# Decide if the series $\sum\frac{4^{n+1}}{3^{n}-2}$ converges or diverges and, if it converges, find its sum

Decide if the series $$\sum_{n=1}^\infty\frac{4^{n+1}}{3^{n}-2}$$ converges or diverges and, if it converges, find its sum.

Is this how you would show divergence attempt:

For $n \in [1,\infty), a_n = \frac{4^{n+1}}{3^n -2} \geq 0$

For $n \in [1,\infty), a_n = \frac{4^{n+1}}{3^n-2} \geq \frac{4^{n+1}}{3^n} = b_n$

Since $\sum_{n=1}^{\infty} \frac{4^{n+1}}{3^n}$ is a geometric series with $r = \frac{4}{3} > 1$. Therefore it diverges by the geometric series test and by the comparison test $\sum a_n$ diverges too.

• Yes you have got it right. One correction $n \in \Bbb N$. – Error 404 Apr 18 '17 at 6:07
• The numerator is bigger than the denominator! – Angina Seng Apr 18 '17 at 6:20
• You can as well say that the general term doesn't go to zero hence the function diverges. – kingW3 May 8 '17 at 18:08

Assume that the sequence converges for some $n$. Then: $$\lim_{n \to \infty} \left| \ \frac{4^{n+2}}{3^{n+1}-2} \cdot \frac{3^n-2}{4^{n+1}}\ \right| <1$$ $$\Rightarrow \lim_{n \to \infty} \left| \ \frac{4^{n+2}}{3^{n+1}} \cdot \frac{3^n}{4^{n+1}} \ \right| < 1$$ $$\Rightarrow \lim_{n \to \infty} \left| \ 4 \cdot \left(\frac{4}{3}\right)^{n+1} \cdot \left(\frac{3}{4}\right)^{n+1} \cdot \frac{1}{3} \ \right| < 1$$
$$\Rightarrow\left| \ \frac{4}{3} \ \right| < 1 \ \text{which is obviously false}$$
Therefore, the sum never converges for any $n$.
For $n\geq 1$ we have $3^n>3^n-2>0$ so $0<3^{-n}<(3^n-2)^{-1}$ so $4^{n+1}(3^n-2)^{-1}>4^{n+1}3^{-n}=4 (4/3)^n>4.$ If the terms of the series do not converge to $0$ then the sum cannot be convergent.