# 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.

• Everyone will ask: what have you attempted so far ? Jul 5 '15 at 8:25
• I have edited my question now to include my attempts. Jul 5 '15 at 21:06
• @K.Nantomah Voted to reopen.
– Eff
Jul 5 '15 at 21:06

## 1 Answer

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)