Determining Divergence

Question

How can I prove that this series diverges? I don't think you can use a comparison test, but maybe I'm mistaken.

$$\sum \dfrac 1{n^{4/5}+10^{10}}$$

Answer 1

Hint: $\dfrac{1}{n^{4/5}+10^{10}}\ge \dfrac{1}{n^{4/5}+n^{4/5}}$, for big enough $n$.

Answer 2

Since we have $ \ \dfrac{1}{n^{4/5} \ + \ 10^{10}} \ $ , with a sum in the denominator, the simple or Direct Comparison Test won't help for checking divergence, since the inequality versus the divergent series $ \ \Sigma \ \frac{1}{n^{4/5}} \ $ (by the "$ p- $test", $ \ p \ = \ \frac{4}{5} \ < \ 1 \ $ ) "runs the wrong way":

$$ \frac{1}{n^{4/5}} \ \ge \ \frac{1}{n^{4/5}+10^{10}} \ \ . $$

(We'd be OK if the series were $ \ \Sigma \ \dfrac{1}{n^{4/5} \ - \ 10^{10}} \ $ . )

Git Gud shows a way to demonstrate divergence applying Direct Comparison. $ ^* $ One may also use the Limit Comparison Test, setting up the limit ratio

$$ \lim_{n \rightarrow \infty} \ \frac{\frac{1}{n^{4/5}}}{\frac{1}{n^{4/5} \ + \ 10^{10} }} \ \ = \ \ \lim_{n \rightarrow \infty} \ \frac{n^{4/5} \ + \ 10^{10} }{n^{4/5}} \ \ = \ \ \lim_{n \rightarrow \infty} \ 1 \ + \ \frac{ 10^{10} }{n^{4/5}} \ = \ 1 \ \ . $$

Since this limit is a positive constant, the LCT tells us that both series have the same convergence/divergence behavior. Thus, $ \ \Sigma \ \dfrac{1}{n^{4/5} \ + \ 10^{10}} \ $ diverges.

$ ^* $ That approach is good to know about. (And occasionally you run into an instructor who doesn't discuss the LCT, or won't let you use it...)