5
$\begingroup$

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 4

6

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.

  • 0
    @Downvoters: if something is not clear, please feel free to ask questions.2012-03-30
5

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?

4

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...

0

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. $

  • 6
    This really doesn’t answer the question.2012-03-29