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.