Divergence of infinite series $(\frac{3k-2}{4k+2})^{2k-3}$

Question

I would like to ask if my solution for testing the divergence of the infinite series below is correct.

$$\sum_{k=1}^{\infty} \left(\frac{3k-2}{4k+2}\right)^{2k-3}$$

I used the Cauchy ratio test.

$$ \lim_{k \to \infty} \left(\frac{3k-2}{4k+2}\right)^{2k-3} = \lim_{k \to \infty} \left(\frac{3k-2}{4k+2}\right)^{2k} \cdot \left(\frac{3k-2}{4k+2}\right)^{-3} = \lim_{k \to \infty} \left(\frac{\left(3k-2\right)^{2}}{\left(4k+2\right)^{2}}\right)^{k} \cdot \left(\frac{3k-2}{4k+2}\right)^{-3} = \lim_{k \to \infty}\left(\frac{9k^{2}-12k+4}{16k^2+16k+4}\right)^{k} \cdot \left(\frac{k\left(3-\frac{2}{k}\right)}{k\left(4+\frac{2}{k}\right)}\right)^{-3} = \lim_{k \to \infty} \left(\frac{k^{2}\left(9-\frac{12}{k}+\frac{4}{k^{2}}\right)}{k^{2}\left(16+\frac{16}{k}+\frac{4}{k^{2}}\right)}\right)^{k} \cdot \left(\frac{3}{4}\right)^{-3} = \lim_{k \to \infty} \frac{64}{27} \cdot \left(\frac{9}{16}\right)^{k} $$ Therefore: $$ \left\lvert \frac{\frac{64}{27} \cdot \left(\frac{9}{16}\right)^{k}}{\frac{64}{27} \cdot \left(\frac{9}{16}\right)^{k+1}} \right\rvert = \left\lvert \frac{\frac{9^{k}}{16^{k}}}{\frac{9^{k+1}}{16^{k+1}}} \right\rvert = \left\lvert \frac{9^{k}\cdot 16^{k+1}}{16^{k}\cdot 9^{k+1}} \right\rvert = \left\lvert \frac{9^{k}}{9^{k+1}} \cdot \frac{16^{k+1}}{16^{k}} \right\rvert = \left\lvert 9^{k-k+1}\cdot16^{k+1-k} \right\rvert = \left\lvert 9\cdot16 \right\rvert = 144 $$

As $144 > 1$, the series is divergent.

Is this a correct solution?

Answer 1

For this problem, the ratio test requires careful calculations with which you made some mistakes. The easiest approach is to note the $k$th term is positive but less than $(3/4)^{2k-3}$, so the series converges.

Answer 2

$$ \left\lvert \frac{9^{k}}{9^{k+1}} \cdot \frac{16^{k+1}}{16^{k}} \right\rvert = \left\lvert 9^{k-\color{red}{(k+1)}}\cdot16^{k+1-k} \right\rvert = \left\lvert 9^{k-k-1}\cdot16^{k+1-k} \right\rvert = \left\lvert \frac {16} 9 \right\rvert $$