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konstantin123 [22]
3 years ago
9

How to find the limit

Mathematics
1 answer:
Ludmilka [50]3 years ago
8 0
\displaystyle\lim_{n\to\infty}\left(k!+\frac{(k+1)!}{1!}+\cdots+\frac{(k+n)!}{n!}\right)=\lim_{n\to\infty}\dfrac{\displaystyle\sum_{i=0}^n\frac{(k+i)!}{i!}}{n^{k+1}}=\lim_{n\to\infty}\frac{a_n}{b_n}

By the Stolz-Cesaro theorem, this limit exists if

\displaystyle\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}

also exists, and the limits would be equal. The theorem requires that b_n be strictly monotone and divergent, which is the case since k\in\mathbb N.

You have

a_{n+1}-a_n=\displaystyle\sum_{i=0}^{n+1}\frac{(k+i)!}{i!}-\sum_{i=0}^n\frac{(k+i)!}{i!}=\frac{(k+n+1)!}{(n+1)!}

so we're left with computing

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)!}{(n+1)!\left((n+1)^{k+1}-n^{k+1}\right)}

This can be done with the help of Stirling's approximation, which says that for large n, n!\sim\sqrt{2\pi n}\left(\dfrac ne\right)^n. By this reasoning our limit is

\displaystyle\lim_{n\to\infty}\frac{\sqrt{2\pi(k+n+1)}\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\sqrt{2\pi(n+1)}\left(\dfrac{n+1}e\right)^{n+1}\left((n+1)^{k+1}-n^{k+1}\right)}

Let's examine this limit in parts. First,

\dfrac{\sqrt{2\pi(k+n+1)}}{\sqrt{2\pi(n+1)}}=\sqrt{\dfrac{k+n+1}{n+1}}=\sqrt{1+\dfrac k{n+1}}

As n\to\infty, this term approaches 1.

Next,

\dfrac{\left(\dfrac{k+n+1}e\right)^{k+n+1}}{\left(\dfrac{n+1}e\right)^{n+1}}=(k+n+1)^k\left(\dfrac{k+n+1}{n+1}\right)^{n+1}=e^{-k}(k+n+1)^k\left(1+\dfrac k{n+1}\right)^{n+1}

The term on the right approaches e^k, cancelling the e^{-k}. So we're left with

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}

Expand the numerator and denominator, and just examine the first few leading terms and their coefficients.

\displaystyle\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{n^{k+1}+(k+1)n^k+\cdots+1+n^{k+1}}=\frac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}

Divide through the numerator and denominator by n^k:

\dfrac{n^k+\cdots+(k+1)^k}{(k+1)n^k+\cdots+1}=\dfrac{1+\cdots+\left(\frac{k+1}n\right)^k}{(k+1)+\cdots+\frac1{n^k}}

So you can see that, by comparison, we have

\displaystyle\lim_{n\to\infty}\frac{(k+n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\lim_{n\to\infty}\frac1{k+1}=\frac1{k+1}

so this is the value of the limit.
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