Difference between revisions of "Euler-Mascheroni constant"

From specialfunctionswiki
Jump to: navigation, search
Line 1: Line 1:
 +
__NOTOC__
 
The Euler-Mascheroni constant is the number $\gamma$ defined by the formula
 
The Euler-Mascheroni constant is the number $\gamma$ defined by the formula
 
$$\gamma = \lim_{m \rightarrow \infty} 1 + \dfrac{1}{2} + \ldots + \dfrac{1}{m}-\log(m) = 0.577215664901532 \ldots.$$
 
$$\gamma = \lim_{m \rightarrow \infty} 1 + \dfrac{1}{2} + \ldots + \dfrac{1}{m}-\log(m) = 0.577215664901532 \ldots.$$
  
 
=Properties=
 
=Properties=
−
<div class="toccolours mw-collapsible mw-collapsed">
+
[[The Euler-Mascheroni constant exists]]<br />
−
<strong>Theorem:</strong> The limit defining $\gamma$ exists.
 
−
<div class="mw-collapsible-content">
 
−
<strong>Proof:</strong> Let $u_n=\displaystyle\int_0^1 \dfrac{t}{n(t+n)} dt$. Clearly,
 
−
$$0 < u_n < \displaystyle\int_0^1 \dfrac{1}{n^2} dt = \dfrac{1}{n^2}.$$
 
−
Now compute
 
−
$$\begin{array}{ll}
 
−
u_n &= \dfrac{1}{n} \displaystyle\int_0^1 1 - \dfrac{n}{t+n} dt \\
 
−
&= \dfrac{1}{n} \left[ 1 - n(\log (1+n)-\log n) \right] \\
 
−
&= \dfrac{1}{n} - \log \left(\dfrac{n+1}{n} \right).
 
−
\end{array}$$
 
−
 
 
−
Since $u_n < \dfrac{1}{n^2}$ and we know that $\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k^2}$ converges (it is the [[Riemann zeta]] function evaluated at $z=2$), we can conclude that $\displaystyle\sum_{k=1}^{\infty} u_k$ also converges.
 
−
 
 
−
Notice that due to [[telescoping]] and the properties of the [[logarithm]],
 
−
$$\displaystyle\sum_{k=1}^m \log \left( \dfrac{k+1}{k} \right) = \log \left(\dfrac{2}{1} \right) + \log\left( \dfrac{3}{2} \right) + \ldots \log \left( \dfrac{m+1}{m} \right)= \log(m+1)$$
 
−
Now we see
 
−
$$\begin{array}{ll}
 
−
\displaystyle\sum_{k=1}^{\infty} u_k &= \displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k} - \log \left( \dfrac{k+1}{k} \right) \\
 
−
&=\displaystyle\lim_{m \rightarrow \infty} \left[ \displaystyle\sum_{k=1}^{m} \dfrac{1}{k} - \log(m+1) \right].
 
−
\end{array}$$
 
−
Since $\displaystyle\lim_{m \rightarrow \infty} \log \left( \dfrac{m+1}{m} \right)=0$, we may rewrite it as $\log(m+1)-\log(m)$ and insert it into the above equation to get
 
−
$$\begin{array}{ll}
 
−
\displaystyle\sum_{k=1}^{\infty} u_k &=\displaystyle\lim_{m \rightarrow \infty} \left[ \displaystyle\sum_{k=1}^{m} \dfrac{1}{k} - \log(m+1) + \log(m+1)-\log(m) \right] \\
 
−
&= \displaystyle\lim_{m \rightarrow \infty} \left( 1 + \dfrac{1}{2} + \dfrac{1}{3} + \ldots + \dfrac{1}{m} - \log(m) \right),
 
−
\end{array}$$
 
−
as was to be shown. █
 
−
</div>
 
−
</div>
 
 
[[Reciprocal gamma written as an infinite product]]<br />
 
[[Reciprocal gamma written as an infinite product]]<br />
 
[[Exponential integral Ei series]]<br />
 
[[Exponential integral Ei series]]<br />

Revision as of 20:30, 20 June 2016

The Euler-Mascheroni constant is the number $\gamma$ defined by the formula $$\gamma = \lim_{m \rightarrow \infty} 1 + \dfrac{1}{2} + \ldots + \dfrac{1}{m}-\log(m) = 0.577215664901532 \ldots.$$

Properties

The Euler-Mascheroni constant exists
Reciprocal gamma written as an infinite product
Exponential integral Ei series

Further properties

The Euler-Mascheroni constant appears in the definition of...

  1. the hyperbolic cosine integral
  2. the Barnes G function

See Also

Meissel-Mertens constant

References

Retrieved from "http://specialfunctionswiki.org/index.php?title=Euler-Mascheroni_constant&oldid=6379"