009C Sample Midterm 2, Problem 2

Determine convergence or divergence:

$\sum _{n=1}^{\infty }{\frac {3^{n}}{n}}$

Foundations:
1. Direct Comparison Test
Let $\{a_{n}\}$ and $\{b_{n}\}$ be positive sequences where $a_{n}\leq b_{n}$
for all $n\geq N$ for some $N\geq 1.$
2. If $\sum _{n=1}^{\infty }b_{n}$ converges, then $\sum _{n=1}^{\infty }a_{n}$ converges.
3. If $\sum _{n=1}^{\infty }a_{n}$ diverges, then $\sum _{n=1}^{\infty }b_{n}$ diverges.

Solution:

Step 1:
First, we note that
${\frac {3^{n}}{n}}>0$
for all $n\geq 1.$
This means that we can use a comparison test on this series.
Let $a_{n}={\frac {3^{n}}{n}}.$

Step 2:
Let $b_{n}={\frac {1}{n}}.$
We want to compare the series in this problem with
$\sum _{n=1}^{\infty }{\frac {1}{n}}.$
This is the harmonic series (or $p$-series with $p=1.$)
Hence, $\sum _{n=1}^{\infty }b_{n}$ diverges.

Step 3:
Also, we have $b_{n}<a_{n}$ since
${\frac {1}{n}}<{\frac {3^{n}}{n}}$
for all $n\geq 1.$
Therefore, the series $\sum _{n=1}^{\infty }a_{n}$ diverges
by the Direct Comparison Test.

Final Answer:
diverges

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