# A dubious proof using Weierstrass-M test for $\sum^n_{k=1}\frac{x^k}{k}$

I have been trying to prove the uniform convergence of the series $$f_{n}(x)=\sum^n_{k=1}\frac{x^k}{k}$$ Obviously, the series converges only for $x\in(-1,1)$. Consequently, I decided to split this into two intervals: $(-1,0]$ and $[0,1)$ and see if it converges on both of them using the Weierstrass M-test.

For $x\in(-1,0]$, let's take $q\in(-1,x)$. We thus have: $$\left|\frac{x^k}{k}\right|\leq\left|x^k\right|\leq\left|q^k\right|$$ and since $\sum|q^n|$ is convergent, $f_n$ should be uniformly convergent on the given interval. Now let's take $x\in[0,1)$ and $q\in(x,1)$. Now, we have: $$\left|\frac{x^k}{k}\right|=\frac{x^k}{k}\leq\ x^k\leq{q^k}$$ and once again, we obtain the uniform convergence of $f_n$.

However, not sure of my result, I decided to cross-check it by checking whether $f_n$ is Cauchy. For $x\in(-1,0]$, I believe it was a positive hit, since for $m>n$ we have: $$\left|f_{m}-f_{n}\right|=\left|f_{n+1}+f_{n+2}+...f_{m}\right|\leq\left|\frac{x^n}{n}\right|\leq\frac{1}{n}$$ which is what we needed. However, I haven't been able to come up with a method to show the same for $x\in[0,1)$. Now, I am not so sure whether $f_n$ is uniformly convergent on $[0,1)$. If it is, then how can we show it otherwise, and if it isn't, then how can we disprove it? Also, what's equally important - what did I do wrong in the Weierstrass-M test?

-
It is a power series, so it converges uniformly on its entire interval of convergence. For the Cauchy part, I don't understand what you did to get $|f_{n+1}+\cdots + f_m|\leq |x^n/n|$. For the M-test, don't break it into the two separate intervals, instead show that for any $0<r<1$ it converges uniformly on $(-r,r)$ using the same proof you gave. – Matt Sep 17 '12 at 13:24
It does not converge uniformly on $[0,1)$. This is because the partial sums are not bounded. Consider $|f_n(x)+\cdots+f_m(x)|$ for large $n$ and $m$ and take the limit as $x\rightarrow1$ to see this. – David Mitra Sep 17 '12 at 13:27
You can alter your (incorrect) argument by fixing $q$ first in $(-1,1)$. Then the $M$-test will show uniform convergence on $[-q,q]$. The argument you used showing that $(f_n)$ is uniformly Cauchy on $(-1,0]$ is correct; so this series does converge uniformly on $(-1,0]$ (I don't think the $M$-test can be used to prove this, however). – David Mitra Sep 17 '12 at 13:35
@Matt It's not true that a power series converges uniformly on the entire interval of convergence; but a power series does converge uniformly on a closed and bounded subset of its interval of convergence. – David Mitra Sep 17 '12 at 13:43
also, it does converge at -1 – mike Sep 17 '12 at 14:39

Weierstrass M-test only gives you uniform convergence on intervals of the form $[-q,q]$, where $0<q<1$. Your proof shows this.

You also get uniform convergence on the interval $[-1,0]$, but to see this you need other methods. For example the standard estimate for the cut-off error of a monotonically decreasing alternating series will work here.

As David Mitra pointed out, the convergence is not uniform on the interval $[0,1)$. Elaborating on his argument: No matter how large an $n$ we choose, the divergence of the harmonic series tells us that $\sum_{k=n+1}^{n+p}(1/k)>2$ for $p$ large enough. We can then select a number $a\in(0,1)$ such that the powers $a^k, n<k\le n+p$ are all larger than $1/2$. Then it follows that for all $x\in(a,1)$ $$\sum_{k=n+1}^{n+p}\frac{x^k}k\ge \sum_{k=n+1}^{n+p}\frac{a^k}k>\frac12\sum_{k=n+1}^{n+p}\frac1k>1.$$ Thus the Cauchy condition fails on the subinterval $(a,1)$.

-
Great point for the lack of uniform convergence on $[0,1)$. As for uniform convergence on $(-1,0]$, what is missing in the second proof I used (i.e. showing that it is Cauchy)? – Johnny Westerling Sep 17 '12 at 15:45
@JohnnyWesterling Your proof on $(-1,0]$ is fine, though it wouldn't hurt to write explicitly that you're using the alternating series bound. – Erick Wong Sep 17 '12 at 16:52

Your choice of $q$ depends on $x$.

-