Upper bound euler constant

Question

Trying to show that the euler constant $\gamma \leq \frac{5}{6}$ .

I'm stuck in this, I was able to bound it with 1 after manipulation with integrals but it doesn't good enough.

then I tried this

$\gamma = \lim _{n\to \infty }\left(\sum _{k=1}^n\frac{1}{k}\:-ln\left(n\right)\right) =\lim _{n\to \infty }\left(\sum _{k=1}^n\frac{1}{k}-\sum _{k=2}^n\ln\left(k\right)-ln\left(k-1\right)\right)$ $= \lim _{n\to \infty }\left(1+\sum _{k=2}^n\frac{1}{k}-\ln\left(\frac{k}{k-1}\right)\right)$

from here I tried to apply taylor series instead of $ln(\frac{k}{k-1})$ but again this doesn't lead anywhere, how can I bound it from above?

Answer 1

One has the series expansion

$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \quad \text{for $\ x\in(-1, 1]$.}$$

Since your limit is equal to the following,

$$\lim_{n \to \infty}{\left( \sum_{k=1}^n{\frac{1}{k}} \ - \ \sum_{k=2}^n{\ln\left(\frac{k}{k-1}\right)} \right)}$$

$$= \lim_{n \to \infty}{\left( \sum_{k=1}^n{\frac{1}{k}} \ - \ \sum_{k=1}^{n-1}{\ln\left(1 + \frac{1}{k}\right)} \right)}$$

we have that

$$\gamma \ = \ \lim_{n \to \infty}{\left( \sum_{k=1}^n{\frac{1}{k}} \ - \ \sum_{k=1}^{n-1}{\left(\frac{1}{k} - \frac{1}{2k^2} + \frac{1}{3k^3} - \frac{1}{4k^4} + \dots\right)} \right)}$$

$$= \lim_{n \to \infty}{\left( \sum_{k=1}^n{\frac{1}{k}} \ - \ \sum_{k=1}^{n-1}{\frac{1}{k}}\right)} \ + \ \lim_{n\to\infty}{\left(\sum_{k=1}^{n-1}{\left(\frac{1}{2k^2} - \frac{1}{3k^3} + \frac{1}{4k^4} - \dots\right)} \right)}$$

$$= \lim_{n \to \infty}{\left( \frac{1}{n} \right)} \ + \ \sum_{k=1}^{\infty}{\left(\frac{1}{2k^2} - \frac{1}{3k^3} + \frac{1}{4k^4} - \dots\right)}$$

$$= \sum_{k=1}^{\infty}{\left(\frac{1}{2k^2} - \frac{1}{3k^3} + \frac{1}{4k^4} - \dots\right)}$$

Hence, one has the bound

$$\gamma \ \lt \ \sum_{k=1}^{\infty}{\left(\frac{1}{2k^2} - \frac{1}{3k^3} + \frac{1}{4k^4}\right)} \ \lt \ \sum_{k=1}^{\infty}{\left(\frac{1}{2k^2} - \frac{1}{4k^3} + \frac{1}{4k^4}\right)}$$

$$= \ \frac{1}{2}\sum_{k=1}^{\infty}{\frac{1}{k^2}} \ - \ \frac{1}{4}\sum_{k=2}^{\infty}{\left(\frac{1}{k^3} - \frac{1}{k^4}\right)}$$

$$\lt \ \frac{1}{2}\sum_{k=1}^{\infty}{\frac{1}{k^2}} - \frac{1}{4}\cdot\left(\frac{1}{2^3} \ - \ \frac{1}{2^4}\right) \ = \ \frac{1}{2}\sum_{k=1}^{\infty}{\frac{1}{k^2}} - \ \frac{1}{64}.$$

