Prove that $f(x) = \sum_{n=1}^\infty \frac{x^2}{x^3+n^3}$ is continuous on $[0,\infty)$

Question

Exercise Prove that $f(x) = \displaystyle\sum_{n=1}^\infty \frac{x^2}{x^3+n^3}$ is continuous on $[0,\infty)$

I am wondering if there is a way we can say this is less than some power series whose radius of convergence is infinite.

If that won't work, is there any way to use the M-Test? The only problem with the M-Test is that $x^2$ varies greatly, which would make choosing a sequence of constants $M_k$ difficult.

Aside from that, the only other thing I could imagine is showing that the sum of the terms (functions) is uniformly convergent (to $f$) and that each term of the series is continuous which would imply that the sum function $f$ is continuous.

Would any of these strategies be useful in this situation? Would one be more efficient than another?

Answer 1

I thank Sangchul Lee for the initial hints in the comments.

The point is that continuity is a local property. The advantage of locality is that one doesn't need to worry about universality of bounds , because confining your input to a certain interval is far more convenient to push through. Let $f(x) = \sum_{n=1}^\infty \frac{x^2}{x^3+n^3}$. Note that $f$ exists as a function, because $$ \frac{x^2}{x^3+n^3} \leq x^2\frac{1}{n^3} \implies f(x)\ \leq x^2\sum_{n=1}^\infty \frac{1}{n^3} \leq Cx^2 $$ for $C = \sum_{n=1}^\infty \frac{1}{n^3} <\infty$.

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Approach 1

• We prove that $h(x) = \sum_{n=1}^{\infty} \frac 1{x^3+n^3}$ is a uniformly convergent series on $[0,\infty)$ using the Weierstrass M-test. Details hidden below.

Note that $h(x) = \sum_{n=1}^\infty f_n(x)$ where $f_n(x) = \frac 1{x^3+n^3}$. Note that $|f_n(x)| \leq M_n = \frac 1{n^3}$ for all $x \in [0,\infty)$ and $n \geq 1$. Furthermore, $\sum_{n=1}^\infty M_n = \sum_{n=1}^\infty \frac 1{n^3}< \infty$. The Weierstrass M-test shows that $h(x)$ is uniformly and absolutely convergent on $[0,\infty)$.

• By the uniform limit theorem, $h(x)$ is a continuous function on $[0,\infty)$. We already know that $x^2$ is continuous on $[0,\infty)$. By the product rule, $x^2h(x) = f(x)$ is continuous on $[0,\infty)$, as desired.

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Approach 2

• Let $x \in [0,\infty)$. We prove that for all $a$, $f$ as a series is uniformly convergent in $[0,a)$.

Once again, let $f(x) = \sum_{n=1}^\infty \frac{x^2}{x^3+n^3} = \sum_{n=1}^\infty g_n(x)$ be defined on $[0,a)$. Then, note that $$g_n(x) = \frac{x^2}{x^3+n^3} \leq \frac{x^2}{n^3} \leq \frac{a^2}{n^3} = M_n$$ and $\sum_{n=1}^\infty M_n = \sum_{n=1}^\infty \frac{a^2}{n^3}<\infty$. The Weierstrass M-test applies.

• Thus, $f$ is continuous , by the uniform limit theorem, in $[0,a)$ for every $a$. In particular, for any $x \in [0,\infty)$, obviously $x \in [0,\lceil x \rceil)$ so $f$ is continuous at $x$, concluding the problem.

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I think I should add some comments around the approaches listed : the point is that the Weierstrass M-test doesn't apply to $f$ if the domain is $[0,\infty)$.

Basically, if you see the second approach, the $M_n$ used there would be infinite if I slid $x$ to infinity. The first approach realizes that one must "remove" the reason (which is $x^2$) why the Weierstrass test fails which is the infinite behaviour, see that the M-test works now, and then put back the reason $x^2$ without compromising on continuity.

The second one is based upon the fact that we can compromise on uniformity by going "local" : the trouble for the M-test is at infinity, but going local means that we avoid infinity, and are therefore able to assert continuity on bounded intervals, which we paste together (and don't need to worry about the $\delta$ being the same $x$ in different patches for a given $\epsilon$ because we aren't looking for uniform continuity!) to get our result.

Answer 2

Uniform Convergence

Since the series $$ g(x)=\sum_{n=1}^\infty\frac1{n^3+x^3}\tag1 $$ converges at $x=0$, it converges uniformly on $[0,\infty)$ by comparison. That is, for all $x\in[0,\infty)$, $$ \sum_{n=k}^\infty\frac1{n^3+x^3}\le\sum_{n=k}^\infty\frac1{n^3}\tag2 $$ which can be made as small as desired, independent of $x$. Therefore, $g(x)$ is continuous.

Then, $f(x)=x^2g(x)$ is continuous, being the product of two continuous functions on $[0,\infty)$.

