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Is there a simple way to show this bound for Euler’s constant:

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

For proving convergence, I showed that $\sum_{k=1}^n 1/k - \ln(n)$ is bounded and monotonic, using the property

$$\frac{1}{k+1} < \int_k^{k+1} x^{-1} dx <\frac{1}{k}$$

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

But this does not help prove the error bounds.

Thanks

scobaco
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1 Answers1

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Use convexity.

Since $f(x) = 1/x$ is both decreasing and convex we have for $k \leqslant x \leqslant k+1$,

$$\frac{f(k+1) - f(k+2)}{k + 2 - (k+1)} \leqslant \frac{f(x) - f(k+1)}{k + 1 - x} \leqslant \frac{f(k) - f(k+1)}{k + 1 - k} $$

Hence,

$$\tag{*}\left(\frac{1}{k+1} - \frac{1}{k+2}\right)(k + 1 - x) \leqslant \frac{1}{x}- \frac{1}{k+1} \leqslant \left(\frac{1}{k} - \frac{1}{k+1} \right) (k+1 - x)$$

Note that

$$\int_{k}^{k+1}(k + 1 - x) \, dx = \left.-\frac{1}{2} (k+1 -x)^2\right|_k^{k+1} = \frac{1}{2}$$

Integrating (*) over $[k,k+1]$ and summing from $k = n$ to $N-1$ we get

$$\frac{1}{2(n+1)} - \frac{1}{2(N+1)}\leqslant \sum_{k=n}^{N-1} \int_{k}^{k+1} \frac{dx}{x}- \sum_{k=n}^{N-1} \frac{1}{k+1} \leqslant \frac{1}{2n} - \frac{1}{2N},$$

which reduces to

$$\frac{1}{2(n+1)} - \frac{1}{2(N+1)}\leqslant \sum_{k=1}^n\frac{1}{k} -\log n - \left(\sum_{k=1}^N \frac{1}{k}- \log N\right) \leqslant \frac{1}{2n} - \frac{1}{2N}$$

Then take the limit as $N \to \infty$.

RRL
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  • I don't understand your first inequality. It seems should be $k+1 \le x \le k+2$ – Mariana Apr 24 '21 at 14:15
  • @Mariana: If $f$ is a convex function and $a < b < c$, then $$ \frac{f(b) - f(a)}{b-a} \leqslant \frac{f(c) - f(a)}{c-a} \leqslant \frac{f(c) - f(b)}{c-b}$$. – RRL Apr 24 '21 at 16:05
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    Use that inequality with $k < x < k+1 < k+2$. We get $ \frac{f(k+1) - f(x)}{k+1 - x} \leqslant \frac{f(k+2) - f(k+1)}{k+2 - (k+1)}$ which implies $\frac{f(k+1) - f(k+2)}{k+2 - (k+1)} \leqslant \frac{f(x) - f(k+1)}{k+1 - x} $. Proceed in a similar way for the second inequality in the answer. – RRL Apr 24 '21 at 16:11
  • Thank you so much! – Mariana Apr 24 '21 at 17:09