# Question about the integral test

by salazar888
Tags: integral, test
 P: 12 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
HW Helper
Thanks
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P: 26,127
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)
 HW Helper P: 4,507 Obviously, since the series $\sum \frac{1}{n^2}$ converges, the sum you wrote, $4 + \sum \frac{1}{n^2}$ converges also. RGV
P: 12

## 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.
 HW Helper P: 4,507 I saw the error and edited it immediately. RGV
Math
Emeritus
 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
$$\sum\left(\frac{1}{n^2}+ 4\right)$$