Uniform convergence of series

[Grigorios Kostakos]

Examine if the series \[\displaystyle\mathop{\sum}\limits_{n=1}^{+\infty}\frac{x^n}{1+x^n}\,,\quad x\in(-1,1)\,,\] converges uniformly in \((-1,1)\).

[Tolaso J Kos]

No, it does not. Restraing the domain in \( [0, 1) \) it is easy to note that \( \displaystyle S_N(x) =\sum_{n=1}^{N}\frac{x^n}{1+x^n} \) which are the partial sums converge point to point on \( [0, 1) \) to \( \displaystyle S(x)=\sum_{n=1}^{\infty}\frac{x^n}{1+x^n} \) by comparison with the geo series \( \displaystyle \sum_{n=1}^{\infty}x^n \).

Hence:
$$S(x)-S_N(x)=\sum_{n=1}^{\infty}\frac{x^n}{1+x^n}-\sum_{n=1}^{N}\frac{x^n}{1+x^n}=\sum_{n=N+2}^{\infty}\frac{x^n}{1+x^n}\geq \frac{x^{N+2}}{1+x^{N+2}}$$

Letting \( x \to 1 \) we get that \( \displaystyle \sup_{x \in [0, 1)}\left | S(x)-S_N(x) \right |\geq \frac{1}{2} \) and the convergence is not uniform.

[Grigorios Kostakos]

A solution which shows that the series is not uniformly convergent neither in \((-1,0]\) :

\(\left({\exists\,\varepsilon=\tfrac{1}{2{\rm{e}}^2}>0}\right)\left({\forall\,n_0\in\mathbb{N}}\right)\left({\exists\,n=2n_0}\right)\left({\exists\,m=2n_0-1}\right)\bigl({\exists\,x=\tfrac{1}{n_0+1}-1\in(-1,1)}\bigr)\) such that \({n}\geqslant{n_0},\,m\geqslant{n_0}\) and \begin{align*}
\displaystyle\left|{\mathop{\sum}\limits_{k=2n_0}^{2n_0}\frac{\bigl({\frac{1}{n_0+1}-1}\bigr)^{k}}{1+\bigl({\frac{1}{n_0+1}-1}\bigr)^{k}}}\right|&=\frac{\bigl({\frac{1}{n_0+1}-1}\bigr)^{2n_0}}{1+\bigl({\frac{1}{n_0+1}-1}\bigr)^{2n_0}}\\
&=\frac{\bigl({1-\frac{1}{n_0+1}}\bigr)^{2n_0}}{1+\bigl({1-\frac{1}{n_0+1}}\bigr)^{2n_0}}\\
&>\frac{1}{2}\,\Bigl({1-\frac{1}{n_0+1}}\Bigr)^{2n_0}\\
&\stackrel{(*)}{>}\frac{1}{2}\,\frac{1}{{\rm{e}}^2}\\
&=\frac{1}{2{\rm{e}}^2}\\
&=\varepsilon\,.
\end{align*} Therefore the series does not converges uniformly in \((-1,1)\).


\((*)\) The sequence \(\Bigl({\bigl({1-\tfrac{1}{n+1}}\bigr)^{n}}\Bigr)_{n\in\mathbb{N}}\) is strictly decreasing with limit the number \({\rm{e}}^{-1}\).

[Grigorios Kostakos]

Let's explore the same question in \((-\infty,-1)\) : Examine if the series \[\displaystyle\mathop{\sum}\limits_{n=1}^{+\infty}\frac{x^n}{1+x^n}\,,\quad x\in(-\infty,-1)\,,\] converges uniformly in \((-\infty,-1)\) .

[Tolaso J Kos]

We see that in the interval \( (-1, 1) \) the series converges absolutely because:
$$\left | \sum_{n=1}^{\infty}\frac{x^n}{1+x^n} \right |\leq \sum_{n=1}^{\infty}\left | \frac{x^n}{1+x^n} \right |\leq \sum_{n=1}^{\infty}\left | x \right |^n$$

the last series converges as a geo series .

We can easily check that in the interval \( (1, +\infty) \) is impossible to converge since that thing inside goes to \( 1 \) as \( n \) goes to \( +\infty \) hence the series does not converge. Simillarly one can show that when \( x \in (-\infty, -1) \) then the sequence inside the series is written as \( \displaystyle \frac{(-x)^n}{1+(-x)^n} \) tends to \( 1 \) not zero (quite easily) and hence it does not converge.

Therefore we cannot discuss uniform convergence.