# Direct Comparison Test

1. May 6, 2013

### Platformance

1. The problem statement, all variables and given/known data

Show that:

$\sum \frac{3}{n^{2} + 1}$

converges from n = 1 to ∞

2. Relevant equations

If Ʃbn converges, and Ʃan < Ʃbn.

Ʃan also converges.

3. The attempt at a solution

$\sum \frac{1}{n^{2}}$ converges

$\sum \frac{3}{n^{2} + 1}$ = 3 * $\sum \frac{1}{n^{2} + 1}$

$\sum \frac{1}{n^{2} + 1}$ < $\sum \frac{1}{n^{2}}$ for all n from 1 to ∞.

Therefore $\sum \frac{1}{n^{2} + 1}$ converges.

Therefore $\sum \frac{3}{n^{2} + 1}$ also converges.

The problem I am having is if the 3 remained in the summation.

$\sum \frac{3}{n^{2} + 1}$ is not less than $\sum \frac{1}{n^{2}}$ for all n from 1 to ∞.

Why does placing the 3 outside the summation make the problem work?

2. May 6, 2013

### SammyS

Staff Emeritus
If you leave the 3 inside the sum, you will have to find a different converging series to compare it to. The method will still work.

3. May 6, 2013

### Platformance

So that means I would need to compare the series to $\frac{3}{n^{2}}$ instead.

I can prove $\frac{3}{n^{2}}$ through the integral test. Though speaking of the integral test, it doesn't seem like the test works for the original series since I am getting tan-1(∞). So would the integral test be inconclusive in this case?

4. May 6, 2013

### Dick

arctan(∞), which is what I assume is what you mean by tan^(-1)(∞), is finite.

5. May 6, 2013

### Ray Vickson

You are making it WAAAAAY too complicated: it is a simple fact---easily proved---that if c is a constant, then $\sum_n c a_n = c \sum_n a_n$, in the sense that if one side converges, so does the other (and the two sides are equal), and if one side diverges so does the other.

6. May 6, 2013

### SammyS

Staff Emeritus
No. You would not necessarily have to compare to $\displaystyle \sum_{n=1}^\infty\frac{3}{n^{2}}\ .$

There are many other possibilities.