How to generalise the limit of the product $\lim _{n \rightarrow \infty} \prod_{k=1}^n\left(1+\frac{1}{n}+\frac{k}{n^2}\right) $?

Question

When I met the result in the post that

$$ \boxed{\lim _{n \rightarrow \infty}\prod_{k=1}^n\left(1+\frac{k}{n^2}\right) =\sqrt{e}}, $$

I appreciated very much the application of $G.M.\leq A.M.$ in the answer and modified the solution as below: $$ P_n^2=\prod_{k=0}^n\left[1+\frac{1}{n}+\frac{k(n-k)}{n^4}\right] $$ Using $G.M.\leq A.M.$, we have

$$ 0 \leqslant k(n-k) \leqslant \frac{n^2}{4} $$ Plugging back yields

$$ \left(1+\frac{1}{n}\right)^n \leqslant p_n^2 \leqslant \prod_{k=0}^n\left(1+\frac{1}{n}+\frac{1}{4 n^2}\right)=\left(1+\frac{1}{2 n}\right)^{2 n} $$

Since $$ \lim _{n \rightarrow \infty}\left(1+\frac{1}{2 n}\right)^{2 n}=e=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n, $$ By Squeezing Theorem, we conclude that

$$ \lim _{n \rightarrow \infty} \prod_{k=1}^n\left(1+\frac{k}{n^2}\right)=\sqrt{e} $$

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I then tried to investigate the limit $$ \lim _{n \rightarrow \infty} \prod_{k=1}^n\left(1+\frac{1}{n}+\frac{k}{n^2}\right) $$ by the same technique.

Let the product be $Q_n$. Firstly we are going to find similarly the upper bound of $Q_n^2$ in terms of $n$. $$\displaystyle \begin{aligned}Q_n^2 & =\prod_{k=1}^n\left[\left(1+\frac{1}{n}\right)^2+\left(1+\frac{1}{n}\right) \frac{1}{n}+\frac{k(n-k)}{n^4}\right]\end{aligned}\tag*{} $$

For the last terms, using $G.M.\leq A.M.$ again, we have $\displaystyle 0\leq k(n-k) \leqslant \frac{n^2}{4}\tag*{} $

Then plugging back yields $$\displaystyle \begin{aligned}Q_n^2 & \leqslant \prod_{k=1}^n\left[\left(1+\frac{1}{n}\right)^2+\left(1+\frac{1}{n}\right) \frac{1}{n}+\frac{1}{4 n^2}\right] \\& =\prod_{k=1}^n\left(1+\frac{1}{n}+\frac{1}{2 n}\right)^2 \\& =\prod_{k=1}^n\left(1+\frac{3}{2 n}\right)^2\\&=\left(1+\frac{3}{2 n}\right)^{2 n} \end{aligned}\tag*{} $$

Next we are going to find the lower bound of $Q_n^2$ in terms of $n$. $$\displaystyle Q_n^2\geqslant \prod_{k=1}^n\left[\left(1+\frac{1}{n}\right)^2+\left(1+\frac{1}{n}\right) \frac{1}{n}\right]= \prod_{k=1}^n \left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\right]\\=\left(1+\frac{1}{n}\right)^n\left(1+\frac{2}{n}\right)^n\tag*{} $$

Now we can find the limits of the bounds of $Q_n^2$ as $n$ tends to $\infty$. $$\displaystyle \lim _{n \rightarrow \infty}\left(1+\frac{3}{2 n}\right)^{2n}=\left[\lim _{n \rightarrow \infty}\left(1+\frac{1}{\frac{2 n}{3}}\right)^{\frac{2 n}{3}}\right]^{3}=e^3\tag*{} $$ and $$\displaystyle \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n\left(1+\frac{2}{n}\right)^n=e \cdot e^2=e^3\tag*{} $$ By the Squeeze Theorem, $\displaystyle \lim _{n \rightarrow \infty} Q_n^2=e^3\tag*{} $ Hence we can conclude that $\displaystyle \boxed{\lim _{n \rightarrow \infty} \prod_{k=1}^n\left(1+\frac{1}{n}+\frac{k}{n^2}\right)=e^{\frac{3}{2}}\,} \tag*{} $

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I am curious about what happens when both the number of terms and the power of $n$ in the product increase. I started to investigate the limit and surprisingly found that

$$\boxed{\lim _{n \rightarrow \infty} \prod_{k=1}^n\left(1+\frac{1}{n}+\frac{1}{n^2}+\cdots +\frac{1}{n^{m-1}} +\frac{k}{n^m}\right)=e\,}$$ where $m\ge 3$. How to prove it?

