Inequality $\frac{1}{k^a} > \frac{1}{(k+1)^a} + \frac{1}{(k+1)^3}$ finding $a$ so inequality holds for $k \gt 2$

On my quest to getting some understanding on how to choose the proper expression to strengthen some inequalities that are proven by induction I've come across following inequality:

$$\frac{1}{k^a} > \frac{1}{(k+1)^a} + \frac{1}{(k+1)^3}$$

I would like to know how to find such $$a$$ for which the whole inequality holds true for every $$k\gt2$$.

Context of this problem

In my course on discrete mathematics we've studied also the topic of mathematical induction, including problems where we are proving an inequality. Those kind of problems are usually a bit harder when you should show that sum of $$n$$ fractions is always bellow certain constant, in other words the sum is converging. One of such problems is for example Basel problem. Those kind of inequalities are solved by strengthening the hypothesis. When I asked my professor how to choose the correct strengthening term, the answer was that it is a mixture of experience and a bit of guesswork. So the inequality above is the result of me seeking better answer to the question: Why can I strengthen the hypothesis with $$\frac{1}{n^2}$$ but not with $$\frac{1}{n^3}$$

• It seems that, when $a\in\left(0.1,2\right]$, the inequality holds for $k>2$. (If you want it to be true for all $k>0$, then every $a\in(0.272,2]$ works.) Oct 15 '19 at 22:50
• Thanks, but is there a way how to discover these values? Oct 16 '19 at 4:33
• I can prove that every $a\in[1,2]$ works. I am not sure how to prove that when $a\in [0.0099889,1]$, the inequality is also true (for $k>2$). Oct 16 '19 at 8:14
• By $k>2$, do you mean $k\in(2,\,\infty)$ or $k\in\Bbb Z_{\ge3}$?
– J.G.
Oct 16 '19 at 8:21
• Lower bound: $~$ smallest value which solves $~3^{-x}=4^{-x}+4^{-3}$ $~$ --- $~$ Upper bound: $~$ We see from $~(1+1/k)^{kx}>(1+1/(k+1)^{3-x})^k~$ that it must be $~3-x\geq 1~$ and therefore maximal $~x=2~$ . Oct 16 '19 at 10:06

Suppose that $$a$$ satisfies the inequality. If $$a\leq 0$$, then $$0<\frac{1}{(k+1)^3}<\frac{1}{k^a}-\frac{1}{(k+1)^a}\leq 0$$ for every $$k>0$$. This is a contraiction. Therefore, $$a>0$$.

For $$a\geq 1$$, we have from Bernoulli's Inequality that $$\left(1-\frac{1}{k+1}\right)^a\geq 1-\frac{a}{k+1}$$ for all $$k\geq 0$$. Therefore, $$\frac{1}{(k+1)^a}\geq \frac{1}{k^a}-\frac{a}{(k+1)k^a}$$ for each $$k>0$$. This means $$\frac{a}{(k+1)k^a}\geq \frac{1}{k^a}-\frac{1}{(k+1)^a}>\frac{1}{(k+1)^3}\,.$$ Hence, $$k^a for every $$k>0$$. This immediately implies that $$a\leq 2$$.

On the other hand, if $$1\leq a\leq 2$$, then Bernoulli's Inequality implies that $$\left(1+\frac{1}{k}\right)^a\geq 1+\frac{a}{k}$$ for every $$k>0$$. Therefore, $$\frac{1}{k^a}\geq \frac{1}{(k+1)^a}+\frac{a}{k(k+1)^a}$$ for all $$k>0$$. This means \begin{align}\frac{1}{k^a}-\frac{1}{(k+1)^a}&\geq \frac{a}{k(k+1)^a}\geq \frac{a}{k(k+1)^2}\\&\geq \frac{1}{k(k+1)^2}> \frac{1}{(k+1)^3}\end{align} for all $$k>0$$. Ergo, when $$a\geq 1$$, the inequality $$\frac{1}{k^a}-\frac{1}{(k+1)^a}>\frac{1}{(k+1)^3}\tag{*}$$ holds for every $$k>0$$ (or for every $$k>2$$) if and only if $$1\leq a\leq 2$$.

