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Difference between revisions of "Euler-Mascheroni constant"

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===Alternate formulation of the limit===
 
===Alternate formulation of the limit===
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The tangent-line approximation (first-degree [[Taylor polynomial]]) of <math>\ln(k + 1)</math> about <math>x = k</math> is <cmath>\ln(k + 1) = \ln(k) + ((k + 1) - k)\ln'(k) + E_k</cmath> for some error term <math>E_k</math>. Using <math>\ln'(x) = \frac{1}{x}</math> and simplifying, <cmath>\ln(k + 1) = \ln(k) + \frac{1}{k} + E_k.</cmath> Applying the tangent-line formula [[Recursion|recursively]] for all <math>k</math> descending from <math>n</math> to <math>1</math>,
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The tangent-line approximation (first-degree [[Taylor polynomial]]) of <math>\ln(k + 1)</math> about <math>x = k</math> is <cmath>\ln(k + 1) = \ln(k) + ((k + 1) - k)\ln'(k) + E_k</cmath> for some error term <math>E_k</math>. Using <math>\ln'(x) = \frac{1}{x}</math> and simplifying, <cmath>\ln(k + 1) = \ln(k) + \frac{1}{k} + E_k.</cmath> Applying the tangent-line formula [[Recursion|recursively]] for all <math>k</math> descending from <math>n - 1</math> to <math>1</math>,
  
 
<cmath>\begin{align*} \ln(n) &= \ln(n-1) + \frac{1}{n-1} + E_{n-1} \\ &= \left( \ln (n-2) + \frac{1}{n-2} + E_{n-2} \right) + \frac{1}{n - 1} + E_{n-1} \\ &= \dots \\ &= \ln(1) + \left( \sum_{k=1}^{n-1} \frac{1}{k} \right) + \left( \sum_{k=1}^{n-1} E_k \right) . \end{align*}</cmath>
 
<cmath>\begin{align*} \ln(n) &= \ln(n-1) + \frac{1}{n-1} + E_{n-1} \\ &= \left( \ln (n-2) + \frac{1}{n-2} + E_{n-2} \right) + \frac{1}{n - 1} + E_{n-1} \\ &= \dots \\ &= \ln(1) + \left( \sum_{k=1}^{n-1} \frac{1}{k} \right) + \left( \sum_{k=1}^{n-1} E_k \right) . \end{align*}</cmath>
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Hence, <math>\gamma = - \sum_{k=1}^{\infty} E_k</math> is a defined constant.
 
Hence, <math>\gamma = - \sum_{k=1}^{\infty} E_k</math> is a defined constant.
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==See also==
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*[[Harmonic series]]
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*[[Natural logarithm]]
  
 
[[Category: Constants]]
 
[[Category: Constants]]

Latest revision as of 16:32, 19 September 2022

The Euler-Mascheroni constant $\gamma$ is a constant defined by the limit \[\gamma = \lim_{n \rightarrow \infty} \left( \left( \sum_{k=1}^n \frac{1}{k} \right) - \ln(n) \right).\] Its value is approximately \[\gamma = 0.577215 \dots .\]

Whether $\gamma$ is rational or irrational and (if irrational) algebraic or transcendental is an open question.

Proof of existence

Alternate formulation of the limit

The tangent-line approximation (first-degree Taylor polynomial) of $\ln(k + 1)$ about $x = k$ is \[\ln(k + 1) = \ln(k) + ((k + 1) - k)\ln'(k) + E_k\] for some error term $E_k$. Using $\ln'(x) = \frac{1}{x}$ and simplifying, \[\ln(k + 1) = \ln(k) + \frac{1}{k} + E_k.\] Applying the tangent-line formula recursively for all $k$ descending from $n - 1$ to $1$,

\begin{align*} \ln(n) &= \ln(n-1) + \frac{1}{n-1} + E_{n-1} \\ &= \left( \ln (n-2) + \frac{1}{n-2} + E_{n-2} \right) + \frac{1}{n - 1} + E_{n-1} \\ &= \dots \\ &= \ln(1) + \left( \sum_{k=1}^{n-1} \frac{1}{k} \right) + \left( \sum_{k=1}^{n-1} E_k \right) . \end{align*}

Because $\ln(1) = 0$, we may rearrange to \[\left( \sum_{k=1}^{n-1} \frac{1}{k} \right) - \ln(n) = -\sum_{k=1}^{n-1} E_k.\] Adding $\frac{1}{n}$ to both sides yields \[\left( \sum_{k=1}^{n} \frac{1}{k} \right) - \ln(n) = \left( -\sum_{k=1}^{n-1} E_k \right) + \frac{1}{n}.\] Taking the limit as $n$ goes to infinity of both sides,

\begin{align*} \lim_{n \rightarrow \infty} \left( \left( \sum_{k=1}^n \frac{1}{k} \right) - \ln(n) \right) &= \lim_{n \rightarrow \infty} \left( \left( -\sum_{k=1}^{n-1} E_k \right) + \frac{1}{n} \right) \\ &= - \lim_{n \rightarrow \infty} \left( \sum_{k=1}^{n-1} E_k \right) + \lim_{n \rightarrow \infty} \frac{1}{n} \\ &= - \lim_{n \rightarrow \infty} \left( \sum_{k=1}^{n-1} E_k \right) \\ &= - \sum_{k=1}^{\infty} E_k. \end{align*}

Thus, $\gamma = - \sum_{k=1}^{\infty} E_k$.

Convergence of the sum of error terms

We have $\ln''(x) = \left(\frac{1}{x} \right)' = -\frac{1}{x^2}$. For $k \geq 1$, the maximum absolute value of $-\frac{1}{x^2}$ for $x \in [k, k+1]$ is $\frac{1}{k^2}$. Therefore, by the Lagrange Error Bound, \[|E_k| \leq \left| \frac{1^2 \left( \frac{1}{k^2} \right) }{2!} \right| = \frac{1}{2k^2}.\]

The series $\sum_{k=1}^{\infty} \frac{1}{k^2}$ famously converges to $\frac{\pi^2}{6}$ by the Basel problem, so $\sum_{k=1}^{\infty} -\frac{1}{2k^2}$ converges to $-\frac{\pi^2}{12}$ and $\sum_{k=1}^{\infty} \frac{1}{2k^2}$ converges to $\frac{\pi^2}{12}$.

Because $E_k \in \left[ -\frac{1}{2k^2}, \frac{1}{2k^2} \right]$ for all $k$, the Series Comparison Test gives that $\sum_{k=1}^{\infty} E_k$ must converge to a value in $\left[-\frac{\pi^2}{12}, \frac{\pi^2}{12} \right]$.

Hence, $\gamma = - \sum_{k=1}^{\infty} E_k$ is a defined constant.

See also

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