What is the series expression for (1+1/x)^x about x = \infty? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:04:11Z http://mathoverflow.net/feeds/question/77389 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77389/what-is-the-series-expression-for-11-xx-about-x-infty What is the series expression for (1+1/x)^x about x = \infty? Chris 2011-10-06T19:39:21Z 2011-10-06T23:06:23Z <p>This seems like it must have been addressed somewhere already, but I cannot find it in any standard series tables.</p> <p>I have the equation:</p> <p>$f(z) = \left(1 + \frac{1}{z}\right)^z$. </p> <p>What is the general form for the $n$th term of the series? That is, if I have</p> <p>$f(z) \sim \sum_{n=0}^{\infty} \frac{c_n}{z^n}$</p> <p>near $z = \infty$, what is the form of $c_n$?</p> http://mathoverflow.net/questions/77389/what-is-the-series-expression-for-11-xx-about-x-infty/77396#77396 Answer by Robert Israel for What is the series expression for (1+1/x)^x about x = \infty? Robert Israel 2011-10-06T20:23:56Z 2011-10-06T20:23:56Z <p>$\log f(z) = z \log(1+1/z) = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} z^{-k}$ as $z \to +\infty$, so $f(z)$ is the exponential of this sum. See <a href="http://oeis.org/A055505" rel="nofollow">http://oeis.org/A055505</a> for the numerators and <a href="http://oeis.org/A055535" rel="nofollow">http://oeis.org/A055535</a> for the denominators of the coefficients.</p> http://mathoverflow.net/questions/77389/what-is-the-series-expression-for-11-xx-about-x-infty/77397#77397 Answer by Gjergji Zaimi for What is the series expression for (1+1/x)^x about x = \infty? Gjergji Zaimi 2011-10-06T20:24:51Z 2011-10-06T23:06:23Z <p>Markus Brede proves the following formula in the paper <a href="http://www.springerlink.com/content/c1q27x0730v30hth/" rel="nofollow">"On the convergence of the sequence defining Eulerâ€™s number"</a>. Let $$\left(1+\frac{1}{z}\right)^z=\sum_{n\geq 0} \frac{a_n}{z^n}$$ then we have $$a_n=e\sum_{v=0}^n \frac{S(n+v,v)}{(n+v)!}\sum_{m=0}^{n-v}\frac{(-1)^m}{m!}$$ where $S(a,b)$ are Stirling numbers of the first kind. This shows that all coefficients are rational multiples of $e$. I found the article through <a href="http://oeis.org/search?q=1%2C11%2C7%2C2447&amp;sort=&amp;language=english&amp;go=Search" rel="nofollow">OEIS</a>.</p>