Prove that $\sum \frac{a_n}{a_n+3}$ diverges

Question

Suppose $a_n>0$ for each $n\in \mathbb{N}$ and $\sum_{n=0}^{\infty} a_n $ diverges. How would one go about showing that $\sum_{n=0}^{\infty} \frac{a_n}{a_n+3}$ diverges?

Answer 1

Then $\sum\limits_n\min\{a_n,1\}$ diverges and $\frac{a_n}{a_n+3}\geqslant\frac14\min\{a_n,1\}$. QED.

Answer 2

Let $b_n=\dfrac{a_n}{a_n+3}$. If the $a_n$ are unbounded, then $b_n$ does not approach $0$, and therefore $\sum b_n$ diverges.

If the $a_n$ are bounded by $B$, then $b_n\ge \dfrac{1}{B+3} a_n$, and $\sum b_n$ diverges by comparison with $\sum a_n$.