How to show that $\sum {2^j + j \over 3^j - j}$ converges

Question

I'm not quite sure how to go about growing the numerator or shrinking the denominator to perform a tricky comparison test, or hashing out the ratio test, so any help would be much appreciated!

Answer 1

Hint:

$$\frac{2^j + j}{3^j - j} \le \frac{2^j + 2^j}{3^j - j} \le \frac{2^j + 2^j}{3^j - \frac{1}{2} 3^j}$$

for all sufficiently large $j$.

Answer 2

$\dfrac{2^j + j}{3^j - j}-\dfrac{2^j}{3^j} =\dfrac{3^j(2^j + j)-2^j(3^j - j)}{3^j(3^j - j)} =\dfrac{j(3^j+2^j)}{3^j(3^j - j)} $, so $\big|\dfrac{2^j + j}{3^j - j}-\dfrac{2^j}{3^j}\big| =\big|\dfrac{j(3^j+2^j)}{3^j(3^j - j)}\big| <\big|\dfrac{2j3^j)}{3^j(3^j - j)}\big| =\big|\dfrac{2j}{3^j - j}\big| $.

All you need now is $\lim_{j \to \infty} \dfrac{j}{3^j}= \ 0 $.

Answer 3

The terms are positive, and asymptotically equal to $$a_j=\frac{2^{j}}{3^j}$$ That is, if $b_j$ is your sequence, $\lim\limits_{j\to\infty}\dfrac{a_j}{b_j}=1$. Can your prove this?