Examine if the series
$$\sum_{n=1}^{\infty}\frac{x^n}{1+x^n}, \;\; x \in (-1, 1)$$
converges uniformly on $(-1, 1)$.
Solution:
Restraining the domain in $[0, 1)$ we note that the partial sums $\displaystyle S_N(x)=\sum_{n=1}^{N}\frac{x^n}{1+x^n}$ converge pointwise (point to point) on $[0, 1)$ to $\displaystyle S(x)=\sum_{n=1}^{\infty}\frac{x^n}{1+x^n}$ with comparison with the geo $\displaystyle \sum_{n=1}^{\infty} x^n$.
However, the convergence is not uniform since:
$$S(x)-S_N(x)=\sum_{n=1}^{\infty}\frac{x^n}{1+x^n}-\sum_{n=1}^{N}\frac{x^n}{1+x^n}=\sum_{n=N+2}^{\infty}\frac{x^n}{1+x^n}\geq \frac{x^{N+2}}{1+x^{N+2}}$$
Letting $x \rightarrow 1$ we note that $\displaystyle \sup_{x \in [0, 1)}\left | S(x)-S_N(x) \right |\geq \frac{1}{2}$ and with this the problem comes to and.
$$\sum_{n=1}^{\infty}\frac{x^n}{1+x^n}, \;\; x \in (-1, 1)$$
converges uniformly on $(-1, 1)$.
Solution:
Restraining the domain in $[0, 1)$ we note that the partial sums $\displaystyle S_N(x)=\sum_{n=1}^{N}\frac{x^n}{1+x^n}$ converge pointwise (point to point) on $[0, 1)$ to $\displaystyle S(x)=\sum_{n=1}^{\infty}\frac{x^n}{1+x^n}$ with comparison with the geo $\displaystyle \sum_{n=1}^{\infty} x^n$.
However, the convergence is not uniform since:
$$S(x)-S_N(x)=\sum_{n=1}^{\infty}\frac{x^n}{1+x^n}-\sum_{n=1}^{N}\frac{x^n}{1+x^n}=\sum_{n=N+2}^{\infty}\frac{x^n}{1+x^n}\geq \frac{x^{N+2}}{1+x^{N+2}}$$
Letting $x \rightarrow 1$ we note that $\displaystyle \sup_{x \in [0, 1)}\left | S(x)-S_N(x) \right |\geq \frac{1}{2}$ and with this the problem comes to and.
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