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The Limit $$ \begin{align} \lim_{x\to\infty}\sum_{n=1}^\infty\frac{x^2}{n^3+x^3} &=\lim_{x\to\infty}\sum_{n=1}^\infty\frac1{(n/x)^3+1}\frac1x\tag{3a}\\[6pt] &=\int_0^\infty\frac{\mathrm{d}t}{t^3+1}\tag{3b}\\[3pt] &=\frac13\int_0^\infty\left[\frac1{t+1}-\frac{\left(t-\frac12\right)-\frac32}{\left(t-\frac12\right)^2+\frac34}\right]\mathrm{d}t\tag{3c}\\ &=\left[\frac16\log\left(\frac{(t+1)^2}{t^2-t+1}\right)+\frac1{\sqrt3}\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right)\right]_0^\infty\tag{3d}\\[6pt] &=\frac{2\pi}{3\sqrt3}\tag{3e} \end{align} $$ Explanation:
$\text{(3a)}$: divide numerator and denominator by $x^3$
$\text{(3b)}$: write the Riemann sum as an integral
$\text{(3c)}$: partial fractions
$\text{(3d)}$: integrate the parts
$\text{(3e)}$: evaluate

If we don't mind using contour integration, we could use this answer to evaluate $\text{(3b)}$ directly.

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A Closed Form

At the risk of bringing some complex functions into the answer, we can give an explicit sum to the series.

Let $\omega=e^{2\pi i/3}=-\frac12+i\frac{\sqrt3}2$, then because $1+\omega+\omega^2=0$, $$ \begin{align} \frac{x^2}{n^3+x^3} &=\frac13\left(\frac1{n+x}+\frac\omega{n+\omega x}+\frac{\omega^2}{n+\omega^2x}\right)\tag{4a}\\[3pt] &=-\frac13\left(\left(\frac1n-\frac1{n+x}\right)+\omega\left(\frac1n-\frac1{n+\omega x}\right)+\omega^2\left(\frac1n-\frac1{n+\omega^2x}\right)\right)\tag{4b} \end{align} $$ Therefore, using the Extended Harmonic Numbers, $H(x)$: $$ \sum_{n=1}^\infty\frac{x^2}{n^3+x^3} =-\frac13\left(H(x)+\omega H(\omega x)+\omega^2H\!\left(\omega^2x\right)\right)\tag5 $$ We can write $(5)$ in terms of the Digamma function, $\psi(x)$, if we note that $H(x)=\psi(x+1)+\gamma$.

The Extended Harmonic numbers are analytic, except at the negative integers. Therefore, the sum is real analytic, thus continuous, on the non-negative real axis.

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Asymptotic Series

Using $(5)$ and the asymptotic series for the Harmonic Numbers $$ H(x)\sim\log(x)+\gamma+\frac1{2x}-\frac1{12x^2}+\frac1{120x^4}-\frac1{252x^6}+\frac1{240x^8}-\frac1{132x^{10}}\tag6 $$ as derived in this answer, we get $$ \begin{align} f(x) &\sim-\frac13\left(\log(x)+\omega\log(\omega x)+\omega^2\log\left(\omega^2x\right)\right)\tag{7a}\\ &\phantom{{}\sim{}}{-\frac13}\left(\gamma+\omega\gamma+\omega^2\gamma\right) &\text{vanishes}\tag{7b}\\ &\phantom{{}\sim{}}{-\frac16}\left(\frac1x+\frac\omega{\omega x}+\frac{\omega^2}{\omega^2x}\right)\tag{7c}\\ &\phantom{{}\sim{}}{+\frac1{36}}\left(\frac1{x^2}+\frac\omega{\omega^2x^2}+\frac{\omega^2}{\omega^4x^2}\right) &\text{vanishes}\tag{7d}\\ &\phantom{{}\sim{}}\vdots\\ &\phantom{{}\sim{}}{+\frac1{396}}\left(\frac1{x^{10}}+\frac\omega{\omega^{10}x^{10}}+\frac{\omega^2}{\omega^{20}x^{10}}\right)\tag{7e}\\ &=-\frac13\left(\log(1)+\omega\log(\omega)+\omega^2\log\left(\omega^2\right)\right)\tag{8a}\\ &\phantom{{}={}}{-\frac1{2x}}\tag{8b}\\ &\phantom{{}={}}{-\frac1{120x^4}}\tag{8c}\\ &\phantom{{}={}}{+\frac1{132x^{10}}}\tag{8d}\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{2\pi}{3\sqrt3}-\frac1{2x}-\frac1{120x^4}+\frac1{132x^{10}}}\tag9 \end{align} $$ Explanation:
$\text{(7a)}$: $\log(x)+\omega\log(\omega x)+\omega^2\log\left(\omega^2x\right)$
$\phantom{\text{(7a):}}$ $=\left(1+\omega+\omega^2\right)\log(x)+\left(\log(1)+\omega\log(\omega)+\omega^2\log\left(\omega^2\right)\right)$
$\text{(7b)}$: $\frac1{x^n}$ in $H(x)$ becomes
$\phantom{\text{(7b):}}$ $-\frac13\left(\frac1{x^n}+\frac{\omega^{1-n}}{x^n}+\frac{\omega^{2(1-n)}}{x^n}\right)=-\frac1{x^n}[n\equiv1\pmod3]$
$\phantom{\text{(7b):}}$ where $[\dots]$ are Iverson Brackets
$\text{(7c)-(7e)}$: same as $\text{(7b)}$
$(8)$: explained in $(7)$
$(9)$: $-\frac13\left(\log(1)+\omega\log(\omega)+\omega^2\log\left(\omega^2\right)\right)$
$\phantom{\text{(9):}}$ $=-\frac13\left(0+\left(\frac{-1+i\sqrt3}2\right)\left(\frac{2\pi i}3\right)+\left(\frac{-1-i\sqrt3}2\right)\left(-\frac{2\pi i}3\right)\right)$
$\phantom{\text{(9):}}$ $=\frac{2\pi}{3\sqrt3}$