# 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 $$3 - \frac{1}{k^2}.$$

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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.

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@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?

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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...

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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 $$\int_1^M\frac1{x^{1+\varepsilon}}\,dx =-\frac1{\varepsilon x^\varepsilon}\biggr|_1^M= \frac1\varepsilon\Bigl(1-\frac1{M^\varepsilon}\Bigr) \le\frac1\varepsilon \quad\text{for all }M\ge1.$$

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