Since the function $$f:\mathbb{R}\to\mathbb{R}$$ defined by $$f(t):=\dfrac{1}{2^t}-\dfrac{1}{3^t}-\dfrac{1}{3^3}$$ for each $$t\in\mathbb{R}$$ is increasing when $$0\leq t\leq 1$$, it has a unique zero on $$[0,1]$$. A numerical solver says that $$b\approx 0.099889$$ is the zero. This shows that, for $$0, if $$\frac{1}{2^a}-\frac{1}{3^a}\geq \frac{1}{3^3}\,,$$ then $$b\leq a\leq 1$$. It seems to be the case that all $$a\in[b,1]$$ works (via WolframAlpha). Therefore, the answer to the OP's question is the inequality (*) is true for all real numbers $$k>2$$ if and only if $$b\leq a\leq 2$$.

If you only care about the integral values of $$k$$, then consider $$g(t):=\dfrac{1}{3^t}-\dfrac{1}{4^t}-\dfrac{1}{4^3}$$ for each $$t\in\mathbb{R}$$. If $$c$$ is the unique zero of $$g$$ on $$[0,1]$$, then all $$a\in(c,1]$$ works. Note that $$c\approx 0.0584002\,.$$ Therefore, the inequality (*) is true for all integers $$k>2$$ if and only if $$c.

If $$k$$ is allowed to be any positive real number, then define $$h(x,t):=\dfrac{1}{x^t}-\dfrac{1}{(x+1)^t}-\dfrac{1}{(x+1)^3}$$ for $$x>0$$ and $$t\in\mathbb{R}$$. Let $$d$$ be the smallest $$t\in(0,1]$$ such that $$h(x,t)\geq 0$$ for every $$x>0$$. Then, $$d\approx 0.2714\,.$$ Ergo, the inequality (*) is true for all real numbers $$k>0$$ if and only if $$d.

• I think this is a complete answer. ;) Oct 16 '19 at 13:19
• @user90369 I think it is a bit incomplete since I am not providing proof that every $a\in [b,1]$ works for real $k>2$, or every $a\in [c,1]$ works for integers $k>2$. I do not plan to either, but this can be seen from WolframAlpha, so I guess you can argue that the work is done. Oct 16 '19 at 14:35
• Yes, I understand. But for an orientation it’s of course enough. If the OP wants to have it proofed in details, I think one have to show with $~a_0>0, k_0>0~$ and $~k_0^{-a_0} = (k_0+1)^{-a_0} + (k_0+1)^{-3}~$ : $~$ For $~a_0<a\leq 2~$ we have $~k_0^{-a} > (k_0+1)^{-a} + (k_0+1)^{-3}~$ and for $~ k_0<k~$ it’s $~k^{-a_0} > (k+1)^{-a_0} + (k+1)^{-3}~$ . With the help of your suggestions it should be solvable for the OP. Did I forget something ? Oct 16 '19 at 15:19

The inequality is equivalent to $$\frac1{k^a}-\frac1{(k+1)^a}\gt\frac1{(k+1)^3}\tag1$$ The Mean Value Theorem says there is a $$\xi\in(k,k+1)$$ so that $$\frac1{k^a}-\frac1{(k+1)^a}=\frac{a}{\xi^{a+1}}\tag2$$ Therefore, $$\frac{a}{k^{a+1}}\gt\frac1{k^a}-\frac1{(k+1)^a}\gt\frac{a}{(k+1)^{a+1}}\tag3$$ Thus, for $$(1)$$ to be true for all $$k$$, we need $$1\le a\le2$$.

For $$0\lt a\lt1$$, $$(3)$$ shows that $$(1)$$ is true for $$k\ge\left(\frac1a\right)^{\frac1{2-a}}-1$$, which is equivalent to $$\newcommand{\W}{\operatorname{W}} a\ge-\frac1{\log(k+1)}\W\left(-\frac{\log(k+1)}{(k+1)^2}\right)\tag4$$ where $$\W$$ is Lambert W. For $$k=2$$, $$(4)$$ gives $$a\ge-\frac1{\log(3)}\W\left(-\frac{\log(3)}9\right)\doteq0.12787\tag5$$ Actually, it appears that $$a\ge\frac1{10}$$ will work for $$k\ge2$$:

Plot of $$\color{#3F3D99}{\frac1{k^{1/10}}-\frac1{(k+1)^{1/10}}}$$ vs $$\color{#993D71}{\frac1{(k+1)^3}}$$: 