Limit of $\sqrt[n]{(x+1)...(x+n)} - x$ as $x \to +\infty$

Question

Let $n \in \mathbb{N}^{\ast}$. I want to determine the following limit :

$$ \lim \limits_{x \to +\infty} \sqrt[n]{(x+1)\ldots(x+n)} - x.$$

Let $x = \frac{1}{t}$ with $t \to 0$. It is equivalent to the following limit :

$$ \lim \limits_{t \to 0} \sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t}. $$

$$ \begin{align*} \sqrt[n]{\displaystyle \frac{(t+1)\ldots(t+n)}{t^{n}}} - \frac{1}{t} &= {} \frac{1}{t} \sqrt[n]{(t+1)\ldots(t+n)} - \frac{1}{t} \\[2mm] &= \frac{1}{t} \sqrt[n]{ n! + u(t) } - \frac{1}{t} \\[2mm] \end{align*} $$

where $u(0)=0$ and $\displaystyle \lim \limits_{t \to 0} u(t) = 0$. Since $\displaystyle \sqrt[n]{1+t} = 1 + \frac{t}{n} + o(t)$ as $t \to 0$, $$\displaystyle \sqrt[n]{n! + u(t)} = \sqrt[n]{n!} \; \sqrt[n]{\displaystyle 1 + \frac{u(t)}{n!} } = \sqrt[n]{n!} \bigg( 1 + \frac{u'(0) t}{n(n!)} + o(t) \bigg).$$

I do not see how to go on from there !

Answer 1

$$ \lim \limits_{x \to +\infty} \sqrt[n]{(x+1)\ldots(x+n)} - x$$

$$=\lim_{t\to0}\frac{\sqrt[n]{(1+t)(1+2t)\cdots(1+nt)}-1}t$$

$$=\lim_{t\to0}\frac{(1+t)(1+2t)\cdots(1+nt)-1}t\frac1{\sum_{r=0}^{n-1}\lim_{t\to0}(\sqrt[n]{(1+t)(1+2t)\cdots(1+nt)})^r}$$

$$=\lim_{t\to0}\frac{(1+2+\cdots+n)t+O(t^2)}t\cdot\frac1{\sum_{r=0}^{n-1}1}$$

Answer 2

There is a very simple trick that involves bounding such products by matching terms of opposite sizes. This idea motivates considering the following inequalities.

$(x+1)(x+n) \le (x+k)(x+n+1-k) \le (x+\frac{n+1}{2})(x+\frac{n+1}{2})$ for any $x \ge 0$ and $k \in [1..n]$.

The above can be proven by many different methods. After this, it is trivial to bound the expression under the limit by $\sqrt{(x+1)(x+n)} - x$ and $(x+\frac{n+1}{2}) - x$, and then taking limits for both yield $\frac{n+1}{2}$ as $x \to \infty$, and hence by squeeze theorem we are done.

Answer 3

$\bf{My\; Solution::}$ Given $$\displaystyle \lim_{x\rightarrow \infty}\sqrt[n]{(x+1)(x+2)(x+3).....(x+n)}-x$$

Using $\bf{A.M\geq G.M},$ Here $$(x+1),(x+2),(x+3),............(x+n)>0\;,$$ when $x\rightarrow \infty$

So $$\displaystyle \frac{(x+1)+(x+2)+..............+(x+n)}{n}\geq \sqrt[n]{(x+1)(x+2)............(x+n)}$$

and equality hold when $$(x+1)=(x+2)=.............=(x+n)\;,$$ bcz here $x\rightarrow \infty$

So all are equal,

So $$\displaystyle \frac{nx+\frac{n(n+1)}{2}}{n} = \sqrt[n]{(x+1)(x+2)...........(x+n)}$$

So $$\displaystyle \lim_{n\rightarrow \infty}\frac{nx+\frac{n(n+1)}{2}}{n}-x = \lim_{n\rightarrow \infty}\sqrt[n]{(x+1)(x+2)...........(x+n)}-x$$

So $$\displaystyle \lim_{n\rightarrow \infty}\sqrt[n]{(x+1)(x+2)...........(x+n)}-x = \frac{n+1}{2}$$

Answer 4

Asymptotic results I proved a number of years ago (when I write $a(n) \approx b(n)-c(n)$ I mean $\lim_{n \to \infty} \dfrac{b(n)-a(n)}{c(n)} =1 $):

$$ x(x+1)...(x+n-1) \approx(x+(n-1)/2)^n -(x+(n-1)/2)^{n-2}\dfrac{n^3-n}{24} $$

and

$$ (x(x+1)...(x+n-1))^{1/n} \approx(x+(n-1)/2) -\dfrac{n^2-1}{24(x+(n-1)/2)} $$

The second result above improves the approximation requested by the OP.

