Show a function defined by summation is increasing, another is decreasing Problem: For real numbers $x\ge1$ and $k>0$, let $f:R\rightarrow R$ and $g:R\rightarrow R$ be defined as follows. 
$f(x) = -\frac{1}{x}+\sum_{n=1}^{\infty}\frac{1}{(nk+x)^2}$ ,
$g(x)=\frac{1}{2x^2}-\frac{1}{x}+\sum_{n=1}^{\infty}\frac{1}{(nk+x)^2}$. 
Show that $f$ is increasing and $g$ is decreasing (that is, $f'(x)\ge0$ and $g'(x)\le0$ ). 
I was given this problem to solve. I tried differentiating each function with respect to $x$ and got the following: 
$f'(x) = \frac{1}{x^2}-2\sum_{n=1}^{\infty}\frac{1}{(nk+x)^3}$ ,
$g'(x)=-\frac{1}{x^3}+\frac{1}{x^2}-2\sum_{n=1}^{\infty}\frac{1}{(nk+x)^3}$. 
I was unable to proceed after this stage. Also i tried the following. 
For $x>y\ge1$, i wanted to show that $f(x)-f(y)\ge0$, however, i got
$f(x)-f(y)=\frac{x-y}{xy}+\sum_{n=1}^{\infty}\left[\frac{1}{(nk+x)^2}-\frac{1}{(nk+y)^2}\right]$. 
I got a similar expression for $g$. Once again, I am unable to proceed from here. 
Your solution(s) to this problem will greatly help me. Thank you in advance.
 A: We can investigate the monotonicity of the functions $f$ and $g$ as you did because of the following
Lemma. [Fich, 435, Th.7 at p. 438] Let $\{u_n(x)\}$ be a sequence of functions on a segment $[a,b]$ such that for each $n$ a function $u_n’(x)$ is continuous. If a series $\sum_{n=1}^\infty u_n(x)$ converges to $u(x)$ on $[a,b]$ and a series $\sum_{n=1}^\infty f’_n(x)$ uniformly converges to $v(x)$ on $[a,b]$ then $u’(x)=v(x)$.
Let’s establish bounds for the function  $$S(k,x)= \sum_{n=1}^{\infty}\frac{1}{(nk+x)^3}=\frac 1{k^3} \sum_{n=1}^{\infty}\frac{1}{(n+x/k)^3}.$$  
An upper bound 
$$S(k,x)=\frac 1{k^3} \sum_{n=1}^{\infty}\frac{1}{(n+x/k)^3}<$$ 
$$\mbox{(by area comparison)}$$ 
$$\frac 1{k^3} \int_{0}^{\infty}\frac{dt}{(t+x/k)^3}=
\frac 1{k^3} \frac{-1}{2(t+x/k)^2}\Big|_0^{\infty}=
\frac 1{k^3}\frac{1}{2(x/k)^2}=\frac{1}{2kx^2}.$$ 
A lower bound 
$$S(k,x)=\frac 1{k^3} \sum_{n=1}^{\infty}\frac{1}{(n+x/k)^3}>$$ 
$$\mbox{(by area comparison)}$$ 
$$\frac 1{k^3} \int_{1}^{\infty}\frac{dt}{(t+x/k)^3}=
\frac 1{k^3} \frac{-1}{2(t+x/k)^2}\Big|_1^{\infty}=
\frac 1{k^3}\frac{1}{2(1+x/k)^2}=\frac{1}{2k(k+x)^2}.$$ 
Therefore since $f'(x) = \frac{1}{x^2}-2S(k,x)$ the function $f$ decreases for $k<1$ and $x>\frac{k^2+k\sqrt{k}}{1-k}$. Nevertheless, for $k\ge 1$ the function $f$  increases, because then $S(k,x)< \frac{1}{2x^2}$.   
Since $$g'(x)= -\frac{1}{x^3}+\frac{1}{x^2}-2S(k,x)>-\frac{1}{x^3}+\frac{1}{x^2}-\frac{1}{kx^2}=\frac{kx-k-x}{kx^3},$$ the function $g$ increases for $x>1$ and $k>\frac1{x-1}+1$. 
Nevertheless, since $$g'(x)= -\frac{1}{x^3}+\frac{1}{x^2}-2S(k,x)<-\frac{1}{x^3}+\frac{1}{x^2}-\frac{1}{k(k+x)^2}=\frac{x^3(k-1)+x^2(2k^2-k)+x(k^3-2k^2)-k^3}{k(k+x)^2x^3},$$ the function $g$ decreases for sufficiently small $k$. 
References 
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, v. II, 7-th edition, M.: Nauka, 1970. (in Russian)
