From working on a problem I was lead to consider the function $\frac{T_n(n)}{e^n}$ where $T_n(x)$ is the $n$'th order Taylor polynomial of $e^x$.

Numerical evidence suggest that

$$\lim_{n\to \infty} \frac{T_n(n)}{e^n} \equiv\lim_{n\to \infty} \frac{\sum_{k=0}^n\frac{n^k}{k!}}{\sum_{k=0}^\infty\frac{n^k}{k!}} = \frac{1}{2}$$

Is there a nice proof for this statement? More generally: is there a 'standard' approach for evaluating limits on the form $\lim_{n\to\infty}\frac{f_n(x_n)}{f(x_n)}$ where $f_n$ is a series converging (uniformly) to $f$ and where $x_n$ is an unbounded sequence? I would also apprechiate refs. to similar questions on this site or in the literature (I could only find this one).

  • 2
    $\begingroup$ The integral form of the remainder might help. $1 - T_n(n)/e^n = \frac{n^{n+1}}{n!e^n}\int_0^1(1-t)^{n-1}e^{tn}dt$. Maybe $1/2$ pops out after taking the limit using Stirling's approximation for $n!$. $\endgroup$ – RRL Feb 26 '15 at 6:39
  • $\begingroup$ Thanks for the suggestion RLL. @Travis thats just what I was looking for, perfect! btw should I just delete it or vote duplicate? $\endgroup$ – Winther Feb 26 '15 at 6:51
  • 1
    $\begingroup$ This question has been asked a few times, though it's very much the sort of question that's hard to search for I think. One nice sketch of a proof is that the expression in the limit is the simply the sum of the probabilities that a given random Poisson variable with mean $n$ has value $\leq n$. But in the limit as $n \to \infty$, the Poisson distribution converges to a normal distribution; such a distribution is symmetric about its mean, so the limit must tends to $\frac{1}{2}$. $\endgroup$ – Travis Willse Feb 26 '15 at 6:55
  • $\begingroup$ @Henrik You're welcome, I'm glad you found my comment useful. Anyway, it's already attracted some attention and a good answer, and I'd anyway just vote to close (rather than deleting) for the latter reason alone. $\endgroup$ – Travis Willse Feb 26 '15 at 6:56

Look here: A limit involves series and factorials

That answer links to here: http://journals.cambridge.org/download.php?file=%2FPEM%2FPEM2_24_03%2FS0013091500016503a.pdf&code=fd828d6902ca6a380244640216120c97

This has a result of (who else) Ramanujan where he proved (in S. RAMANUJAN, J. Ind. Math. Soc. 3 (1911), 128; ibid. 4 (1911), 151-152; Collected Papers (Chelsea, New York; 1962), 323-324) that

$$e^n/2 = \sum_{k=0}^{n-1} n^k/k! + (n^n/n!) r(n)$$

where, for large $n$, $r(n) \approx 1/3 + 4/(135n) + O(1/n^2)$.

  • $\begingroup$ Thank you Marty, that is just what I was looking for! $\endgroup$ – Winther Feb 26 '15 at 6:53
  • 1
    $\begingroup$ Yup, most things asked here were done by either Euler or Ramanujan. The advantage of it being done by Ramanujan is that his collected works are much shorter and in English. $\endgroup$ – marty cohen Feb 26 '15 at 6:58

Not the answer you're looking for? Browse other questions tagged or ask your own question.