2
$\begingroup$

How can I prove or disprove uniform convergence $\sum\limits_{n=1}^\infty (-x)^n\cdot \ln x$ for $x\in (0,1]$

  • 1
    Are those brackets $[]$ just to indicate the sum is over the product of those two terms or are you rounding to the nearest integer? Considering that $\ln x$ is not a function of $n$ why not factor it out? $\sum_{n=1}^\infty (-x)^n$ is geometric and can easily be worked out explicitely2012-05-07
  • 0
    You're right. Just that simple. Thanks, I missed it.2012-05-07
  • 1
    Presumably you mean $x \in (0,1]$?2012-05-07

2 Answers 2

2

A bit of calculus will reveal that $-x^n \ln x$ has a maximum value of ${1\over ne}$ (obtained at $x=e^{-1/n})$ over $(0,1]$. The terms of the series thus converge uniformly to zero. And, since the sequence $( -x^n\ln x)$ is monotone and has nonnegative terms, the series $\sum\limits_{n=1}^\infty (-1)^{n+1} (-x^n\ln x)$ converges uniformly on $(0,1]$ since it is "alternating" (we have $\Bigl| \sum\limits_{n=k}^m (-1)^{n+1}(-x^n\ln x)\Bigr|\le |x^{k }\ln(x)|\le1/(ke)$).

2

In order to prove or disprove uniform convergence you need to check whether $$ \lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}\left|\sum\limits_{k=n}^\infty(-x)^k\ln(x)\right|=0 $$ But this series can be computed explicitly $$ \sum\limits_{k=n}^\infty(-x)^k\ln(x)=\ln(x)\sum\limits_{k=n}^\infty(-x)^k=\ln(x)\frac{(-x)^n}{1+x} $$ The rest is clear.

  • 0
    You want $n$, not $k$, in that last exponent.2012-05-07
  • 0
    *The rest is clear*: I agree, for example because $|x\log x|$ is uniformly bounded on $[0,1]$.2012-05-07
  • 0
    @BrianM.Scott thanks!2012-05-07