Prove using induction principles

Question

$$\forall{n,a>1}:\;\sum\limits_{k=1}^{2^n-1}\frac{1}{k^a}\;\leq\left(\frac{1-2^{n(1-a)}}{1-2^{1-a}}\right)$$

For any fixed value of $a > 1$.

Induction step:

$$\sum_{k=1}^{2^{n+1} - 1} \frac{1}{k^a} = (\sum\limits_{k=1}^{2^n-1}\frac{1}{k^a}) + \frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + \frac{1}{(2^n + 2)^a} + ... + \frac{1}{(2^{n+1} -1)^a} \leq\left(\frac{1-2^{n(1-a)}}{1-2^{1-a}}\right) + \frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + \frac{1}{(2^n + 2)^a} + ... + \frac{1}{(2^{n+1} -1)^a}$$

I need help from Induction step and on. So if someone would help me, that would be greatly appreciated! People on this website keep putting this problem on hold even though I have clarified it as much as I can.

I need to prove P(n+1) is true:

$$\forall{n,a>1}:\;\sum\limits_{k=1}^{2^{n+1}-1}\frac{1}{k^a}\;\leq\left(\frac{1-2^{(n+1)(1-a)}}{1-2^{1-a}}\right)$$

Answer 1

Hint. $$\sum\limits_{k=2^n}^{2^{n+1}-1} \frac{1}{k^a} \leq \sum\limits_{k=2^n}^{2^{n+1}-1}\frac{1}{2^{na}}=\frac{1}{2^{n(a-1)}}$$ and $$ \frac{1 - 2^{n(1-a)}}{1-2^{1-a}} + \frac{1}{2^{n(a-1)}} = \cdots $$ By the way, a circuitous route might by using the geometric series formula (you certainly don't need to do this, but it's interesting), you know that $$ \sum\limits_{k=0}^{n-1} \frac{1}{2^{k(a-1)}} = \frac{1-2^{n(1-a)}}{1-2^{1-a}} $$ and so $$ \frac{1 - 2^{n(1-a)}}{1-2^{1-a}} + \frac{1}{2^{n(a-1)}} = \sum\limits_{k=0}^{n-1} \frac{1}{2^{k(a-1)}} + \frac{1}{2^{n(a-1)}} = \sum\limits_{k=0}^n \frac{1}{2^{k(a-1)}} = \cdots $$

Answer 2

Ellaborating on the hint:

Notice that $$\frac{1 - 2^{n(1-a)}}{1 - 2^{1-a}} = 1 + 2^{1-a} + \ldots + 2^{(n-1)(1-a)}$$ and that $$\frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + \ldots + \frac{1}{(2^{n+1} - 1)^a} \leq 2^n \cdot \frac{1}{(2^n)^a}.$$

Now $$\sum_{k=1}^{2^{n+1}-1}\frac{1}{k^a} = \sum_{k=1}^{2^{n}-1}\frac{1}{k^a} + \sum_{k=2^n}^{2^{n+1}-1}\frac{1}{k^a}$$ By the induction hypothesis,
\begin{align*} \sum_{k=1}^{2^{n+1}-1}\frac{1}{k^a} &\leq \frac{1 - 2^{n(1-a)}}{1 - 2^{1-a}} + \sum_{k=2^n}^{2^{n+1}-1}\frac{1}{k^a} \\ &\leq 1 + 2^{1-a} + \ldots + 2^{(n-1)(1-a)} + 2^{n(1-a)}. \end{align*} Now what does this final expression evaluate to? It is the sum of a certain geometric progression.