# Simple proof Euler–Mascheroni $\gamma$ constant

I'm searching for a really simple and beautiful proof that the sequence $(u_n)_{n \in \mathbb{N}} = \sum\nolimits_{k=1}^n \frac{1}{k} - \log(n)$ converges.
At first I want to know if my answer is OK.

My try:
$\lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} - \log (n) = \lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} + \sum\limits_{k=1}^{n-1} \log(k)-\log(k+1) = \lim\limits_{n\to\infty} \frac{1}{n} + \sum\limits_{k=1}^{n-1} \log(\frac{k}{k+1})+\frac{1}{k}$

$= \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$
Now we prove that the last sum converges by the comparison test:
$\frac{1}{k}-\log(\frac{k+1}{k}) < \frac{1}{k^2} \Leftrightarrow k<k^2\log(\frac{k+1}{k})+1$
which surely holds for $k\geqslant 1$

As $\sum\limits_{k=1}^{\infty} \frac{1}{k^2}$ converges $\Rightarrow \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$ converges and we name this limit $\gamma$
q.e.d

• Hint: look at the function $y=1/x$ compared to "step functions" above and below that curve, and look at what happens when you integrate. – Old John Jan 6 '14 at 23:59
• math.stackexchange.com/questions/306371/… – Will Jagy Jan 7 '14 at 0:01
• the diagrams are not very good, but you can get the idea here. – Old John Jan 7 '14 at 0:05
• @OldJohn, I see, actually a link to a Macalester College discussion of the harmonic series. Natural mistake. – Will Jagy Jan 7 '14 at 0:09
• @OldJohn, true, never met her. There is a running joke on a TV series called NCIS, in which FBI agent Fornell married a woman whom NCIS agent Gibbs had divorced. At one point Fornell says "In my defense, I didn't believe him." – Will Jagy Jan 7 '14 at 0:28

## 2 Answers

One elegant way to show that the sequence converges is to show that it's both decreasing and bounded below.

It's decreasing because $u_n-u_{n-1} = \frac1n - \log n + \log(n-1) = \frac1n + \log(1-\frac1n) < 0$ for all $n$. (The inequality is valid because $\log(1-x)$ is a concave function, hence lies beneath the line $-x$ that is tangent to its graph at $0$; plugging in $x=\frac1n$ yields $\log(1-\frac1n) \le -\frac1n$.)

It's bounded below because $$\sum_{j=1}^n \frac1j > \int_1^{n+1} \frac{dt}t = \log (n+1) > \log n,$$ and so $u_n>0$ for all $n$. (The inequality is valid because the sum is a left-hand endpoint Riemann sum for the integral, and the function $\frac1t$ is decreasing.)

Upper Bound

Note that \begin{align} \frac1n-\log\left(\frac{n+1}n\right) &=\int_0^{1/n}\frac{t\,\mathrm{d}t}{1+t}\\ &\le\int_0^{1/n}t\,\mathrm{d}t\\[3pt] &=\frac1{2n^2} \end{align} Therefore, \begin{align} \gamma &=\sum_{n=1}^\infty\left(\frac1n-\log\left(\frac{n+1}n\right)\right)\\ &\le\sum_{n=1}^\infty\frac1{2n^2}\\ &\le\sum_{n=1}^\infty\frac1{2n^2-\frac12}\\ &=\sum_{n=1}^\infty\frac12\left(\frac1{n-\frac12}-\frac1{n+\frac12}\right)\\[9pt] &=1 \end{align}

Lower Bound

Note that \begin{align} \frac1n-\log\left(\frac{n+1}n\right) &=\int_0^{1/n}\frac{t\,\mathrm{d}t}{1+t}\\ &\ge\int_0^{1/n}\frac{t}{1+\frac1n}\,\mathrm{d}t\\[3pt] &=\frac1{2n(n+1)} \end{align} Therefore, \begin{align} \gamma &=\sum_{n=1}^\infty\left(\frac1n-\log\left(\frac{n+1}n\right)\right)\\ &\ge\sum_{n=1}^\infty\frac1{2n(n+1)}\\[3pt] &=\sum_{n=1}^\infty\frac12\left(\frac1n-\frac1{n+1}\right)\\[6pt] &=\frac12 \end{align}

A Better Upper Bound

Using Jensen's Inequality on the concave $$\frac{t}{1+t}$$, we get \begin{align} \frac1n-\log\left(\frac{n+1}n\right) &=\frac1n\left(n\int_0^{1/n}\frac{t\,\mathrm{d}t}{1+t}\right)\\ &\le\frac1n\frac{n\int_0^{1/n}t\,\mathrm{d}t}{1+n\int_0^{1/n}t\,\mathrm{d}t}\\ &=\frac1{n(2n+1)} \end{align} Therefore, since the sum of the Alternating Harmonic Series is $$\log(2)$$, \begin{align} \gamma &=\sum_{n=1}^\infty\left(\frac1n-\log\left(\frac{n+1}n\right)\right)\\ &\le\sum_{n=1}^\infty\frac1{n(2n+1)}\\ &=\sum_{n=1}^\infty2\left(\frac1{2n}-\frac1{2n+1}\right)\\[6pt] &=2(1-\log(2)) \end{align}

• can you copy the last part of your answer here ? – Gabriel Romon Nov 10 '17 at 9:52