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I have to use induction to prove the following inequality:

$\sum_{k=1}^{2^{n}} \frac{1}{k} \geq 1 + \frac{n}{2}$.

So the base case is $n=1$ and it's true because both sides will equal $\frac{3}{2}$.

If we assume $n=m+1$ we get

$\sum_{k=1}^{2^{m+1}} \frac{1}{k} \geq 1 + \frac{m+1}{2}$.

But I'm not sure how to prove that the LHS=RHS.

Is it true to say this:

$\sum_{k=1}^{2^{m}} \frac{1}{k} + \frac{1}{m+1} \geq 1 + \frac{m+1}{2}$?

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    Note that $\sum\limits_{k=1}^{2^{m+1}} \frac{1}{k} = \left(\sum\limits_{k=1}^{2^m}\frac{1}{k}\right)+\left(\sum\limits_{k=2^m+1}^{2^{m+1}}\frac{1}{k}\right)$. Notice what happens... an example: $\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\geq \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}$ – JMoravitz Apr 08 '17 at 23:46
  • @JMoravitz That's a pretty weak duplicate – mrnovice Apr 08 '17 at 23:55
  • @mrnovice the first answer of the linked question gives all of the insight one needs to complete this proof, noting that in the $n$'th "group" there are $2^{n-1}$ terms and each of the terms are greater than $\frac{1}{2^n}$. The only missing step is formalizing the argument with induction and recognizing that the original form of the question here is exactly the same observation made in that answer. – JMoravitz Apr 08 '17 at 23:57

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To finish the proof it is enough to note that

$$\sum\limits_{k=1}^{2^{m+1}} \dfrac{1}{k} = \sum\limits_{k=1}^{2^{m}} \dfrac{1}{k} + \sum\limits_{k=2^{m} + 1}^{2^{m+1}} \dfrac{1}{k}$$

The first part of the sum in the right side is $\geqslant 1+\dfrac{m}{2}$, the other one is $\geqslant \sum\limits_{k=2^{m} + 1}^{2^{m+1}} \dfrac{1}{k} \geqslant 2^m \cdot \dfrac{1}{2^{m + 1}} = \dfrac{1}{2}$, so, overall, $$\sum\limits_{k=1}^{2^{m+1}} \dfrac{1}{k} \geqslant 1+ \dfrac{m + 1}{2}$$ which completes your proof by induction.

Swistack
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  • You have an error. $\sum\limits_{k=2^{m} + 1}^{2^{m+1}} \dfrac{1}{k} < 2^m \cdot \dfrac{1}{2^m}$. The summand you want to replace by is not $\frac{1}{2^m}$ but rather $\frac{1}{2^{m+1}}$ – JMoravitz Apr 08 '17 at 23:53
  • @JMoravitz, good spot, thanks! Fixed now. – Swistack Apr 08 '17 at 23:55