To bound the summation term, you can do the following:

$$\sum_{k=1}^{\infty}{\frac{1}{k^2}} = \ \left(1 + \frac{1}{4} + \frac{1}{9} + \ \sum_{k=4}^{\infty}{\frac{1}{k^2}}\right)$$

$$= \ \left(\frac{49}{36} + \ \sum_{k=4}^{\infty}{\frac{1}{k^2}}\right) \ \lt \ \left(\frac{49}{36} \ + \ \sum_{k=4}^{\infty}{\frac{1}{k(k-1)}}\right)$$

$$= \ \frac{49}{36} \ + \ \sum_{k=4}^{\infty}{\left(\frac{1}{k-1} - \frac{1}{k}\right)}$$

$$= \ \left(\frac{49}{36} + \frac{1}{3}\right) = \frac{61}{36}$$

Finally, we have

$$\gamma \ \lt \ \frac{1}{2}\cdot\frac{61}{36} - \ \frac{1}{64} = \frac{479}{576}.$$

Indeed, integration would be an easier and much cleaner way to solve this problem, but the algebraic method still gives the desired result.

Answer 2

if we have $g(x) > 0$ but $g'(x) < 0,$ then $$ \int_n^{n+1} \; g(x) \; dx \; < \; g(n) \; < \; \int_{n-1}^n \; g(x) \; dx $$

Here is a diagram

Sum this up from $n=a$ to $n=b.$ $$ \int_a^{b+1} \; g(x) \; dx \; < \; \sum_{n=a}^b \; g(n) \; < \; \int_{a-1}^b \; g(x) \; dx $$

With $g(x) = \log x$ we have

$$ \log(b+1) - \log a < \; \sum_{n=a}^b \; \frac{1}{n} \; < \; \log b - \log(a-1) $$
Clearly we need to take $a>1.$ If we take $a=3$ the accuracy is good enough for your bound; Here we imagine $b$ a large integer.

$$ \log(b+1) - \log 3 < \; \sum_{n=3}^b \; \frac{1}{n} \; < \; \log b - \log 2 $$ and so

$$ \frac{3}{2} + \log(b+1) - \log 3 < \; \sum_{n=1}^b \; \frac{1}{n} \; < \; \frac{3}{2} + \log b - \log 2 $$

Subtract $\log b$ to find

$$ \frac{3}{2} + \log(1 + \frac{1}{b}) - \log 3 < \; \left(\sum_{n=1}^b \; \frac{1}{n} \right) - \log b\; < \; \frac{3}{2} - \log 2 $$

Letting $b$ grow without bound we find $$ \frac{3}{2} - \log 3 < \; \gamma \; < \; \frac{3}{2} - \log 2 $$

If we had taken $a=4$ we would have found
$$ \frac{11}{6} - \log 4 < \; \gamma \; < \; \frac{11}{6} - \log 3 $$

It comes out $$ H_n - \log (1+n) < \; \gamma \; < \; H_n - \log n $$ which is another satisfying approach: The expression $ H_n - \log (1+n)$ starts low and increases as $n$ increases. The expression $ H_n - \log n$ starts high and decreases as $n$ increases. They crash together at $\gamma.$ See What are "suites adjacentes" called in English?

Answer 3

The function $x \mapsto 1/x$ is convex and decreasing. Hence, for $k \leqslant x \leqslant k+1$ where $k \in \mathbb{N}$ we have $$\frac{\frac{1}{k+1}-\frac{1}{x}}{k+1 - x}\leqslant \frac{\frac{1}{k+2}-\frac{1}{k+1}}{k+2 - k+1}$$

and, thus,

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

Integrating over $[k,k+1]$ we get

$$\frac{1}{k+1} - \log \frac{k+1}{k} \leqslant \left(\frac{1}{k+2}-\frac{1}{k+1} \right)\cdot \frac{1}{2}$$

Summing from $1$ to $m-1$ we get

$$\sum_{k=1}^m \frac{1}{k} - \log m -1 \leqslant \frac{1}{2m+2} - \frac{1}{4}$$

Taking the limit of both sides as $m \to \infty$ yields

$$\gamma - 1 \leqslant -\frac{1}{4} \implies \gamma \leqslant 0.75 < \frac{5}{6}$$