1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Because $$\left| \sum\limits_{k=1}^{n} (-1)^k \right |\leqslant 1$$ and functional sequence $\frac{1}{x+n}$ uniformly with respect to $x$ tends to zero, then according to Dirichlet test series is uniformly converged, which gives continuity for every $x \in (0, +\infty)$.

Then, as formal derivative $$\sum\limits_{k=1}^{n} \frac{(-1)^{k+1}}{(x+n)^2} $$ also is converged uniformly by $\left| \frac{(-1)^{k+1}}{(x+n)^2}\right | \leqslant \frac{1}{n^2}$, then we can state that $f$ have derivative for every $x \in (0, +\infty)$ and holds $$f'(x)=\sum\limits_{k=1}^{\infty} \frac{(-1)^{k+1}}{(x+n)^2}$$ (btw last gives continuity as well).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.