# Determining a particular limit and its asymptotic behaviour

### Problem description

I have the limit $$\lim_{n \to \infty} \sum_{k=1}^{n-2}\left( \frac{(n-3)!}{(n-k-2)!} n^{-k+1}a^k\frac{1}{k+2}\right),$$ with $$a \in \mathbb{R}_{>0}$$ (at Attempt 2 I have typed out the first few terms). I have already shown this limit converges for any $$a < 1$$, but I hope to show that it diverges for any $$a > 1$$ (and hopefully determine its behaviour for $$a = 1$$) as well. I think Attempt 2 could be a way to show this, but I am not sure if the steps I take are actually allowed. Even if they are, I would like a proof more similar to Attempt 1, since it may show how quickly the sum goes to infinity (so that I could quickly see if e.g. $$a^{n/2}$$ times this sum would also go to infinity). Any help checking if my proof is correct in Attempt 2, establishing the behaviour for $$a = 1$$ or giving a more insightful proof is appreciated.

### Attempt 1

My first idea was looking at the last term of the sum, which is given by $$(n-3)! n^{3-n} \frac{a^{n-2}}{n}.$$ I know that in general the inequality $$m! \geq e\left( \frac{m}{e} \right)^m$$ holds (I think I could use Stirling for a better estimate, but the reasoning should be the same). Substituting this we see that the last term is at least $$\left(\frac{n}{n-3}\right)^{3-n}e^{4-n}\frac{a^{n-2}}{n}.$$ For $$n \to \infty$$ the first factor should be finite, which means the last term on itself already clearly converges to infinity for any $$a > e$$. I then noted that if we look at earlier terms, we can estimate the fraction of factorials by saying for $$k\geq 2$$ it is at least $$\frac{(n-3)!}{n^{n-2-k}}$$ and we can apply the same method as for the last term now which yields something similar for every term, just that we take a lower power of $$\frac{a}{e}$$. Even though we have $$x$$ terms which have at least this value, it is clearly not enough, since for $$a < e$$ the convergence will be too quick. It seems this method as I applied it right now only works for $$a > e$$.

### Attempt 2

If we just look at the terms of the sum, we get $$\frac{a}{3} + \frac{n-3}{n}\frac{a^2}{4} + \frac{(n-3)(n-4)}{n^2}\frac{a^3}{5} + \frac{(n-3)(n-4)(n-5)}{n^3}\frac{a^4}{6} + \cdots$$ I am very tempted to say that the limit indeed goes to infinity. I know I cannot just swap around sum and limit and say that all fractions containing $$n$$ go to $$1$$, but I think I can do something very similar. For any $$N \in \mathbb{R}$$ we simply take an l such that $$\frac{a^l}{l+2} > N$$, which is clearly possible. Now the term with index $$k = l$$ on its own will already be at least $$N$$, since for $$n \to \infty$$ the fraction containing $$n$$ will go to $$1$$, so the term will be at least $$N$$. Since this is possible for any real $$N$$ the limit clearly does not converge, and thus diverges. I don't see why this reasoning is wrong, but it feels very weird (and as mentioned, I don't think it's insightful in the behaviour of the function).

Your sum may be written in the form $$\frac{{n^2 }}{{(n - 2)(n - 1)}}\frac{1}{{a^2 }}\left[ { - a - \frac{{n - 1}}{{2n}}a^2 + \sum\limits_{k = 1}^n {\binom{n}{k}(k - 1)!\left( {\frac{a}{n}} \right)^k } } \right].$$ Therefore it is enough to study the sum $$S(a,n)\; \mathop = \limits^{\small{\rm def}} \;\sum\limits_{k = 1}^n {\binom{n}{k}(k - 1)!\left( {\frac{a}{n}} \right)^k } .$$ Since $$(k - 1)!\left( {\frac{a}{n}} \right)^k = \int_0^{ + \infty } {{\rm e}^{ - (n/a)t} t^{k - 1}\, {\rm d}t} ,$$ we find that $$S(a,n) = \int_0^{ + \infty } {{\rm e}^{ - (n/a)t} \frac{{(1 + t)^n - 1}}{t}{\rm d}t} = \int_0^{ + \infty } {{\rm e}^{ - s/a} \frac{{(1 + s/n)^n - 1}}{s}{\rm d}s}.$$ If $$0, an application of the dominated convergence theorem shows that $$\mathop {\lim }\limits_{n \to + \infty } S(a,n) = \int_0^{ + \infty } {{\rm e}^{ - s/a} \frac{{{\rm e}^s - 1}}{s}{\rm d}s} =-\log(1-a).$$ (This can also be shown by applying Tannery's theorem to the definition of $$S(a,n)$$.) Consider now the case that $$a\ge 1$$. Clearly, $$S(a,n) \ge \int_1^{\sqrt n } {{\rm e}^{ - s/a} \frac{{(1 + s/n)^n - 1}}{s}{\rm d}s} .$$ Now for $$1\le s\le \sqrt{n}$$, $$(1 + s/n)^n = \exp (n\log (1 + s/n)) \ge \exp (s - s^2 /(2n)) \ge \exp (s - 1/2).$$ Furthermore $$\exp (s - 1/2) - 1 \ge \frac{1}{5}{\rm e}^s$$ for $$s\ge 1$$. Thus, if $$a\ge 1$$, $$S(a,n) \ge \frac{1}{5}\int_1^{\sqrt n } {\frac{{{\rm e}^{(1 - 1/a)s} }}{s}{\rm d}s} \ge \frac{1}{5}\int_1^{\sqrt n } {\frac{{{\rm d}s}}{s}} = \frac{1}{{10}}\log n$$ and the right-hand side diverges as $$n\to +\infty$$.