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I am familiar with the proof using Bertrand's Postulate as well as the proofs highlighted in this discussion:

Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer?

However, I wanted to try my hand at proving $\sum_{k=1}^n \frac{1}{k}$ is never an integer using a different approach. However, I think I may have hit a dead end. Is this approach leading me nowhere?

Proof so far:

The integral $\int_1^n \frac{1}{x}dx$ is bounded by the sums $\sum_{k=2}^n \frac{1}{k}$ and $\sum_{k=1}^{n-1} \frac{1}{k}$ like so:

$$\sum_{k=2}^n \frac{1}{k} < \int_1^n \frac{1}{x}dx < \sum_{k=1}^{n-1} \frac{1}{k}$$ (for n $\ge$ 2)

Evaluating the integral gives:

$$\sum_{k=2}^n \frac{1}{k} < \ln(n) < \sum_{k=1}^{n-1} \frac{1}{k}$$

Noting: $$\sum_{k=2}^n \frac{1}{k} = \sum_{k=1}^n \frac{1}{k} - 1$$ and $$\sum_{k=1}^{n-1} \frac{1}{k} = \sum_{k=1}^n \frac{1}{k} - \frac{1}{n}$$

We rewrite the inequality as:

$$\sum_{k=1}^n \frac{1}{k} - 1 < \ln(n) < \sum_{k=1}^n \frac{1}{k} - \frac{1}{n}$$

Subtracting the sum from the inquality gives:

$$- 1 < \ln(n) - \sum_{k=1}^n \frac{1}{k} < \frac{1}{n}$$

Next multiply by -1:

$$\frac{1}{n} < \sum_{k=1}^n \frac{1}{k} - \ln(n) < 1$$

and add the $\ln(n)$ term:

$$\frac{1}{n} + \ln(n) \le \sum_{k=1}^n \frac{1}{k} \le 1 + \ln(n)$$

for n $\ge$ 1

The range of this inequality is:

$$ 1+\ln(n) - (\frac{1}{n} + ln(n)) = 1 - \frac{1}{n}$$

Since the bottom bound is irrational and the range is less than 1, there is at most 1 integer between $\frac{1}{n} + \ln(n)$ and $1 + \ln(n)$

However, this is where I get stuck. Is there some way I can reach the correct conclusion from here, or is this attempt for naught?

Community
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user373763
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  • I don't think you can prove this using bounds, especially when the partial sums can get arbitrarily close to an integer. – Alexis Olson Sep 30 '16 at 05:37
  • @alexis, I see. I also find it interesting that as n goes to infinity, the range of the inequality goes to 1. – user373763 Sep 30 '16 at 05:40

1 Answers1

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This approach probably won't work.

For example $2 \in [1/4 + \ln(4), 1 + \ln(4)]$. In fact the fractional part of $ln(n)$ is dense on $[0,1]$, so there are arbitrarily large $n$ such that $[1/n + \ln(n), 1 + \ln(n)]$ contain an integer.

Nick R
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  • Oh that is a pity. I thought that I could somehow find the integer which was the floor of ln(n) + 1 and prove that the harmonic series was not equal to that specific integer, but maybe this is not so. – user373763 Sep 30 '16 at 06:02