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Does this series converge or diverge? If it converges, determine its limit. $$\sum_{n=1}^{\infty}\frac{1}{2^n} + \frac{3}{n}$$

So far I said that $\frac{1}{2^n}$ is a geomotric series that converges, and $\frac{3}{n}$ diverges since its the harmonic series (I think), but I don't know where to go from that! (sorry I'm a beginner)

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    $\begingroup$ Your reason for divergence is correct. $\endgroup$ – Tom Himler Aug 13 '18 at 0:42
  • $\begingroup$ Your reasoning is correct, but it's expressed poorly. Note that "$\frac{1}{2^n}$" isn't a series at all, but $\sum\frac{1}{2^n}$ is. $\endgroup$ – zipirovich Aug 13 '18 at 2:01
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Hint: The series diverges. Give a divergent minorant. You already mentioned, that $\sum_{n=1}^\infty \frac3n$ is a harmonic series, which diverges.

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If $\sum a_n$ diverges, $\sum b_n$ converges, then $\sum (a_n \pm b_n)$ must diverges, otherwise $a_n = (a_n \pm b_n) \mp b_n$ forms a convergent series [sum of two convergent series is also convergent].

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Alternatively, if the sum is finite and is equal to $L$, then by rearranging, $$\sum_{n=1}^\infty \frac{3}{n} = L - \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \ldots.$$ Note that the left side is divergent, but the right side (a geometric series with a constant term added) is convergent. This is a contradiction.

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