Proving Convergence of Direct Comparison Test for \sum \frac{3}{n^{2} + 1}

[Platformance]

Homework Statement

Show that:

[itex]\sum \frac{3}{n^{2} + 1}[/itex]

converges from n = 1 to ∞

Homework Equations

If Ʃb_n converges, and Ʃa_n < Ʃb_n.

Ʃa_n also converges.

The Attempt at a Solution

[itex]\sum \frac{1}{n^{2}}[/itex] converges

[itex]\sum \frac{3}{n^{2} + 1}[/itex] = 3 * [itex]\sum \frac{1}{n^{2} + 1}[/itex]

[itex]\sum \frac{1}{n^{2} + 1}[/itex] < [itex]\sum \frac{1}{n^{2}}[/itex] for all n from 1 to ∞.

Therefore [itex]\sum \frac{1}{n^{2} + 1}[/itex] converges.

Therefore [itex]\sum \frac{3}{n^{2} + 1}[/itex] also converges.

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

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

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

 

[SammyS]

Homework Statement

Show that:

[itex]\displaystyle \sum_{n = 1}^\infty \frac{3}{n^{2} + 1}\ [/itex] converges .

...

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

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

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

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.

 

[Platformance]

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

I can prove [itex]\frac{3}{n^{2}}[/itex] 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?

 

[Dick]

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

I can prove [itex]\frac{3}{n^{2}}[/itex] 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?

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

 

[Ray Vickson]

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

I can prove [itex]\frac{3}{n^{2}}[/itex] 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?

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.

 

[SammyS]

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

No. You would not necessarily have to compare to [itex]\displaystyle \sum_{n=1}^\infty\frac{3}{n^{2}}\ .[/itex]

There are many other possibilities.