It is known around here that $$\lim_{n\to\infty }e^{-n}\left(1+n+\frac{n^2}{2!}+…+\frac{n^n}{n!}\right)=\frac12$$ So what about the following limit:
$$L=\lim_{n \to \infty}n \left(\frac{1}{2}-\frac{1}{e^n}\left(1+n+\frac{n^2}{2!}+…+\frac{n^n}{n!}\right)\right)$$ My thought was to apply Stolz-Cesaro theorem, after rewritting as $$L=\lim_{n \to \infty}\frac{ \left(\frac{1}{2}-\frac{1}{e^n}\left(1+n+\frac{n^2}{2!}+…+\frac{n^n}{n!}\right)\right)}{\frac1n}$$ would produce: $$L=\lim_{n \to \infty}\frac{e^{-(n+1)}\left(1+(n+1)+\frac{(n+1)^2}{2!}+…+\frac{(n+1)^{n+1}}{(n+1)!}\right)-e^{-n}\left(1+n+\frac{n^2}{2!}+…+\frac{n^n}{n!}\right)}{\frac1{n(n+1)}}$$
$$=\lim_{n \to \infty}\frac{e^{-n}\left(\frac{1-e}{e} +\frac1e\frac{(n+1)^{n+1}}{(n+1)!} +n\left(\frac{\left(1+\frac1n\right)}{e}-1\right) +\frac{n^2}{2!}\left(\frac{\left(1+\frac1n\right)^2}{e}-1\right)+\cdots +\frac{n^n}{n!}\left(\frac{\left(1+\frac1n\right)^n}{e}-1\right)\right)}{\frac1{n(n+1)}}$$ But I dont see how can I go further, can you help me evaluate it? Also, are limits of this form already known?
Limit of $n \left(\frac{1}{2}-e^{-n}\left(1+n+\frac{n^2}{2!}+…+\frac{n^n}{n!}\right)\right)$
exponential functionlimits
Related Solutions
The limit of the ratio in your title is nonzero. Give me a few minutes to type up the following, based on your funtion $g(x) = x^{-1/3}$
if we have $g(x) > 0$ but $g'(x) < 0,$ then $$ \int_a^{b+1} \; g(x) \; dx \; < \; \sum_{j=a}^b \; g(j) \; < \; \int_{a-1}^b \; g(x) \; dx $$
Here is a drawing I made, using the letter $f$ rather than $g$
Well, $g$ is integrable at the origin. Let's try $a=1.$ If that is not satisfactory we can just switch to $a=2$ by putting in some extra terms.
$$ \int_1^{n+1} \; x^{-1/3} \; dx \; < \; \sum_{j=a}^b \; j^{-1/3} \; < \; \int_{0}^n \; x^{-1/3} \; dx $$
an antiderivative of $g$ is $G(x) = \frac{3}{2} x^{2/3}$
$$ \frac{3}{2} \left( (n+1)^{2/3} - 1 \right) \; < \; \sum_{j=1}^n \; j^{-1/3} \; < \; \frac{3}{2} n^{2/3} $$
Good enough. Your denominator is simply $ n^{2/3}.$ We see that $$ L = \frac{3}{2} $$
As pointed out by other users, you computed the limit incorrectly. Note that
\begin{align*} \lim_{n \to \infty} n \cdot \frac{1\cdot n + 2 \cdot (n-1) + \cdots + n \cdot 1}{1^3 + 2^3 + \cdots + n^3} &= \lim_{n \to \infty} \cdot \frac{\sum_{k=1}^{n} \frac{k}{n}\left(1-\frac{k-1}{n}\right) \frac{1}{n}}{\sum_{k=1}^{n} \left(\frac{k}{n}\right)^3 \cdot \frac{1}{n}} \\ &= \frac{\int_{0}^{1} x(1-x) \, \mathrm{d}x}{\int_{0}^{1} x^3 \, \mathrm{d}x} \\ &= \frac{1/6}{1/4} = \frac{2}{3}. \end{align*}
Note. It is not impossible to apply Stolz–Cesaro theorem here, but it requires a bit nasty computation. Indeed, set
$$ A_n = n (1 \cdot n + 2 \cdot (n-1) + \cdots + n \cdot 1), \qquad B_n = 1^3 + 2^3 + \cdots + n^3. $$
Writing $A_n$ in summation notation, we get $A_n = n \sum_{k=1}^{n} k (n+1 - k)$. Hence,
\begin{align*} A_n - A_{n-1} &= n^2 + \sum_{k=1}^{n-1} [nk(n+1-k) - (n-1)k(n-k)] \\ &= n^2 + \sum_{k=1}^{n-1} (2nk - k^2). \end{align*}
Using this, we find
$$ \frac{A_n - A_{n-1}}{B_n - B_{n-1}} = \frac{1}{n} + \sum_{k=1}^{n-1} \left( \frac{2k}{n} - \frac{k^2}{n^2} \right) \, \frac{1}{n} \to \int_{0}^{1} (2x-x^2) \, \mathrm{d}x = \frac{2}{3} $$
and therefore $A_n/B_n \to \frac{2}{3}$ by Stolz–Cesaro theorem.
Best Answer
The follow equation is from Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$.
\begin{align} e^{-n}\sum_{k=0}^n\frac{n^k}{k!} &=\frac{1}{n!}\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t\\ &=\frac12+\frac{2/3}{\sqrt{2\pi n}}+O(n^{-1})\tag{11} \end{align} Use this equation we know: $$\lim_{n\to\infty}\sqrt{n} \left(e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}-\frac{1}{2}\right)=\frac{2}{3\sqrt{2\pi }}.$$ So your limit is $\infty$.