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Let $$R_n=\prod_{k=1}^n\left(1+\frac{1}{n}+\frac{1}{n^2}+\cdots +\frac{1}{n^{m-1}} +\frac{k}{n^m}\right)$$ and $$ A=1+\frac{1}{n}+\frac{1}{n^2}+\ldots+\frac{1}{n^{m-1}} $$ Then $$R_n^2 =\prod_{k=1}^n\left[\left(A+\frac{k}{n^m}\right)\left(A+\frac{n-k}{n^m}\right) \right] =\prod_{k=1}^n\left[A^2+\frac{A}{n^{m-1}}+\frac{k( n-k)}{n^{2m}}\right] $$ Using $ 0 \leqslant k(n-k) \leqslant \frac{n^2}{4}, $ we have

$$ A^n\left(A+\frac{1}{n^{m-1}}\right)^n \leqslant R_n^2 \leqslant \prod_{k=1}^n\left(A+\frac{1}{2 n^{m-1}}\right)^2=\left(A+\frac{1}{2 n^{m-1}}\right)^{2 n} $$ Then we finish the proof if we can prove that

$$ \lim _{n \rightarrow \infty} A^n\left(A+\frac{1}{n^{m-1}}\right)^n=e^2= \lim _{n \rightarrow \infty}\left(A+\frac{1}{2 n^{m-1}}\right)^{2 n} $$

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My attempt:

$$\left(1+\frac{1}{n}\right)^n \leqslant A^n=\left(1+\frac{1}{n}+\frac{1}{n^2}+\ldots+\frac{1}{n^{m-1}}\right)^n \leqslant\left(\frac{1}{1-\frac{1}{n}}\right)^n \Rightarrow \lim _{n \rightarrow \infty} A^n=e $$ My Question: How to prove that $$ \lim _{n \rightarrow \infty}\left(A+\frac{1}{n^{m-1}}\right)^n=\lim _{n \rightarrow \infty}\left(A+\frac{1}{2 n^{m-1}}\right)^n=e? $$

Answer 1

Using your notation and Pochhammer's symbols $$R_m=n^{-m n} \left(\frac{n^m-1}{n-1}\right){}_n$$ which is also $$R_m=n^{-m n}\,\,\frac{\Gamma \left(\frac{n^{m+1}+n^2-n-1}{n-1}\right)}{\Gamma \left(\frac{n^{m+1}-1}{n-1}\right)} $$

Taking logarithms and using Stirling approximation, for $m>3$ $$\log(R_m)=1+\frac {1} {2n}+\mathcal{O}\left(\frac{1}{n^2}\right)$$

What is interesting to notice is that, taking much more terms, some interesting patters appear. For example $$\log(R_6)=\color{blue}{1+\frac{1}{2 n}+\frac{1}{3 n^2}+\frac{1}{4 n^3}}+\frac{7}{10 n^4}+\mathcal{O}\left(\frac{1}{n^5}\right)$$ $$\log(R_7)=\color{blue}{1+\frac{1}{2 n}+\frac{1}{3 n^2}+\frac{1}{4 n^3}+\frac{1}{5 n^4}}+\frac{2}{3 n^5}+\mathcal{O}\left(\frac{1}{n^6}\right)$$ $$\log(R_8)=\color{blue}{1+\frac{1}{2 n}+\frac{1}{3 n^2}+\frac{1}{4 n^3}+\frac{1}{5n^4}+\frac{1}{6 n^5}}+\frac{9}{14 n^6}+\mathcal{O}\left(\frac{1}{n^7}\right)$$ This can then write $$\log(R_m)=n \log \left(\frac{n}{n-1}\right)-n^{2-m}\,\Phi \left(\frac{1}{n},1,m-1\right)+\mathcal{O}\left(\frac{1}{n^{m-2}}\right)$$ where $\Phi(.)$ is the Lerch transcendent function.

