Proof by induction: $\sum_{j=1}^{n-1}{j^k}<\frac{n^{k+1}}{k+1}$ with $n,k \in \mathbb{N}$ and $n\geq 2$

Question

Proof by induction: $\sum_{j=1}^{n-1}{j^k}<\frac{n^{k+1}}{k+1}$ with $n,k \in \mathbb{N}$ and $n\geq 2$

Inductive step: $$\sum_{j=1}^{n}{j^k}<\frac{(n+1)^{k+1}}{k+1}$$$$\Bigg(\sum_{j=1}^{n-1} j^k \Bigg)+n^k<\frac{(n+1)^{k+1}}{k+1}$$$$\frac{(n+1)^{k+1}}{k+1}+n^k<\frac{(n+1)^{k+1}}{k+1}$$ How can I continue???

Answer 1

Actually, you made a small error, when you replaced the sum from the second line to the third. The reasoning should be the following, I'll let you find your error:

We have: $$\sum_{j=1}^{n} j^k = \sum_{j=1}^{n-1} j^k + n_k <\frac{n^{k+1}}{k+1} + n^k$$

We want: $$ \frac{n^{k+1}}{k+1} + n^k < \frac{(n+1)^{k+1}}{k+1} \\ \leftrightarrow n^k < \frac{(n+1)^{k+1} - n^{k+1}}{k+1}$$

and by expanding the exponent, we get:

$$ (n+1)^{k+1} = \sum_{i=0}^{k+1} \pmatrix{k+1\\i}n^i = n^{k+1} + (k+1) n^k + \sum_{i=0}^{k-1} \pmatrix{k+1\\i}n^i $$

Therefore:

$$\frac{(n+1)^{k+1} - n^{k+1}}{k+1} = n^k + \frac{1}{k+1} \sum_{i=0}^{k-1} \pmatrix{k+1\\i}n^i > n^k$$

And it's done!

Answer 2

Final line should have $n^{k+1}$ as numerator not $n+1$. After that you should check $n^{k+1}+n^k(k+1) < (n+1)^{k+1}$. If you binomially expand the right hand side you can see the left hand side is like the final 2 terms in the series. The inequality follows