Show the series $ \sum_{j=1}^{\infty} \frac{(2^j)+ j}{(3^j) - j} $ converges Show the following series converges

$$ \sum_{j=1}^{\infty} \frac{(2^j)+ j}{(3^j) - j} . $$

I tried to use the comparison test and tried to compare it with the series of $\dfrac{2^j}{3^j}$ because this is the geometric series. However, $\dfrac {2^j}{3^j}$ is smaller than $\dfrac{2^j+j}{3^j-j}$, so by comparison test, the series $\sum_{j=1}^{\infty}\dfrac{2^j}{3^j}$ converges does NOT indicate that the series $\sum_{j=1}^{\infty}\dfrac{2^j+j}{3^j-j}$ converges... Thus I'm confused. Thanks!
Thanks!
 A: A related problem.
The answer is correct and the downvote is misleading
Hint: You can make comparison test with the series

$$ \sum_{j=1}^{\infty} \left(\frac{2}{3}\right)^{j}. $$

Added: Here is the result you need

Suppose $\sum_{n} a_n$ and $\sum_n b_n $ are series with positive terms, then
if $\lim_{n\to \infty} \frac{a_n}{b_n}=c>0$, then either both series converge or diverge.

In your case the limit

$$ \lim_{n\to \infty} \frac{a_n}{b_n} = 1 > 0. $$

So, you can conclude.
A: I think a simple way of doing this would be to use the fact that exponential functions grow must faster linear ones. So notice that $2^j>j$ for $j\geq1$ and $3^{j-1}>j$ for $j\geq 2$. But $\sum_{j=1}^\infty stuff$ converges only if $\sum_{j=2}^\infty stuff$ converges. So we will deal with the second summation. We can make the summation even bigger by adding $2^j$ in the top instead of $j$ and subtracting $3^{j-1}$ instead of just $j$. 
$$
\sum_{j=2}^\infty\frac{2^j+2^j}{3^j-3^{j-1}}=\sum_{j=2}^\infty \frac{2^{j+1}}{2\cdot 3^{j-1}}=\sum_{j=2}^\infty\frac{2^j}{3^{j-1}}
$$ 
Now we have
$$
0\leq\sum_{j=2}^\infty \frac{2^j+j}{3^j-j} \leq \sum_{j=2}^\infty \frac{2^j}{3^{j-1}}=4
$$
(the sum on the right is simple to calculate as it is a geometric series). This shows your original sum converges. 
EDIT: You asked how to show $3^{j-1}\geq j$ for $j \geq 2$. Well, you can just graph $3^{j-1}$ and $j$ and notice that it is obvious that $3^{j-1}$ is bigger than $j$ for all $j \geq 2$. If you wanted to show it with rigor, you could show this easily using induction. Notice when $j=2$, we have $3^1>2$. So it works when $j=2$. We show that this works for all $j$ by assuming it works for $j=2,3,4,5,\cdots k$ and show it works for $k+1$. Since it works for $j=k$, we know we have 
$$
3^{k-1}>k
$$
Let's add $1$ to both sides
$$
3^{k-1}+1>k+1
$$
Now let's make the really simple observation that $1+1+1=3$. So notice that
$$
3^{k-1}+1+1+1>3^{k-1}+1>k+1
$$
Let's clean up the left hand side, we then have
$$
3^{k-1}+3=3^k>3^{k-1}+1>k+1
$$
So 
$$
3^{k}>k+1
$$
But this is our original formula $3^{j-1}>j$ when $j=k+1$. What does this all mean? Well, we just showed that if this inequality is true for $j=k$ then it is true for $j=k+1$. Big deal! So what?
Ah, there is the magic. We showed that $3^{j-1}>j$ when $j=2$. But we just showed that if this is true for $j=k$ then it is true for $j=k+1$. So since $3^{j-1}>j$ is true when $j=2$, it is true when $j=3$ by what we showed above. But it is true for $j=3$, then it is true for $j=4$. AH HA! But it being true for $j=4$ means it is true for $j=5$. We can continue this forever. So it is true for all $j\geq 2$ as promised. This is what is called mathematical induction. Hopefully, this made sense to you. If not, stick to doing it graphically as above. Since you are probably learning calculus at the moment. Note that $3^{j-1}>j$ when $j=2$. Notice that the derivative of $3^{j-1}$ is always bigger than the derivative of $j$ for $j\geq 2$, so $3^{j-1}$ is bigger than $j$ at $j=2$ and always increase faster for $j>2$. This also shows this.
