## Question about the integral test

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

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hi salazar888! welcome to pf!
 Quote by salazar888 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
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)

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

## Question about the integral test

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

 Recognitions: Homework Help I saw the error and edited it immediately. RGV

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 Quote by 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
$$\sum\left(\frac{1}{n^2}+ 4\right)$$