# Need series convergence help

1. May 19, 2013

### Al3x L3g3nd

1. The problem statement, all variables and given/known data
Does the series
$$\Big( \sum_{n=1}^\infty\frac{1}{(3^n)*(sqrtn)} \Big)$$
Converge or Diverge? By what test?

2. Relevant equations
1/n^p
If p<1 or p=1, the series diverges.
If p>1, the series converges.

If bn > an and bn converges, then an also converges.

3. The attempt at a solution
I use 1/(sqrtn) since it is bigger than 1/((3^n)(sqrtn)).
Since sqrtn is n^1/2 I use the p test.
Since 1/2<1, the series 1/sqrtn diverges and so does the original.
This is wrong

The answer uses 1/3^n as the comparison and it just says that it converges with no explanation.
Also, the ratio test was used and it converged.

Why doesn't my reasoning work?

2. May 19, 2013

### haruspex

You showed that a series with larger terms, and therefore less inclined to converge, doesn't converge. What can you deduce from that?

3. May 19, 2013

### CAF123

You are misapplying the Comparison Test. It is correct to state that $$\frac{1}{3^n \sqrt{n}} < \frac{1}{\sqrt{n}},$$ however while the the sum of the RHS diverges (by p test), you cannot conclude anything about the convergence/divergence of LHS. As a counterexample to your line of reasoning , see for example the following:
$$n^2 > n\,\,\text{for}\,\,n > 1 \Rightarrow \frac{1}{n^2} < \frac{1}{n}.$$ The infinite sum of RHS diverges (by p test) while the LHS converges (by p test).

4. May 19, 2013

### sweet springs

Hi

$$\sum_{n=1}^\infty\frac{1}{3^n\cdot\sqrt{n}} < \sum_{n=1}^\infty\frac{1}{3^n}=\frac{1}{2}$$

Last edited: May 19, 2013
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