How can you prove $1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$ by induction?

Question

Can you provide the steps and corresponding explanations to prove the following predicate by induction?

$$P(n) := 1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$$

I've done some work on it myself by attempting to show that $\frac{k^4}{4} < \frac{(k + 1)^4}{4}$ for the RHS, but I don't understand exactly what I am doing.

Thank you.

Notice: This is not a homework question. I'm attempting to self-study Calculus over the Summer.

Answer 1

HINT

By induction we have

• Base case: $n=1\implies 0<\frac14<1$
• Induction step, we assume true

$$P(n) := 1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$$

and we need to prove

$$P(n+1) := 1^3 + 2^3+\cdots+(n-1)^3+n^3 < \frac{(n+1)^4}{4} < 1^3 + 2^3 + \cdots + n^3+(n+1)^3$$

then we have

$$1^3 + 2^3+\cdots+(n-1)^3+n^3 < n^3 +\frac{n^4}{4}$$

$$\frac{n^4}{4}+(n+1)^3< 1^3 + 2^3 + \cdots + n^3+(n+1)^3$$

then it suffices to prove that

$$n^3 +\frac{n^4}{4}< \frac{(n+1)^4}{4}<\frac{n^4}{4}+(n+1)^3$$

that is

$$n^3 < \frac{(n+1)^4}{4}-\frac{n^4}{4}<(n+1)^3$$

Answer 2

Hint: For the left hand side inequality we have $$ 1^3 + 2^3 +\ldots +n^3 < \frac{n^4}4 + n^3 < \frac{n^4}4 + \frac14 (4n^3 + 6n^2 + 4n + 1) = \frac{(n+1)^4}{4} $$

Answer 3

Let's try a generalization.

$P(n) := \sum_{k=1}^{n-1} k^{m-1} < \frac{n^{m}}{m} < \sum_{k=1}^{n} k^{m-1} $

Base case.

For $n=1$ this is $0 < \frac1{m} < 1 $ which is true.

Induction step.

Suppose true for $n$.

$P(n) := \sum_{k=1}^{n-1} k^{m-1} < \frac{n^{m}}{m} < \sum_{k=1}^{n} k^{m-1} $

Want to show that $P(n) \implies P(n+1)$.

First, the left inequality of $P(n+1)$, which is $\sum_{k=1}^{n} k^{m-1} < \frac{(n+1)^{m}}{m} $.

$\begin{array}\\ \sum_{k=1}^{n} k^{m-1} &=\sum_{k=1}^{n-1} k^{m-1}+n^{m-1}\\ &<\frac{n^m}{m}+n^{m-1}\\ &=\frac{n^{m}+mn^{m-1}}{m}\\ \end{array} $

so we are done if $n^m+mn^{m-1} \lt (n+1)^m $ or, dividing by $n^m$, $1+m/n \lt (1+1/n)^m $ and this follows from Bernoulli's inequality.

Next, the right inequality of $P(n+1)$, which is $\sum_{k=1}^{n+1} k^{m-1} > \frac{(n+1)^{m}}{m} $.

$\begin{array}\\ \sum_{k=1}^{n+1} k^{m-1} &=\sum_{k=1}^{n} k^{m-1}+(n+1)^{m-1}\\ &>\frac{n^m}{m}+(n+1)^{m-1}\\ \end{array} $

so we are done if $\frac{n^m}{m}+(n+1)^{m-1} \ge \frac{(n+1)^m}{m} $ or $n^m \ge (n+1)^m-m(n+1)^{m-1} =(n+1)^{m-1}(n+1-m) $. or $1 \ge (1+1/n)^{m-1}(1-(m-1)/n) $ or $(1+1/n)^{m-1} \le \frac1{1-(m-1)/n} $.

This requires its own proof. which we will do by induction on $m$.

For $m=1$ this is $1 \le 1$ which is true.

Suppose it is true for $m$ where $m < n-1$. Then $(1+1/n)^{m} =(1+1/n)^{m-1}(1+1/n) \le \frac1{1-(m-1)/n}(1+1/n) $. We want $\frac1{1-(m-1)/n}(1+1/n) \le \frac1{1-m/n} $ or $(1-m/n)(1+1/n) \le 1-(m-1)/n $ or $1-(m-1)/n \ge 1-(m-1)/n+m/n^2 $ which is true.

Therefore, if $m < n-1$, or $n > m+1$, the right side is true.

Answer 4

Following up on my comment from above, I'm setting out to prove that $$ \frac{(n-1)^4}4\leq 1+\cdots +(n-1)^3\leq\frac{n^4}4 $$ for suitable $n$ (say $n\geq 1$). (I feel that this ought to be easier as I only have one long sum of cubes to contend with rather than two.)

The base case is easily shown: $0\leq 0\leq \frac14$.

As for the induction step, assume $k\geq 1$ and that $$ \frac{(k-1)^4}4\leq 1+\cdots +(k-1)^3\leq\frac{k^4}4 $$ Adding $k^3$ to all sides, we get $$ \frac{(k-1)^4 + 4k^3}4\leq 1+\cdots +(k-1)^3 + k^3\leq\frac{k^4+4k^3}4 $$ Now in the numerator on the right-hand side we have $$ k^4 + 4k^3\leq k^4 + 4k^3+6k^2+4k+1 = (k+1)^4 $$ and on the left-hand side we have $$ (k-1)^4 + 4k^3 = k^4 + 6k^2-4k+1 = k^4 + 1 + 2k(3k-2)\geq k^4 $$ (where that last inequality is true because both $1$ and $2k(3k-2)$ are positive).

Gathering it all up, this yields $$ \frac{k^4}4\leq \frac{(k-1)^4 + 4k^3}4\leq 1+\cdots +(k-1)^3 + k^3\leq\frac{k^4+4k^3}4\leq \frac{(k+1)^4}4 $$ and this finishes the induction step.