Describe the interval where $f_n(x) = \sum_{k=1}^n \frac{x^k}{x^k + 1}$ uniformly converges

Question

Find the interval $f_n(x) = \sum_{k=1}^n \frac{x^k}{x^k + 1}$ uniformly converges.

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Try

First, note that $f_n(x)$ converges pointwise when on $(-1,1)$.

I claim, $\forall \epsilon \in (0, 1)$, $f_n(x)$ uniformly converges on $[-1 + \epsilon, 1 - \epsilon]$.

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Note : $\forall x \in [-1 + \epsilon, 1-\epsilon]$, we have $|\frac{x^k}{x^k+1}| \le (-1 + \epsilon)^k$, thus

Since $\lim_{n\to\infty} \left(\sum_{k=1}^n (1-\epsilon)^k\right) < +\infty$, we have $f_n$ : uniformly converges (by M-test).

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Is my try on the right way?

Answer 1

What you have tried so far is correct.

If you have to find all intervals where $f_n$ uniformly converges you also have to prove that if $|x|\ge 1$ then it doesn't converge.
Finding a function $g_n$ such that $\forall n \in \Bbb N, g_n\le f_n$ and $g_n$ diverges would be the way to go.

Hint:

Try $g_n(x) = \frac 12$, to show that $f_n$ diverges if $|x| \ge 1$