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Is $f(x)=\sum_{n=2}^{\infty} \frac{1}{n\ln(n)^x}$ continuous on $(1,\infty)$?

I have proven that the infinite series converges on $(1,\infty)$. I want to use the Weierstrass M-test to prove this infinite series uniformly converges and hence continuous. Since $f_n(x)=\frac{1}{n\ln{n}^x} \leq \frac{1}{n}$, I let $M_n = \frac{1}{n}$. However, $\sum M_n$ is a harmonic series.

I want to know if there is any way to set $M_n$ be other values? Or is there any other way to prove $f(x)$ is continuous?

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    $\begingroup$ You can fix $\epsilon>0$ and then use Weierstrass M-test to get uniform convergence on $[1+\epsilon,\infty)$ since $\sum \frac{1}{n\ln(n)^{1+\epsilon}}$ converges. $\endgroup$ – Winther May 3 '16 at 18:02
  • $\begingroup$ it converges uniformly only locally, but not on $(1,1+\epsilon)$. and all you have to do here is proving that for $x \in (a,b)$ with $a > 1$ : $\displaystyle\sum_{n < N} \frac{1}{n \ln(n)^{x+\epsilon}}-\frac{1}{n \ln(n)^{x}}$ is bounded by $C \epsilon$ independently of $N$. since (when $\epsilon \to 0$) : $\quad\ln(n)^{x + \epsilon} = \ln(n)^x e^{\epsilon \ln \ln(n)} \sim \ln(n)^x (1+\epsilon \ln \ln(n))$ and that $\frac{1}{1+\epsilon \ln \ln(n)} - 1 < \epsilon \ln\ln(2)$ ... $\endgroup$ – reuns May 3 '16 at 18:05
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    $\begingroup$ Better to write that as $(\ln n)^x$ $\endgroup$ – zhw. May 3 '16 at 18:24
  • $\begingroup$ @user1952009 Not sure why you're doing all that. Why not ust use $1/[n\ (\ln n)^x)]\le 1/[n\ (\ln n)^a], n>2$ on $[a,\infty)?$ $\endgroup$ – zhw. May 3 '16 at 18:27
  • $\begingroup$ @zhw : sorry, I don't understand what you mean. I'm basically proving that $\displaystyle\sum_n \frac{1}{n\ln(n)^{x+\epsilon}} \to\sum_n \frac{1}{n\ln(n)^{x}}$ when $\epsilon \to 0$. (I don't use any M-test or whatever, because I don't think those are useful if you didn't get how to prove everything first ) $\endgroup$ – reuns May 3 '16 at 18:29
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$f$ is continuous on $(1,+\infty)$, but, since $\lim\limits_{x \to 1^+} f(x) = +\infty$ and every term of the series has a finite limit there, the convergence of the series is not uniform on $(1,+\infty)$.

So you cannot use the Weierstraß $M$-test on the whole interval. But you can use it on each of a family of subintervals exhausting $(1,+\infty)$. For that, however, it is convenient to separate the function $g_2 \colon x \mapsto \frac{1}{2(\ln 2)^x}$ from the others. Since $\ln 2 < 1$, $g_2$ is an increasing and unbounded function on $(1,+\infty)$. We can't bound it on intervals of the form $[a,+\infty)$, whereas for $n \geqslant 3$ the function $g_n \colon x \mapsto \frac{1}{n(\ln n)^x}$ is decreasing and bounded, so we can use $g_n(a)$ to bound it on $[a,+\infty)$. Since you have proved that $\sum_{n = 3}^{\infty} g_n(a)$ converges for $a > 1$, by the $M$-test we know that for an arbitrary $a > 1$ the series

$$\sum_{n = 3}^{\infty} g_n(x)$$

converges uniformly on $[a,+\infty)$ and therefore its sum - call it $f_3$ - is continuous there. Since $a > 1$ was arbitrary, it follows that $f_3$ is continuous on

$$\bigcup_{a > 1} [a,+\infty) = (1,+\infty).$$

Since $g_2$ is continuous, it then follows that $f = g_2 + f_3$ is continuous on $(1,+\infty)$.

If we hadn't separated $g_2$, we would have needed to use intervals $[a,b]$ with $1 < a < b < +\infty$ and use $g_2(b)$ as the bound for $g_2$. No big deal, but slightly less convenient.

There are of course more ways to show that $f$ is continuous:

a) For example, we can note that every $g_n \colon x \mapsto \frac{1}{n(\ln n)^x}$ is a positive convex function on $(1,+\infty)$. So

$$f(x) = \lim_{N\to\infty} \sum_{n = 2}^N \frac{1}{n(\ln n)^x}$$

is the limit of a monotonic sequence of convex functions, and therefore itself convex. But a convex function on an open interval is continuous.

b) We can show that $f$ is an integral (and that implies continuity). We find $g_n'(x) = -g_n(x)\cdot \ln (\ln n)$, and in a way similar to showing the convergence of the series $\sum g_n$ one can show the convergence of $\sum g_n'$ on $(1,+\infty)$. For every $n$, $g_n'$ is increasing, so

$$\sum_{n = 2}^{\infty} g_n'(x)$$

is increasing, and thus Riemann integrable over all intervals $[a,b]$ with $1 < a < b < +\infty$. By the monotone convergence theorem (which holds for the Riemann integral if the pointwise limit function is Riemann integrable; if you already know the Lebesgue integral, the [Lebesgue-] integrability of the limit function is part of the conclusion), since $g_n'(x) < 0$ for $n \geqslant 3$, it thus follows that

$$f(b) - f(a) = \lim_{N\to\infty} \sum_{n = 2}^N \bigl(g_n(b) - g_n(a)\bigr) = \lim_{N\to\infty} \sum_{n = 2}^N \int_a^b g_n'(x)\,dx = \int_a^b \sum_{n = 2}^{\infty} g_n'(x)\,dx,$$

i.e. $f$ is the integral of $\sum g_n'$. If you already know, or prove in some manner, that $\sum g_n'$ is continuous, this shows that $f$ is continuously differentiable. One can iterate the argument to deduce that $f$ is in fact infinitely often continuously differentiable (it's even real-analytic).

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