We have $f:(0,\infty) \rightarrow \mathbb{R}$ defined by infinite series $f(x)=\sum_{n=1}^{\infty} (\frac{x}{n}-log(1+\frac{x}{n}))$

Prove that $f$ is continuous and can be differentiated infinitely.

I have problem right in the beginning. I do not know how to prove that above series in uniformly convergent. I was trying to use series expansion of natural logarithm, but I get this $f(x)=\sum_{n=1}^{\infty}\sum_{k=2}^{\infty} \frac{(-1)^kx^k}{kn^k}$ and I don't know how to proceed.

  • $\begingroup$ I don't think it's uniformly convergent. Do you have a reason to believe that it is? $\endgroup$ – Arthur Apr 24 '16 at 9:40
  • $\begingroup$ @Arthur : the Taylor expansion tell us that when $|x| < C < n$ : $\log(1+x/n) = \frac{x}{n} - \frac{x^2}{2 n^2} + h(x/n)$ where $|h(x/n)| < x^3/n^3$ $\endgroup$ – reuns Apr 24 '16 at 9:43
  • $\begingroup$ You may be right. But we still can use this series to define continuous function if we prove that it is uniformly convergent for any $[a.b] \subseteq (0,\infty)$, right? $\endgroup$ – SekstusEmpiryk Apr 24 '16 at 9:44
  • $\begingroup$ @SekstusEmpiryk : can you prove that $f(x) = \sum_{n=N}^\infty \sum_{k=2}^\infty \frac{(-x)^k}{k n^k}$ is an absolutely convergent double series for $|x| < N$ ? can you prove that its $m$th derivative also is an absolutely convergent power series, hence that $f(x)$ is analytic on $]0,\infty[$ ? $\endgroup$ – reuns Apr 24 '16 at 9:48
  • $\begingroup$ @user1952009 So, it's not universally convergent on $(0, \infty)$, but it is pointwise convergent, and it converges uniformly on any bounded set, which means that results that use uniform convergence (such as the resulting function being continuous and so on) still holds. Is that what you mean? $\endgroup$ – Arthur Apr 24 '16 at 9:55

By "$\log$" you mean the "natural logarithm".
From Wiki of "Natural logarithm", we have the following inequality for all $x\in(0,\infty)$, $$\frac{x}{x+1}\le\log(1+x)\le x.$$ Then we see that each term $u_n(x):=\frac{x}{n}-\log(1+\frac{x}{n})$ satisfies $$0\le u_n(x)\le\frac{x}{n}-\frac{\frac{x}{n}}{\frac{x}{n}+1}=\frac{x^2}{nx+n^2}.$$ Now for $\forall M>0$ and $\forall x\in(0,M]$, we have $$0\le u_n(x)\le\frac{M^2}{n^2},$$ and then $$0\le f(x)=\sum_{n=1}^\infty u_n(x)\le M^2\bigg(\sum_{n=1}^\infty\frac{1}{n^2}\bigg).$$ Hence by Weierstrass criterion, $f$ converges uniformly on $(0,M]$ for all $M>0$, and thereby converges (but may not uniformly) on $(0,\infty)$ by the arbitrariness of $M$.

Therefore, the continuity of $f$ follows from this property of each term $u_n$ and the uniform convergence of $f$ on $(0,M]$ for all $M>0$.

Similarly, for $\forall M>0$ and $\forall x\in(0,M]$, $$0\le u_n'(x)=\frac{1}{n}-\frac{1}{n+x}\le\frac{M}{n^2},$$ $$0\le|u_n^{(k)}(x)|=\bigg|\frac{(-1)^k}{(n+x)^k}\bigg|\le\frac{1}{n^k},$$ thus the infinite differentiability of $f$ follows from the continuity and infinite differentiability of each term $u_n$ and the uniform convergence of all-order derivatives $u_n^{(k)}$ on $(0,M]$ for all $M>0$.

  • $\begingroup$ Infinite differentiability + uniform convergence does not imply that the limit is infinitely differentiable (see: Weierstrass' functions). You also need uniform convergence of the derivatives. $\endgroup$ – D. Thomine Apr 24 '16 at 10:52
  • $\begingroup$ Can we say that since $\sum_{n=1}^\infty u_n(x)$, can be presented as convergent power series using taylor expansion, it is infinitely differentiable (because any convergent power series is)? $\endgroup$ – SekstusEmpiryk Apr 24 '16 at 11:05
  • $\begingroup$ @D.Thomine Yes, you are right. Sorry for my negligence. I will reedit it, adding this part. Thank you. $\endgroup$ – Q. Huang Apr 24 '16 at 11:16
  • $\begingroup$ @SekstusEmpiryk It is true that all convergent power series are infinitely differentiable within their interval of convergence. But in your case, $f(x)=\sum_{n=1}^{\infty}\sum_{k=2}^{\infty} \frac{(-1)^kx^k}{kn^k}$ are not power series, since there are two summation indexes. Only for fixed $n$, $\sum_{k=2}^{\infty} \frac{(-1)^kx^k}{kn^k}$ are power series. $\endgroup$ – Q. Huang Apr 24 '16 at 11:52

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.