# Show $\sum^\infty_{n=1}(\frac{x}{n^{0.6}(1+nx^2)})$ converges uniformly on $\mathbb{R}$

$\sum^\infty_{n=1}\frac{x}{n^{0.6}(1+nx^2)}$ converges uniformly on $\mathbb{R}$

Is $x\rightarrow\sum^\infty_{n=1}(\frac{x}{n^{0.6}(1+nx^2)})$ continuous at all points of $\mathbb{R}$?

I'm stuck on the first step, I feel I want to try and bound the expression inside the sum above. I've tried that and I'm not sure how.

I'm not that confident it is right either, because $x\sum^\infty_{n=1}(\frac{1}{n^{0.6}})$ doesn't converge.

IIRC $\frac{1}{x^p}$ converges for |p|>1, needs investigation for |p|=1 and doesn't converge for |p|<1 - from the ratio test.

• Hint: you can prove the first assertion with Weierstrass M-test (find the maximum of the absolute value of the functions being summed by a simple derivative study: you will see that this boils down to $0.6+0.5>1$) and the answer to the question (yes) follows from that assertion. – Julien Nov 17 '13 at 2:07
• @julien I know this sounds weird but the book prides itself on being in order, I officially don't know that test yet. Could I have a hint without? – Alec Teal Nov 17 '13 at 2:08
• Hmm... then estimate the remainder with the same technique (function study: the max is attained at $x=1/\sqrt{n}$ for each summand). This will give you a uniform estimate hence uniform convergence. – Julien Nov 17 '13 at 2:13
• @julien uniform estimate? I'm new at real analysis sorry. I'd be very thankful if you could post these hints in an answer and flesh them out a bit. – Alec Teal Nov 17 '13 at 2:14
• Ok... but is this homework? – Julien Nov 17 '13 at 2:15

Let $$f_n(x):=\frac{x}{n^{0.6}(1+nx^2)}\qquad f(x):=\sum_{n\geq 1}f_n(x)$$

• Check that this series of functions converges pointwise over $\mathbb{R}$ (this means that for each $x\in\mathbb{R}$ fixed, the series converges: you'll have to treat the case $x=0$ which is trivial separately; for the case $n\neq 0$, note that $|f_n(x)|\leq \frac{1}{xn^{1.6}}$)

• Check that $\sup_{x\in\mathbb{R}} |f_n(x)|=f_n(1/\sqrt{n})=\frac{1}{2n^{1.1}}$.

• Deduce from the latter that

$$\sup_{x\in\mathbb{R}} \big| f(x)-\sum_{n=1}^kf_n(x)\big|\leq \sum_{n\geq k+1} \frac{1}{2n^{1.1}}\longrightarrow 0$$ as $k\rightarrow +\infty$ which is precisely uniform convergence of this series of functions over $\mathbb{R}$ (we say that a sequence of functions $g_k$ converges to $g$ uniformly over a set $S$ if $\sup_{x\in S}|g_k(x)-g(x)|\longrightarrow 0$ as $k$ tends to $+\infty$; we say a series of functions converges uniformly if the sequence of partial sums converges uniformly).

• Now you have probably seen that the uniform limit of a sequence (or series) of continuous functions is continuous. Otherwise, this is a classical $\epsilon/3$ exercise which is most likely a theorem in your book.
• @Alec Teal Ok, I have posted a more detailed sketch. Do you know the result of the last bullet? – Julien Nov 17 '13 at 2:25
• I hate to post again (the 10 or 25 rep I can give you for this is a tiny fraction of 29k) but what's the definition for pointwise convergence? The second one is that upper bound I couldn't find! (how did you find it) the $\rightarrow 0$ follows. What is the definition of "uniform convergence" (I keep seeing this word "uniform" in front of things) (more to come) – Alec Teal Nov 17 '13 at 2:28
• @AlecTeal Boy that's not an easy exercise if you don't know all these terms..;-) – Julien Nov 17 '13 at 2:29
• With that last step as 0<=|x| always, you have sandwhiched the supremum between 0 and something that tends to 0, so it must tend to zero, doesn't that show that f(x)-sum ->0 which is what we started with. UNLESS that has something to do with pointwise convergence, which'd make sense (the f(n) being the points) – Alec Teal Nov 17 '13 at 2:29
• I have seen most of them, I'm reading up on the definitions, I suppose I'm just not that confident with them, I long for more exercises to hone myself on. Regardless this "uniform [something]" pops up a lot, what does uniform mean in maths? – Alec Teal Nov 17 '13 at 2:31