How to prove this series for convergence?

Question

Let $(a_n)$ be a sequence of real numbers satisfying $$a_1 \geq 1 \;\;\;\text{and}\;\;\;a_{n+1}\geq a_n+1$$ for all $n \geq 1$. Then which one of the following is necessarily true?

a) The series $\sum \frac{1}{(a_n)^2}$ diverges.

b) The sequence $a_n$ is bounded.

c) The series $\sum \frac{1}{(a_n)^2}$ converges.

d) The series $\sum \frac{1}{a_n}$ converges.

Here $(a_n)=(n)$ eliminates a), b) and d). So c) is true.

How to prove c) Mathematically? Any hint?

Answer 1

Since $a_{n+1}\geq a_n+1$, $$a_{n}-a_1 = \sum_{k=1}^{n-1}a_{k+1}-a_k \geq n-1$$hence $a_n\geq n$.

By comparison, $\sum_n \frac{1}{(a_n)^2}$ converges.

Answer 2

HINT: $a_n\geq n$, hence $$ \sum_{n=1}^{\infty}\frac{1}{a_n^2}\leq\sum_{n=1}^{\infty}\frac{1}{n^2} $$