I am trying to solve the following exercise and I would like to have a hint about the continuity part:
Let $f(x)=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+4}-\dots$.
Show $f$ is defined for all $x>0$. Is $f$ continuous on $(0,\infty)$? How about differentiable?
My solution (NOTE: completed the exercise):
f defined We can rewrite $f$ as $f(x)=\sum_{n=0}^{\infty} (-1)^n \frac{1}{x+n}$ which converges by the Alternating Series Test so the function is surely defined for all $x>0$.
f continuous Now, let $\varepsilon>0$: then if we take $N>\frac{1}{\varepsilon}$ we have, for all $n,m\geq N, n>m$ and $x>0$ $|\frac{(-1)^{m+1}}{x+(m+1)}+\frac{(-1)^{m+2}}{x+(m+2)}+\dots+\frac{(-1)^n}{x+n}|\leq\frac{1}{x+(m+1)}<\frac{1}{x+m}<\frac{1}{m}<\varepsilon$
(note that wheter $m$ is even or odd makes no difference since $|-x|=|x|$)
so the series converges uniformly Cauchy Criterion for Uniform Convergence of Series and $f$ is thus continuous by Term-by-Term Continuity Theorem.
f differentiable $|f'_n(x)|=|\frac{(-1)^{n+1}}{(x+n)^2}|=\frac{1}{(x+n)^2}\leq\frac{1}{n^2}$ and $\sum_{n=0}^{\infty}\frac{1}{n^2}$ is convergent so the series of derivatives $\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(x+n)^2}$ is uniformly convergent on $(0,\infty)$ by Weierstrass M-Test and in particular on any interval $[a,b]\subset (0,\infty)$; also, since for any $x_0\in [a,b]$, $\sum_{n=0}^{\infty}f_n(x_0)$ is convergent by the Alternating Series Test we can conclude, by the Term-by-Term Differentiability Theorem, that on any interval $[a,b]\subset (0,\infty)$, $\sum_{n=0}^{\infty}f_n(x)$ converges uniformly to a differentiable function $f(x)=\sum_{n=0}^{\infty}f_n(x)=\sum_{n=0}^{\infty} (-1)^n \frac{1}{x+n}$ such that $f'(x)=\sum_{n=0}^{\infty}f'_n(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(x+n)^2}$.
Thank you.
Best Answer
Note for $x>0$, $$ 0< \frac{1}{x+n}-\frac{1}{x+n+1}+\frac{1}{x+n+2}-... \, ... \, ... <\frac{1}{x+n}, $$ so $$ \Big|\frac{1}{x+n}-\frac{1}{x+n+1}+\frac{1}{x+n+2}-... \, ... \, ... \Big|<\frac{1}{x+n}\leq \frac 1n. $$ So this series converges uniformly for $0\leq x<\infty$, so the sum is continuous.