I'm here again trying to solve a problem about series of functions. I have to study the pointwise and uniform convergence of the following series:

$$\sum_{n=1}^\infty n\log(1+\frac{\vert\sin(x)\vert^n}{1+x^n})$$

I tried to study the pointwise convergence first, but there's a problem: I need to find the domain of the functions, so : $1+\frac{\vert\sin(x)\vert^n}{1+x^n}\gt0$ and $1+x^n\neq0$

The problem is that I can't solve the first inequality at all and the second one clearly depends on n. I'm not even sure if I should solve them in order to fin the pointwise convergence .

Help please.

  • 1
    $\begingroup$ "I can't solve the first inequality at all"? Take some baby steps. What if $x\ge 0$? $\endgroup$
    – zhw.
    Aug 27 '21 at 14:31

In this case, it doesn't really matter that the inequation depends on $n$. For the series to be well defined, every term in the sum should be well defined. That means you need $1 + x^n \neq 0 \iff x^n \neq -1$ for every $n$. For even $n$, this always holds and for odd $n$, it holds for $x \neq -1$. Again, as the term should be well defined for every $n$, it means $x = -1$ cannot be in the domain.

The rest of the domain has to be determined by the inequality.

For $x > -1$, $1 + x^n > 0$ trivially, so there are no issues with the domain there.

For $x < -1$, note that $1/|1 + x^n|$ and $|\sin(x)^n|$ both decrease with n, so if your inequality holds for $n = 1$, it will hold for all larger $n$ as well. So all that's left to solve is $1 + |\sin(x)|/(1 + x) > 0 \iff |\sin(x)| < -1 - x$ (noting that $1 + x < 0$). As far as I know, this equation can only be solved numerically.

But anyway, that fully constrains the domain of the functions. That doesn't yet say anything about pointwise or uniform convergence, but that wasn't your question.


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