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.
 A: 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.
A: A simple proof:
Clearly:
$$
n^3+x^3\ge \frac{1}{4}(n+x)^3 \tag{1}
$$
and hence
$$
0\le \frac{nx}{n^3+x^3} \le \frac{4nx}{(n+x)^3}\le\frac{4(n+x)x}{(n+x)^3}=\frac{4x}{(n+x)^2}.
$$
Hence
$$
\sum_{n=1}^\infty\frac{nx}{n^3+x^3}\le 
\sum_{n=1}^\infty\frac{4x}{(n+x)^2}=\frac{4x}{(1+x)^2}+\sum_{n=2}^\infty\frac{4x}{(n+x)^2}\le \frac{4x}{(1+x)^2}+\int_1^\infty \frac{4x\,ds}{(s+x)^2} \tag{2}\\=
\frac{4x}{(1+x)^2}+\frac{4x}{(1+x)}.
$$
Meanwhile
$$
\frac{x^4n}{(x^3+n^3)^2}=\frac{nx}{n^3+x^3}\cdot\frac{x^3}{x^3+n^3}\le 
\frac{nx}{n^3+x^3}.
$$
Notes. Inequality $(1)$ holds since
$$
4(n^3+x^3)=4(n+x)(n^2-nx+x^2)=(n+x)(4n^2-4nx+4x^2)
=(n+x)(n^2+2nx+x^2+3n^2-6nx+3x^2)=(n+x)^3+3(n+x)(n-x)^2\ge(n+x)^3.
$$
The last inequality in $(2)$ holds since if $f: [1,\infty)\to (0,\infty)$ is decreasing, then $\sum_{n=2}^\infty f(n)\le \int_1^\infty f(x)\,dx.$
A: This is what is commonly known as Big O notation, it basically means that asympotically the function behaves like $\frac{1}{x}$ for large x, in this concrete example its used to denote that $x \cdot \sum_{n=1}^{\infty} \frac{n}{n^3 + x^3}$ converges to some finite number, therefore is bounded for all $x \in [0, \infty)$, same for the other series and $O\left(\frac{1}{x^4}\right)$.
