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I need to prove that $$\frac{1}{\left(2^k\right)^a}+\frac{1}{\left(2^k+1\right)^a}+\frac{1}{\left(2^k+2\right)^a}+\dots+\frac{1}{\left(2^{k+1}-1\right)^a}\le \left(\frac{1}{2^{a-1}}\right)^k$$ for $k\ge1, a \gt 1$.

I get stuck after the induction step $$\frac{1}{\left(2^{k+1}\right)^a}+\frac{1}{\left(2^{k+1}+1\right)^a}+\frac{1}{\left(2^{k+1}+2\right)^a}+\dots+\frac{1}{\left(2^{k+2}-1\right)^a}\le \left(\frac{1}{2^{a-1}}\right)^{k+1}$$

McLovin
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  • It's not clear to me what exactly is the full sequence, I dont' understand the oart aftrer the dots $$\frac{1}{\left(2^{k+1}-1\right)^a}$$ which seems to be not correlated to the previous. – user Dec 08 '17 at 17:48
  • The sequence is meant to be $2^k, 2^k + 1, \dots, 2^k + 2^k - 1 = 2^{k+1} - 1$. – Qi Zhu Dec 08 '17 at 17:56

1 Answers1

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Do you have to prove it with induction? Observe $$ \frac{1}{\left(2^k \right)^a} + \dots + \frac{1}{\left(2^{k+1} - 1 \right)^a} \leq \underbrace{\frac{1}{\left(2^k \right)^a} + \frac{1}{\left(2^k \right)^a} + \dots + \frac{1}{\left(2^k \right)^a}}_{2^k \text{ times}} = 2^k \cdot \frac{1}{\left(2^k \right)^a} = \left( \frac{1}{2^{a-1}} \right)^k.$$

Qi Zhu
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  • Actually I don't. Thanks! – McLovin Dec 08 '17 at 17:50
  • Come to think of it.. why is it $2^k$ times and not $k$ times? – McLovin Dec 08 '17 at 18:18
  • Look at the comment I posted below your question, the sequence is $2^k, 2^k+1, \dots, 2^k+2^k - 1$. You add $0, 1, 2, 3, 4, \dots, 2^k - 1$ to $2^k$. It should be clear now that we have $2^k$ numbers. – Qi Zhu Dec 08 '17 at 18:39