# Show that the sequence $\{T_{n}\}$ with $T_n := -a_1^2 - a_2^2 -\dotsc - a_n^2$ converges

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?

• 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 $n$. Apr 22 '20 at 15:06

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 $$\sum_{n=1}^{\infty} -a_n^2$$.
$$|-\sum_{i=m}^{n} a_{n}^2| \leq \Big|\sum_{i=m}^{n} a_n \Big|^2 < \epsilon$$