Interval of convergence of the infinite series $\sum_{n=1}^{\infty} \frac{x^n}{1+x^n}$

Question

I am self-learning Real Analysis and solving some exercise problems from Problems in Mathematical Analysis, by Kaczor and Nowak. I would like someone to verify if the arguments below are technically correct and rigorous.

[K&N 3.2.1] Find where the following series converges pointwise:

(b) $$\sum_{n=1}^{\infty}\frac{x^n}{1+x^n}$$

Proof.

Fix $\displaystyle x=x_{0}$.

Case 1. Let's quickly dispose off the case when $\displaystyle |x_{0} | >1$. We have:

\begin{equation*} \begin{array}{ c c } \lim f_{n}( x_{0}) & =\lim \frac{1}{\left(\frac{1}{x_{0}}\right)^{n} +1} =\frac{1}{0+1} =1\\ & \end{array} \end{equation*}

We know that, if the infinite series $\displaystyle \sum_{n=1}^{\infty} a_n$ converges, then $(a_{n})\to 0$. Thus, $\displaystyle \sum_{n=1}^{\infty } f_{n}( x)$ does not converge for $|x_{0}|>1$.

Case 2. Also, if $\displaystyle x_{0}=1$, then $\displaystyle \lim f_{n}( x_{0}) =\frac{1}{2}$. Thus, $\displaystyle f_{n}( x_{0})$ does not converge for $\displaystyle x_{0} =1$.

Case 3. Assume that $\displaystyle 0< x_{0} < 1$. Since $\displaystyle x_{0} >0$, $\displaystyle x_{0}^{n} >0$ and thus $\displaystyle 1+x_{0}^{n} >1$. Consequently,

\begin{equation*} 0< f_{n}( x_{0}) =\frac{x_{0}^{n}}{1+x_{0}^{n}} < x_{0}^{n} =g_{n}( x_{0}) \end{equation*}

We know that, $\displaystyle \sum_{n=1}^{\infty } g_{n}(x_{0})$ is a convergent series. Hence, by the comparison test, $\displaystyle \sum_{n=1}^{\infty } f_{n}( x_{0})$ converges for $\displaystyle 0< x_{0} < 1$.

Case 4. If $\displaystyle x_{0} =0$, the series converges to $\displaystyle 0$.

Case 5. Now, assume that $\displaystyle -1< x_{0} < 0$. We have:

\begin{equation*} \left| \frac{a_{n+1}}{a_{n}}\right| =\left| \frac{x_{0}^{n+1}}{x_{0}^{n}}\right| \cdot \left| \frac{1+x_{0}^{n}}{1+x_{0}^{n+1}}\right| =|x_{0} |\cdot \left| \frac{1+x_{0}^{n}}{1+x_{0}^{n+1}}\right| \end{equation*}

So,

\begin{equation*} \lim \left| \frac{a_{n+1}}{a_{n}}\right| =|x_{0} |\cdot \lim \left| \frac{1+x_{0}^{n}}{1+x_{0}^{n+1}}\right| \end{equation*}

Since $\displaystyle \lim $$\displaystyle |a_{n} |=|\lim a_{n} |$, we have:

\begin{equation*} \lim \left| \frac{a_{n+1}}{a_{n}}\right| =|x_{0} |\cdot \left| \frac{1+\lim x_{0}^{n}}{1+\lim x_{0}^{n+1}}\right| =|x_{0} |< 1 \end{equation*}

Hence, by the ratio test, $\displaystyle \sum_{n=1}^{\infty } a_{n}$ converges absolutely over $\displaystyle -1< x_{0} < 0$.

Answer 1

1. Correct. Alternatively, 1 is irrelevant as $n \to \infty$...but eh not sure how to make that precise other than what you already said.

2. Correct. Alternatively, any non-zero number added to itself infinitely is $\pm \infty$. Cf Case 4.

3. Correct because $1 > 1+x_0^n > 0$.

4. Correct. The 1 thing outside the scope of Case 2.

5. Looks correct.

5.1 - I haven't done this in awhile but come how you can't do ratio test for case 3 the way you did for case 5?

5.2 - It looks like by

$\displaystyle \lim $$\displaystyle |a_{n} |=|\lim a_{n} |$

5.3 - You mean to say 'limits pass through absolute values'. In this case, you should use a different variable.

$\displaystyle \lim $$\displaystyle |b_{n} |=|\lim b_{n} |$

And then say something like 'choose $b_n = \frac{1+x_{0}^{n}}{1+x_{0}^{n+1}} $.'

Anyway, I don't think you need to say this unless you also need to say limit of ratio is ratio of limits. So just do say like

$$\lim \left| \frac{1+x_{0}^{n}}{1+x_{0}^{n+1}}\right| $$

$$ = \left| \lim \frac{1+x_{0}^{n}}{1+x_{0}^{n+1}}\right| $$

$$ = \left| \frac{\lim(1+x_{0}^{n})}{\lim(1+x_{0}^{n+1})}\right| $$

$$ = \left| \frac{1+\lim(x_{0}^{n})}{1+\lim(x_{0}^{n+1})}\right| $$

$$ = \left| \frac{1+\lim(x_{0}^{n})}{1+\lim(x_{0}^{n+1})}\right| $$

$$ = \left| \frac{1+0}{1+0}\right| $$

$$ = 1 $$