# Question about the integral test

1. Jul 24, 2011

### salazar888

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

We have to determine whether $\sum 1/n^2 + 4$
is convergente or divergent
2. Relevant equations

I'm trying to work the problem through trigonometric substitution. I was wondering if I could just determine that by the P-series test, the function 1/n^2 will always be larger than the other one, since p is greater than 1 in this case, both are convergent.
3. The attempt at a solution

2. Jul 24, 2011

### tiny-tim

welcome to pf!

hi salazar888! welcome to pf!
yes, always 0 < 1/(n2+4) < 1/n2,

the latter sum converges (from the p-series test), so so must the former (from the direct comparison test)

3. Jul 24, 2011

### Ray Vickson

Obviously, since the series $\sum \frac{1}{n^2}$ converges, the sum you wrote, $4 + \sum \frac{1}{n^2}$ converges also.

RGV

Last edited: Jul 24, 2011
4. Jul 24, 2011

### salazar888

The series 1/n does not converge. It's the harmonic series. You read it wrong. I get what you're saying though.

5. Jul 24, 2011

### Ray Vickson

I saw the error and edited it immediately.

RGV

6. Jul 24, 2011

### HallsofIvy

But the series
$$\sum\left(\frac{1}{n^2}+ 4\right)$$
does NOT converge!

7. Jul 24, 2011

### Ray Vickson

I agree, but that is not what he wrote. We all know he meant sum 1/(n^2 + 4), but he wrote sum (1/n^2) + 4, which is very different according to standard math expression padding rules. Since he was using 'tex' anyway, he should have been able to enter "{n^2+4}" as the second argument of the '\frac' command.

RGV

8. Jul 24, 2011

### salazar888

Yes it was my fault. I've only been on the forum for a couple of days. Thanks for the help guys. I will improve at typing the commands.