Find a n.s.c. on $(a_n)_{n \in \mathbb{N}}$ s.t. $\sum_{n=1}^{\infty}\left(\log(1+a^2_n)-\arctan^2(a_n)\right)$ converges

Question

Let be $(a_n)_{n \in \mathbb{N}}$ a sequence s.t. $a_n \to 0$ for $n \to +\infty$. Find a necessary and sufficient condition on $(a_n)_{n \in \mathbb{N}}$ s.t. that following series converges: $$\sum_{n=1}^{\infty}\left(\log(1+a^2_n)-\arctan^2(a_n)\right)$$

Using Taylor-McLaurin's expansions we have:

$\log(1+a^2_n) = a^2_n -\frac{1}{2}a^4_n + o(a^4_n)\,\,\,\,\,,n\to+\infty$

$\arctan^2(a_n)=a^2_n-\frac{1}{3}a^6_n+o(a^{10}_n)\,\,\,\,,n\to+\infty$

replacing these expressions on the original series is it easy to see that a n.s.c. is that $|a_n|<1\,\,\,\,\forall n \in \mathbb{N}$.

By the way, considering the partial sums $S_n=\sum_{k=1}^{n}a_k\,\,$, is it enough asking that $\exists\,\, M\in \mathbb{R}$ s.t. $|S_n|<M\,\,\,\forall n\in \mathbb{N}$? Is that a n.s.c.?

Thank you

Answer 1

Your condition doesn't really makes sense... The condition $|a_n|<1$ will of course hold when $n$ big enough. What you have is $$6\big(\log(1+a_n^2)-\arctan^2(a_n)\big)=a_n^4+o(a_n^4).$$ Therefore your series converges $\iff$ $\sum_{n=1}^\infty a_n^4$ converges.