series comparison

Question

Consider the series ∑^∞ _n=1 (1/n+a), where a > 0 is a constant . use the comparison test to see whether it converges or not?

What i was thinking is if we compare it 1/n, we can say that with the p-series test 1/n diverges since p is smaller than 1. and since 1/n would be bigger than (1/n+a) which means this function will also diverge.

This is just a theory.

Answer 1

Consider the series ∑^∞ _n=1 (1/(n+a)), where a > 0 is a constant . use the comparison test to see whether it converges or not?

1/(n+a)<1/n, using the p-series test we know right away that n^p is p=1. If p=1 or <1 then the series will diverge

We are given a_n=1/(n+a) and b_n=1/n, we can use limit of a_n/b_n to divide to determine convergence rule

(1/(n+a))/ (1/n), we need to divide the complex fraction and change it to a reciprocal 1/(n+a) *n

next we will simplify the fraction as n/(n+a)

then we determine if series converges using lim n-->infinity

since n/(n+a) approaches n/n=1 then the limit becomes 1. If series is less than or equal to 1, then series diverges

So series diverges