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Find the least Positive Integer Satisfying $1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} \ge 4$

My try:

I actually applied a Brute Force taking:

$$\int \frac{dx}{x} \ge 4$$ Then we get: $$\ln x \ge 4$$ Hence $$x \ge e^4$$

So approximately $n=55$

Is there any better approach?

Umesh shankar
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    A quick code shows that $ n \ge 31$ satisfies the conditions. – The Demonix _ Hermit Oct 21 '19 at 16:36
  • Maybe use the trapezoidal rule to get $$\sum_{k=1}^n\frac{1}{k}-\frac{1}{2}-\frac{1}{2n}\approx \ln n$$ or something like that? At least this means you are looking for $n$ such that $\ln n\approx 3.5$. This gives $n\approx 33$. I don't know how to improve this yet. – Batominovski Oct 21 '19 at 16:37

1 Answers1

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This is a slight variation from Peter's answer. Since at small $n$ we cannot use the form
$$1+\frac12+\frac13+\dots+\frac1n=\gamma+\ln{(n)}+O\left(\frac1n\right)$$ directly unless we can control the error term hence we need explicit bounds on the estimate before approximating the series with $\log n$ and $\gamma$. We have

$$ \log(n+1/2)+\frac{1}{24n^2}-\frac{1}{24n^3}\leq H_n-\gamma \leq \log(n+1/2)+\frac{1}{24n^2} $$

Now we can replace $H_n$ with $4$ and use the upper bound to get the minimum lower integer bound as $n_{\min} = 31$.

Nilotpal Sinha
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    I get what you mean but how exactly are you solving the inequality$$\log(n+1/2)+\frac{1}{24n^2}-\frac{1}{24n^3}\leq 4-\gamma \leq \log(n+1/2)+\frac{1}{24n^2}$$ – Peter Foreman Oct 21 '19 at 17:08
  • If you mean to prove the value of $n$ must be an integer $\ge31$ then it is sufficient to use$$\gamma+\ln{(n)}\lt1+\frac12+\frac13+\dots+\frac1n\lt\gamma+\ln{(n)}+\frac1{2n}$$to get that $n_{min}=31$. – Peter Foreman Oct 21 '19 at 17:17