009C Sample Midterm 2, Problem 2 Detailed Solution

Determine convergence or divergence:

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

Foundations:
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  $N\geq 1.$
1. If  $\sum _{n=1}^{\infty }b_{n}$  converges, then  $\sum _{n=1}^{\infty }a_{n}$  converges.
2. 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 (by the Direct Comparison Test)

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