Converge relationship between two series

Suppose $$\{a_n\}_{n=1}^{\infty}$$ is a non-negative sequence. I'm considering the relationship between the following two series. $$\sum_{n=1}^{\infty} \left[ a_n - log(1+a_n) \right] \quad \textit{and} \quad \sum_{n=1}^{\infty} a_n^{2}$$

My question is does the first one converge imply the second one converges? How about reverse direction?

I try to expand $$f(x) = x - log(1+x)$$ by Taylor expansion, which is $$f(x) = x - log(1+x) = \frac{x^2}{2} - \frac{x^3}{3} + \frac{x^4}{4} - \cdots$$ and try to use Limit comparison test, where we let $$g(x) = x^2$$.

However, I stuck at how check if $$\lim_{n\rightarrow\infty} \frac{f(a_n)}{g(a_n)}$$ exist and not equal to zero.

Note that $$f(0)=0$$ and that $$f(x)>0$$ if $$x\ne0$$. Besides, it's not hard to prove that $$\lim_{n\to\infty}a_n=0$$ if and only if $$\lim_{n\to\infty}f(a_n)=0$$. So:
• If $$\displaystyle\sum_{n=1}^\infty\bigl(a_n-\log(1+a_n)\bigr)$$ converges, then $$\lim_{n\to\infty}f(a_n)=0$$, and so $$\lim_{n\to\infty}a_n=0$$. But then$$\lim_{n\to\infty}\frac{a_n-\log(1+a_n)}{a_n^{\,2}}=\frac12$$and therefore the series $$\displaystyle\sum_{n=1}^\infty a_n^{\,2}$$ converges too (not that $$a_n-\log(1+a_n)=f(a_n)\geqslant0$$ for every $$n\in\Bbb N$$).
• If $$\displaystyle\sum_{n=1}^\infty a_n^{\,2}$$ converges, then the same argument as above shows that the series $$\displaystyle\sum_{n=1}^\infty\bigl(a_n-\log(1+a_n)\bigr)$$ converges too.