Here's a question that I've come across: Prove by induction that for every integer $n$ greater than or equal to $1$,

$${\sum_{i=1}^{2^n}} \frac{1}{i} \ge 1 +\frac{n}{2}.$$

Now I know how to prove by induction, but wouldn't this fail $p(1)$ since

$$\frac{1}{1} \ge 1 +\frac{1}{2}$$ is not true?

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    $\begingroup$ Note that $2^1=2$. $\endgroup$ – Balloon Jan 11 '17 at 23:02
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    $\begingroup$ The case $n=1$ is $$\sum_{i=1}^2\frac1i=\frac11+\frac12\ge\cdots$$ $\endgroup$ – user228113 Jan 11 '17 at 23:02

Check the upper bound of the summation;

If $n = 1$, then we have

$$\sum_{i=1}^2 \frac1i = 1 + \frac12 \geq 1 + \frac12$$

which is definitely true.

Given the discussion in the comments, I expand my answer to include the induction proof as I would do it:


We already have the base case for $n = 1$; Assume it is true up to $k$. We show it is true for $k + 1$.

The summation becomes

$$\sum_{i=1}^{2^{k+1}} \frac1i = \sum_{i=1}^{2^k} \frac1i + \sum_{i=2^k+1}^{2^{k+1}} \frac1i \geq 1 + \frac{k}{2} + \sum_{i=2^k+1}^{2^{k+1}} \frac1i$$

which holds because of the induction step. We want to check if

$$1 + \frac{k}{2} + \sum_{i=2^k+1}^{2^{k+1}} \frac1k \geq 1 + \frac{k+1}{2} = 1 + \frac{k}{2} + \frac12$$

it suffices to show that

$$\sum_{i=2^k+1}^{2^{k+1}} \frac1i \geq \frac12$$

which is, indeed, true. Notice that every term on that series is greater than, or equal to, $\frac1{2^{k+1}}$:

$$\sum_{i=2^k+1}^{2^{k+1}} \frac1i \geq \sum_{i=2^k+1}^{2^{k+1}} \frac1{2^{k+1}} = 2^k\cdot \frac1{2^{k+1}} = \frac12$$

which concludes the proof.

  • $\begingroup$ Now I'm curious how this is going to end. During the induction step, does the following reasoning hold: ${\sum\limits_{i=1}^{2^{n+1}}} \frac{1}{i} \ge 1 +\frac{n}{2}+\frac{1}{2^{n+1}}\geq 1+\frac{n}{2}$ $\endgroup$ – emka Jan 11 '17 at 23:30
  • $\begingroup$ I don't think you wrote the inequality you want to prove! check that ;) $\endgroup$ – RGS Jan 11 '17 at 23:33
  • $\begingroup$ I got ya, there should be an $n+1$ in that middle piece. $\endgroup$ – emka Jan 12 '17 at 0:05
  • $\begingroup$ @emka indeed; did you manage to finish it? $\endgroup$ – RGS Jan 12 '17 at 0:29
  • $\begingroup$ I'm wondering if you did as well $\endgroup$ – Shawn Javadi Jan 12 '17 at 0:30

Neither: the question is not wrong and you are also not going insane ... You just made a little mistake. :)

  • $\begingroup$ It was a binary question; are you sure we should disregard the Law of the Excluded Middle? ;) $\endgroup$ – Wildcard Jan 12 '17 at 3:09
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    $\begingroup$ @Wildcard Well, either we disregard it or we don't. :) $\endgroup$ – Bram28 Jan 12 '17 at 3:23

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