$$ which converges if and only if $p>1$.

hamam_Abdallah
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  • Thank you! I always try to solve these with convergence tests. Is there a different way? – Rabia K. Apr 27 '21 at 22:47
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    @RabiaK. Set $a_n=\frac{1}{p^n-q^n}$, $m=p/q$. Then $\frac{a_{n+1}}{a_n}=\frac{1}{q}\cdot\frac{m^n-1}{m^{n+1}-1}\rightarrow \frac{1}{mq}=\frac{1}{p}$ as $n\rightarrow \infty$. Hence our series diverges if $p<1$ and converges if $p>1$ by ratio test. When $p=1$ we get the series $\sum_{n-1}^{\infty}\frac{1}{1-q^n}$ which diverges by $n^{\text{th}}$ term test (since $0<q<1$) – Matthew H. Apr 27 '21 at 22:53