Let $g \colon \mathbb{N} \rightarrow \{0,1\}$ such that $g(n)=0$, if $n \equiv 0$ or $n \equiv 1 ~(\bmod~4)$; or $g(n)=1$, if $n \equiv 2$ ou $n \equiv 3 ~(\bmod~4)$.
And let $s_n$ = $\sum_{n=1}^\infty \frac{(-1)^{g(n)}}{\sqrt{n}}$. With the help from my previous question, we know the series is convergent.
By QC_QAOA's answer:
We have
$$\sum_{n=1}^\infty > \frac{(-1)^{g(n)}}{\sqrt{n}}=\frac{1}{1}-\frac{1}{\sqrt{2}}-\frac{1}>>{\sqrt{3}}+\frac{1}{\sqrt{4}}+\cdots$$
Now, note that for for $n\equiv 2\ (\text{mod}\ 4)$ we have the terms
$$\cdots-\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\frac{1}{\sqrt{n+3}}-\cdots$$
$$\cdots-\frac{\sqrt{n}+\sqrt{n+1}}{\sqrt{n(n+1)}}+\frac{\sqrt{n+2}+\sqrt{n+3}}{\sqrt{(n+2)(n+3)}}$$
That is, the sum can be rewritten as
$$\sum_{n=1}^\infty \frac{(-1)^{g(n)}}{\sqrt{n}}=1+\sum_{n=1}^\infty (-1)^n\frac{\sqrt{2n}+\sqrt{2n+1}}{\sqrt{2n(2n+1)}}$$
But we see that the term in this sum is decreasing. We conclude by the alternating series test that the sum converges
The obstacle I encontered lies in the last equality. If we consider the lhs as $s_n$ and rhs as $t_n$, the equality holds for all $n \in \mathbb{N}$ but only when $s_{2n+1} = t_{n}$. Since we are dealing with a conditional convergent series, then it is not always true that if $t_{n}=s_{2n+1}$ converges, then $s_n$ also converges.
My question finally is: how can I overcome this obstacle and complete this proof?
What I thought of doing?
- Let $L$ be the limit of $t_n$, let $j\in \mathbb{N}$, let $\varepsilon >0$ and consider $s_{2n+1+j}$. We must prove for all $j$ that $|s_{2n+1+j}-L|\leq \varepsilon$. By triangle's inequality, $$|s_{2n+1+j}-L|= |s_{2n+1+j}-s_{2n+1}+s_{2n+1}-L|\leq |s_{2n+1+j}-s_{2n+1}|+|s_{2n+1}-L|\leq\varepsilon$$
the $|s_{2n+1}-L|$ part we can easily choose a $n_0$ and make it less than $\frac{\varepsilon}{2}$ for all $n\geq n_0$, since we know it converges. But my problem with this way of trying to overcome the obstacle lies in the $|s_{2n+1+j}-s_{2n+1}|$ part.
Let me know of any mistakes, I sinceraly thank you.
Define
$$s_n=\frac{(-1)^{g(n)}}{\sqrt{n}}$$
$$t_n=(-1)^n\frac{\sqrt{2n}+\sqrt{2n+1}}{\sqrt{2n(2n+1)}}$$
We have that
$$s_{4n+2}+s_{4n+3}=-\frac{1}{\sqrt{4n+2}}-\frac{1}{\sqrt{4n+3}}=-\frac{\sqrt{4n+2}+\sqrt{4n+3}}{\sqrt{(4n+2)(4n+3)}}=t_{2n+1}$$
$$s_{4n+4}+s_{4n+5}=\frac{1}{\sqrt{4n+4}}+\frac{1}{\sqrt{4n+5}}=\frac{\sqrt{4n+4}+\sqrt{4n+5}}{\sqrt{(4n+4)(4n+5)}}=t_{2n+2}$$
Also
$$s_1=1$$
Then
$$\sum_{n=1}^\infty s_n=1+\sum_{n=0}^\infty (s_{4n+2}+s_{4n+3}+s_{4n+4}+{s_{4n+5}})=1+\sum_{n=0}^\infty (t_{2n+1}+t_{2n+2})=1+\sum_{n=1}^\infty t_n$$