# Induction Proof: Proof of Strict Inequality involving Exponents of 3 in the Denominator.

I have been trying to do this problem by using induction but I became stuck halfway through:

Use induction to show that $$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{n^3}\right) \lt 3 - \frac{1}{n^2}$$ for $n\geq 2$. Does the series $$\sum_{n=1}^{\infty}\frac{1}{n^3}$$ converge? Justify your conclusions.

So far I have this:

Base Case of Induction, $n=2$ \begin{align*} \frac{1}{n^3} &\lt 3- \frac{1}{n^2}\\ \frac{1}{8} &\lt 3 - \frac{1}{4}\\ \frac{1}{8} &\lt \frac{11}{4} \end{align*}

Induction Step: Assume true for some $k \geq 2 :$ $$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) \lt 3 - \frac{1}{k^2}$$

Show true with $n= k+1$ $$2 + 4 + \frac{2}{27} + \cdots + \frac{2}{k^3} + \frac{2}{(k+1)^3} \lt 3-\frac{1}{(k+1)^2}.$$

From there, I have no idea what to do.

I was planning to have $$3 - \frac{1}{k^2} + \frac{2}{(k+1)^3} \lt 3 - \frac{1}{(k+1)^2}$$ but I feel like it won't work since $$2\left(1 + \frac{1}{8} + \frac{1}{27}+\cdots+\frac{1}{k^3}\right)$$ does not equal $$3 - \frac{1}{k^2}.$$

## 4 Answers

Hint $\$ It is simple to prove by induction that an increasing function remains $\ge$ its initial value, i.e. that $\rm\ f(n+1) \ge f(n)\:$ for $\rm\:n\ge 2\:$ $\:\Rightarrow\:$ $\rm\:f(n) \ge f(2)\:$ for $\rm n\ge 2$.

Now apply that to $\rm\:f(n) =$ RHS $-$ LHS of your inequality, which is increasing since

$$\rm f(n+1) - f(n)\: =\ \frac{3\:n+1}{n^2 (n+1)^3} > 0\ \ \ for\ \ \ n\ge 2\qquad \bf QED$$

Remark $\$ It's worth mention that this may be viewed as prototypical example of telescopy. Indeed, viewing $\rm\:f(n)\:$ as the sum of its first differences $\rm\:f(n) = f(2) + \sum_{k=2}^{n-1} (f(k+1)-f(k)),\:$ the proof reduces to a trivial induction that a sum is $> 0\:$ if each summand is. In a similar way one may exploit telescopy to simplify many inductive proofs to trivial inductions (e.g. the trivial induction that $\rm 1^n = 1$). For many examples of this technique see my prior posts on telescopy.

• @Downvoters: if something is not clear, please feel free to ask questions. – Bill Dubuque Mar 30 '12 at 23:51

When you go from

$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) \lt 3 - \frac{1}{k^2}\tag{1}$$

to

$$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}+\frac1{(k+1)^3}\right) \lt 3 - \frac{1}{(k+1)^2}\;,\tag{2}$$

the righthand side increases by

$$\left(3-\frac1{(k+1)^2}\right)-\left(3-\frac1{k^2}\right)=\frac1{k^2}-\frac1{(k+1)^2}\;,$$

while the lefthand side increases by $\dfrac2{(k+1)^3}$. If you can show that $$\frac2{(k+1)^3}\le\frac1{k^2}-\frac1{(k+1)^2}\;,\tag{3}$$ you’ve shown that $(1)$ implies $(2)$, which is your induction step; do you see why?

Now $(3)$ can by multiplied through by $k^2(k+1)^3$ to yield the equivalent inequality $$2k^2\le (k+1)^3-k^2(k+1)\;,$$ which in turn simplifies to $0\le 3k+1$. This is certainly true for $k\ge 2$, and it’s equivalent to $(3)$, so $(3)$ is true for $k\ge 2$, and your induction step is done.

For the second part of the question, note that the $n$-th partial sum of the infinite series is $$\sum_{k=1}^n\frac1{k^3}\;,$$ and from the first part of the question you know that $$2\sum_{k=1}^n\frac1{k^3}<3-\frac1{n^2}\;,$$ so $$\sum_{k=1}^n\frac1{k^3}<\frac32-\frac1{2n^2}\;.$$ Thus, the partial sums form an increasing sequence bounded above by ... what? And what can you conclude from that about the convergence of the infinite series?

It doesn't matter that the sum does not equal $3- \frac{1}{k^3}$. What matters is that the sum is less than $3 - \frac{1}{k^3}$.

We have: $$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) \lt 3 - \frac{1}{k^2}.$$ Adding $\frac{2}{(k+1)^3}$ to both sides does not change the inequality, so we have $$2\left(1+ \frac{1}{8} + \frac{1}{27} + \cdots + \frac{1}{k^3}\right) + \frac{2}{(k+1)^3} \lt 3 - \frac{1}{k^2} + \frac{2}{(k+1)^3}.$$ The left hand side is the quantity you want.

With the right hand side, we have $$\frac{1}{k^2}\gt \frac{1}{(k+1)^2},$$ so $$-\frac{1}{k^2} \lt -\frac{1}{(k+1)^2},$$ hence $$\frac{2}{(k+1)^3} - \frac{1}{k^2} \lt \frac{2}{(k+1)^3} - \frac{1}{(k+1)^2} = \frac{1}{(k+1)^2},$$ so $$3 - \frac{1}{k^2}+\frac{2}{(k+1)^3} \lt 3 - \frac{1}{(k+1)^2}.$$ And since $a\lt b$ and $b\lt c$ implies $a\lt c$, then...

For the last part of the question, note that the quantity on the left hand side is twice the $n$th partial sum of $$\sum_{n=2}^{\infty} \frac{1}{n^3}.$$ So you should be able to conclude that the partial sums of $$\sum_{n=1}^{\infty}\frac{1}{n^3}$$ are bounded above; since all terms are positive, that means that...

Your sum converges to $$\sum_{n=1}^\infty \frac{1}{n^3} = \zeta(3),$$ because the series $\sum_{n=1}^\infty \frac1{n^{1+\varepsilon}}$ (cf. Riemann zeta function) converges for every ε > 0, because $$\int_1^M\frac1{x^{1+\varepsilon}}\,dx =-\frac1{\varepsilon x^\varepsilon}\biggr|_1^M= \frac1\varepsilon\Bigl(1-\frac1{M^\varepsilon}\Bigr) \le\frac1\varepsilon \quad\text{for all }M\ge1.$$

• This really doesn’t answer the question. – Brian M. Scott Mar 29 '12 at 22:01