Converge relationship between two series

Question

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.

Answer 1

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.