I am working on the same problem as in this question: Inequality proof in Tom Apostol's calculus book (Thm 1.15)
But I am unable to understand how to get to the first step in the recommended answer.
Specifically, it suggests that in trying to prove this inequality:
$$ \sum_{k=1}^{n-1}k^p < \frac{n^{p+1}}{p+1} \tag{1} $$
We need to show:
$$ \frac{n^{p+1}}{p+1}+n^p\leq \frac{(n+1)^{p+1}}{p+1} \tag{2} $$
Through further manipulations, we then arrive at a form that is equivalent to Bernoulli's inequality:
$$ (1+x)^k \gt 1 + kx \tag{3}$$ where $k = p+1$ and $x= \frac{1}{n} $
However, I am completely lost for how I can begin transforming the original inequality towards equation. I've noted that the proof of Bernoulli's inequality itself uses induction for the variable p. Therefore, my thinking is that there are exist either some theorems of which I am unaware that are useful in the beginning manipulations, OR the initial step is to begin with induction for the variable n. In trying the latter, I began by showing the base case n=1 is valid:
$$ \sum_{k=1}^{0}k^p = 0 < \frac{1}{p+1} \tag{4} $$
so in assuming the nth case, we try to show that the statement holds for n+1:
$$ \sum_{k=1}^{n}k^p < \frac{(n+1)^{p+1}}{p+1} \tag{5} $$
From here, there needs to be some way of converting the summation into a form that uses an n or n+1 base, but I am stuck on this. Online, I've found that Faulhaber's formula says:
$$ \sum_{k=1}^{n}k^p = \frac{1}{p+1}\sum_{k=0}^{p} \binom{p+1}{k} B_k n^{p-k+1} $$
where $B_k$ are the Bernoulli numbers. However, these haven't been introduced in the book yet so I think there should be another way