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