I am looking to show pointwise convergence and (potentially) uniform convergence of the following:


I know (from my book) this converges for my given values of $x \in (0,1)$, but I can't figure out how to do this. I tried using the ratio test, but I wasn't able to get an answer that made sense($x$ is what I kept getting). I also tried Weierstrass M-Test, but could only to think to compare it to $$\sum_{k=1}^\infty\frac{1}{k}$$ which doesn't work either. Wolfram says this can be shown using the ratio test.

Can somebody give me an idea of what to use for the M-Test or maybe do the ratio test so I can see if I am doing something wrong?

Edit: I think my pointwise convergence to $x$ is correct, I just want to double-check that this is not uniform convergence. Am I correct?

  • $\begingroup$ Easiest way for pointwise convergence would be $\frac{x^k}{k} \le x^k$ and $\sum_{k=1}^\infty x^k = \frac{1}{1-x} -1$. $\endgroup$ – Hetebrij May 3 '16 at 22:25
  • 2
    $\begingroup$ "x is what I kept getting" well you meant $|x|$ -- but that is exactly what you want! $\endgroup$ – zhw. May 3 '16 at 22:27
  • $\begingroup$ Correct, I recognize that now, but what about the uniformity? $\endgroup$ – BridgeSkier May 3 '16 at 22:30

As a start, the partial sum of your series is $$ \sum_{k=1}^n {x^k\over k}=\int_0^x{1-t^n\over 1-t}\,dt=-\log(1-x)-\int_0^x{t^n\over 1-t}\,dt. $$ Using this it is easy to show that the series converges uniformly on $[-r,r]$ for each $r\in(0,1)$, and with a little more work that it even converges uniformly on $[-1,r)$ for each $r\in(0,1)$.


We have $$ \lim_{k\to\infty}\sqrt[k]{\frac{1}{k}}=1$$ hence according to Hadamard's theorem the radius of convergence of $\sum_k\frac{x^k}{k}$ is $1$.

This means that the series converges absolutely for all $x$ with $|x|<1$, and therefore (by the Weierstrass $M$-test) converges uniformly on $[-r,r]$ for all $0\leq r<1$. The series diverges at $x=1$ and converges conditionally at $x=-1$.

  • $\begingroup$ Unfortunately, while I recognize that this works, I'm unable to use that theorem. Do you know another way to show this? $\endgroup$ – BridgeSkier May 3 '16 at 22:39
  • $\begingroup$ Actually the series converges uniformly on $[-1,r)$ for $0<r<1.$ $\endgroup$ – zhw. May 3 '16 at 22:46

Added later: For $x\in (0,1),$ set $f(x)=\sum_{k=1}^{\infty} x^k/k$ (we already know the series converges pointwise on $(0,1)$, so $f$ is well defined). Let $S_n(x) = \sum_{k=1}^{n} x^k/k.$ Note that $f > S_n$ on $(0,1).$ Note also that each $S_n$ is bounded on $(0,1).$ If $S_n\to f$ uniformly on $(0,1),$ then $f$ is itself bounded on $(0,1),$ say by $M.$ Choose $n$ such that $\sum_{k=1}^{n} 1/k> M.$ (Possible, since the harmonic series diverges). Then

$$M = \sup_{(0,1)} f \ge \sup_{(0,1)} S_n = \sum_{k=1}^{n} 1/k> M,$$

contradiction. Thus $\sum_{k=1}^{\infty} x^k/k$ does not converge uniformly on $(0,1).$

Hint: The uniform limit of bounded functions is bounded. Is that true here?

  • $\begingroup$ The function values are bounded between $0$ and $1$, correct?.......So, yes, it is uniform on $(0,1)$? $\endgroup$ – BridgeSkier May 3 '16 at 22:50
  • $\begingroup$ What function values? No, you're off a bit here. $\endgroup$ – zhw. May 3 '16 at 22:51
  • $\begingroup$ "Is that true here?" You tell us. It was your attempt. $\endgroup$ – hardmath May 3 '16 at 23:55

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