$a_{n+1}=a_{n}+\frac{1}{n^{2}} a_{n-1}$, for $n \geq 2$

Question

Let $a_{1}=2020, \\ a_{2}=2021,\\ a_{n+1}=a_{n}+\frac{1}{n^{2}} a_{n-1}$, for $n \geq 2$.

Then which of the following is correct?

A) $\sum_{n=1}^{\infty} \frac{1}{a_{n}}$ converges to $\frac{1}{2020}$.

B) $\sum_{n=1}^{\infty} \frac{1}{a_{n}}$ is divergent.

C) $\sum_{n=1}^{\infty} \frac{1}{a_{n}}$ converges to $\frac{1}{2021}$.

D) None of the above.

In this question option A and C are incorrect. Since the first term of the sequence is $\frac{1}{2020}$.

What about the option $B$?

Answer 1

The series is divergent. Let $C=2021$. Notice that $a_n\leq C n$ for $n\in \{1, 2\}$. We want to prove this for every integer $n$. The base case of the induction is done and, the induction step follows easily as $$a_{n+1}=a_n+\frac{1}{n^2}a_{n-1}\leq Cn+ \frac{C(n-1)}{n^2}\leq Cn+C\frac{1}{4}<C(n+1).$$

So $$ \sum_n \frac{1}{a_n} \geq \frac{1}{C}\, \sum_{n} \frac{1}{n}=\infty,$$ since the harmonic series is divergent.