Here is a transcription from part of my PDF:

Let $P_n(x) =\prod_{i=0}^{n-1} (x+i) $, $r =\frac{n-1}{2} $ and $u =x+r $.

Note that $r$ is a function of $n$, and $u$ and $v$ (defined below) are functions of $n$ and $x$.

If $S_i(n) =\frac1{2i}\sum_{j=0}^{n-1} (r-j)^{2i} $ then, for any $k \ge 1$, $$\sum_{i=1}^k \dfrac{S_i(n)}{u^{2i}} \le \ln\left(\dfrac{u^n}{P_n(x)}\right) \le \sum_{i=1}^{k-1} \dfrac{S_i(n)}{u^{2i}} +\dfrac{S_k(n)}{u^{2k}(1-r^2/u^2} $$

This follows from $P_n^2(x) =\prod_{j=0}^{n-1} (u^2-(r-j)^2) $ and the result valid for $k \ge 1$ and $0 \le z < 1$ that $\sum_{i=1}^k \dfrac{z^i}{i} \le -\ln(1-z) \le \sum_{i=1}^{k-1} \dfrac{z^i}{i} +\dfrac{z^k}{k(1-z)} $.

Explicit expressions are $S_1(n) =\dfrac{n^3-n}{24} $ and $S_2(n) =\dfrac{(n^3-n)(3n^2-7)}{960} $.

Letting $k=1$ in this result and letting $v =(P_n(x))^{1/n} $, this becomes $\dfrac{n^3-n}{24u^2} \le \ln\left(\dfrac{u^n}{P_n(x)}\right) \le \dfrac{n^3-n}{24(u^2-r^2)} $ and $\dfrac{n^2-1}{24u^2} \le \ln\left(\dfrac{u}{v}\right) \le \dfrac{n^2-1}{24(u^2-r^2)} $.

To convert these inequalities involving logs to inequalities using differences, we will use the following result.

If $c/u^2 \le \ln(a/b) \le c/(u^2 - r^2)$ where $a, b,$ and $c$ are all positive and c are all positive, then

A. $a - b \ge bc/u^2$;

B: $a - b \le ac/(u^2 - r^2)$;

for any $d > 0$,

C: if $u \ge r\sqrt{1 + 1/d} $ then $a - b \le ac(1 + d)/u^2 $;

D if $u \ge r\sqrt{1 + c/(r^2d)} $ then $a - b \ge ac(1 - d)/u^2 $.

Proof. Follows from the inequality true for all $z > 0$ that $1 - 1/z \le \ln z \le z - 1 $.

From this, after a moderate amount of algebra, we get these:

$$ \lim_{n \to \infty}\dfrac{u^n}{P_n(x)} = \lim_{n \to \infty}\dfrac{u}{v} =1 , $$ $$ \lim_{n \to \infty}\dfrac{u^n-P_n(x)}{u^{n-2}} = \dfrac{n^3-n}{24} , $$ and $$ \lim_{n \to \infty} u(u-v) =\dfrac{n^2-1}{24} . $$

These give the result I quoted at the begining.

Answer 5

We have $$\sqrt[n]{a} - \sqrt[n]{b} = \dfrac{a-b}{\sum_{k=0}^{n-1} a^{k/n}b^{(n-1-k)/n}}$$ In our case, $a = (x+1)(x+2)\cdots(x+n)$ and $b=x^n$. We have $a-b = \frac{n(n+1)}2x^{n-1} + O(x^{n-2})$. The denominator is $nx^{n-1} + O(x^{n-2})$. Hence, we have the expression as $$\frac{n(n+1)/2 + O(1/x)}{n + O(1/x)} \to \frac{(n+1)}2 \text{ as }x \to \infty$$