Let $\{a_{n}\}$ with $a_n \leq 0$. Let $\{S_{n}\}$ with $S_n := a_1 + a_2+ \dotsc + a_n$. Assume that $\{S_{n}\}$ is bounded. Let $T_n := -a_1^2 - a_2^2- \dotsc - a_n^2$. Prove the following:

  1. $\lim_{n \to \infty} a_n = 0$.

  2. $\{T_{n}\}$ converges. [Hint: compare $a_n$ with $-a_n^2$ for $n$ large]

What I have done for part $1$: notice that $ a_n=S_n -S_{n-1} $. Moreover, $S_n \leq S_{n-1}$ since $a_n \leq 0$. $\{S_{n}\}$ is bounded and decreasing so it converges to a limit $l$. Thus, $\lim_{n \to \infty} a_n=l-l = 0$.

I am stuck on part $2$. What exactly is meant by the hint? Is the goal to show that $\{T_{n}\}$ is bounded and decreasing?

  • $\begingroup$ Possibly helpful: note that if $(S_n)$ converges then we necessarily have $a_n\to0$ as $n\to\infty$. Therefore, for large enough $n$ we have $|a_n|<1$ and so $|a_n^2|<|a_n|$ for large enough $n$. $\endgroup$
    – csch2
    Apr 22 '20 at 15:06

I am just putting full proof what @csh2 hinted at.Note that we will be use cauchy convergence test.

First since $\sum_{n=1}^{\infty} a_n < \infty$ there exists $N$ s.t $$\forall m,n >N |\sum_{i=m}^n a_n| < \sqrt{\epsilon}$$ and we will show that this $N$ suffices for $\sum_{n=1}^{\infty} -a_n^2$.

$$|-\sum_{i=m}^{n} a_{n}^2| \leq \Big|\sum_{i=m}^{n} a_n \Big|^2 < \epsilon$$

and Hence the series conveges


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