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

Question

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

I have looked at an answer here, but I do not understand what these results give us. For example, in the first answer:

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

What do we do with the last expression? I understand we want to compare it with something, but don't see what.

Here is the same problem. – Mhenni Benghorbal Apr 15 '14 at 04:10 — Mhenni Benghorbal, Apr 15 '14 at 04:10

MT_ · Accepted Answer · 2014-04-15T03:57:17.467

2

Hint: $$\frac{2^j + 2^j}{3^j - \frac 12 3^j} = \frac{2^{j+2}}{3^j} = 4 (\frac{2}{3})^j$$

What kind of series is this?

Regarding the inequality:

The first is clearly true since $2^j > j$ for all $j \geq 1$. The second is true for $j < \frac{1}{2} 3^j$ (which is true for $j \geq 1$) since the sum is larger when the denominator is less. Then by comparison, since this geometric series converges, so does the original series.

edited Apr 15 '14 at 03:57

answered Apr 15 '14 at 03:07

MT_

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Hmm, I understand why the final series converges, but could you please elaborate on the steps more? – kiwifruit Apr 15 '14 at 03:10
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@kiwifruit I have updated the post to answer – MT_ Apr 15 '14 at 04:08

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

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