Show $ f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$ ,$\ g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}$ are bounded on $[0, \infty)$.

If $f(x), g(x)$ are defined as following on $[0 , \infty)$, $$\tag 1 f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$$

$$\tag 2 g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}.$$ . Then how to show that $f,g$ are bounded function on $[0 , \infty)$?

I found this problem on this Is $f(x)=\sum_{n=1}^\infty\frac{nx^2}{n^3+x^3}$ uniformly continuous on $[0,\infty)$?.

In solution's last steps, i don't know why this is correct answer.

Could you explain for me to elaborate?

Thank you.

First note that for $x \ge 0$ and $n \in \Bbb N$ $$ \frac{x^4n}{(n^3 + x^3)^2} =\frac{x^3}{n^3+x^3}\cdot\frac{nx}{n^3 + x^3} < \frac{nx}{n^3 + x^3} $$ and therefore $g(x) \le f(x)$, so that it suffices to show that the function $f$ is bounded on $[0, \infty)$.

For $0 \le x \le 1$ we have $$ \frac{nx}{n^3 + x^3} \le \frac{1}{n^2} \implies f(x) \le \sum_{n=1}^\infty \frac{1}{n^2} . $$

For fixed $x >1$ and $m =1, 2, 3, \ldots$ consider all $n$ with $x(m-1) \le n < xm$. For each of these $n$, $$ \frac{nx}{n^3 + x^3} \le \frac{mx^2}{(m-1)^3x^3 + x^3} = \frac 1x \cdot \frac{m}{(m-1)^3+1} $$ and there are at most $\lfloor x \rfloor +1$ of such $n$. Therefore $$ f(x) \le \frac{\lfloor x \rfloor +1}{x} \sum_{m=1}^\infty \frac{m}{(m-1)^3+1} \le 2 \sum_{m=1}^\infty \frac{m}{(m-1)^3+1} $$ for $x > 1$.

Previous solution (more complicated): The idea for estimating $f(x)$ is to replace the infinite sum by a “similar” integral: $$ \sum_{n=1}^\infty \frac{nx}{n^3+x^3} \lessapprox \int_0^\infty \frac{ux}{u^3+x^3} \, dx = \int_0^\infty \frac{v}{v^3+1} \, . $$ The two integrals are equal via the substitution $u=xv$, and the last integral is independent of $x$, so that we get a uniform upper bound. Of course the “approximate inequality” must be stated and proved precisely, so here are the gory details:

For fixed $x>0$ we consider the function $\varphi$ defined on $[0, \infty)$ by $$ \varphi(t) = \frac{tx}{t^3 + x^3} \,. $$ It is easy to see (by calculating the derivative) that $\varphi $ is increasing on $[0, \frac{x}{2^{1/3}}]$ and decreasing on $[\frac{x}{2^{1/3}}, \infty)$.

If $\frac{x}{2^{1/3}} \le 1$ then $\varphi$ is decreasing on $[1, \infty)$, so that each term in the sum of $f(x)$ (with the exception of the first term) can be estimate above by an integral over $\varphi$: $$ f(x) = \varphi(1) + \sum_{n=2}^\infty \varphi(n) \\ \le \varphi(1) + \sum_{n=2}^\infty \int_{n-1}^n \varphi(u) \, du = \varphi(1) + \int_1^\infty \varphi(t) \, dt \le 1 + \int_0^\infty \frac{ux}{u^3+x^3} \, du \, $$ and with the substitution $u = xv$ we get $$ \tag{*} f(x) \le 1 + \int_1^\infty \frac{v}{v^3+1} \, dv \, . $$

If $\frac{x}{2^{1/3}} > 1$ then we can proceed similarly. With $m = \lfloor \frac{x}{2^{1/3}} \rfloor$ we estimate $$ f(x) = \sum_{n=1}^{m-1} \varphi(n) + \varphi(m) + \varphi(m+1) + \sum_{n=m+2}^{\infty} \varphi(n) \\ \le \int_0^\frac{x}{2^{1/3}} \varphi(u) \, du + 2 \varphi(\frac{x}{2^{1/3}}) + \int_\frac{x}{2^{1/3}}^\infty \varphi(u) \, du \\ = \frac{4}{ 2^{1/3} \cdot 3x} + \int_0^\infty \frac{ux}{u^3+x^3} \, du \\ < 1 + \int_0^\infty \frac{ux}{u^3+x^3} \, du = 1 + \int_0^\infty \frac{v}{v^3+1} \, dv $$ so that $(*)$ holds as well.