Answer 2

Notice that $A + \frac{1}{n^{m - 1}} = 1 + \frac{1}{n} + \mathrm{o}\!\left(\frac{1}{n}\right)$ and $A + \frac{1}{2n^{m - 1}} = 1 + \frac{1}{n} + \mathrm{o}\!\left(\frac{1}{n}\right)$ when $m \geqslant 3$. Therefore, it is enough to show that for any $\varepsilon_n = \mathrm{o}\!\left(\frac{1}{n}\right)$, $\left(1 + \frac{1}{n} + \varepsilon_n\right)^n \rightarrow e$.

And indeed, for $n$ large enough, $1 + \frac{1}{n} + \varepsilon_n > 0$ and using $\ln(1 + x) \sim x$ when $x \rightarrow 0$, we obtain, $$ n\ln\!\left(1 + \frac{1}{n} + \varepsilon_n\right) \sim n\left(\frac{1}{n} + \varepsilon_n\right) \rightarrow 1. $$ Taking the exponential of this limit, we obtain $\left(1 + \frac{1}{n} + \varepsilon_n\right)^n \rightarrow e$. QED.

Answer 3

Other Methode $$ \text{L}＝\lim_{n\to＋\infty}\left({\prod_{k＝1}^{n}\left({1＋\frac {1}{n}＋\frac {k}{n^2}}\right)}\right)\implies\;\ln(\text{L})＝\lim_{n\to＋\infty}\sum_{k＝1}^{n}\ln\left({1＋\frac {n＋k}{n^2}}\right)\\~\\\text{And we know that}\;\\~\\\forall\;x>0\;\;x-\frac {x^2}{2}<\ln(1＋x)<\;x\\~\\\text{Let x＝}\frac {n＋k}{n^2}\implies\;\frac {n＋k}{n^2}-\frac {(n＋k)^2}{2n^4}<\ln\left({1＋\frac {n＋k}{n^2}}\right)<\frac {n＋k}{n^2}\\~\\\implies\;\frac {1}{n^2}\sum_{k＝1}^{n}(n＋k)\;-\frac {1}{2n^4}\sum_{k＝1}^{n}(n^2＋k^2＋2kn)<\sum_{k＝1}^{n}\ln\left({1＋\frac {n＋k}{n^2}}\right)<\frac {1}{n^2}\sum_{k＝1}^{n}(n＋k)\\~\\\implies\frac {3n^2＋n}{2n^2}-\frac {14n^3＋9n^2＋n}{12n^4}<\sum_{k＝1}^{n}\ln\left({1＋\frac {n＋k}{n^2}}\right)<\frac {3n^2＋n}{2n^2}\\~\\\text{and therfore }\\~\\\lim_{n\to\infty}\left({\frac {3n^2＋n}{2n^2}-\frac {14n^3＋9n^2＋n}{12n^4}}\right)＝\lim_{n\to\infty}\left({\frac {3n^2＋n}{2n^2}}\right)＝\frac {3}{2}\\~\\\implies\;\lim_{n\to\infty}\left({\sum_{k＝1}^{n}\ln\left({1＋\frac {n＋k}{n^2}}\right)}\right)＝\frac {3}{2}\implies\;\ln(\text{L})=\frac {3}{2}\;\\~\\\implies\;\text{L}＝e^{\frac {3}{2}}＝\sqrt {e^{3}}\\~\\\text{Finally}\;\lim_{n\to＋\infty}\left({\prod_{k＝1}^{n}\left({1＋\frac {1}{n}＋\frac {k}{n^2}}\right)}\right)＝\sqrt {e^{3}} $$

Answer 4

Using the inequality : $\ln (1+x)<x$, we have $$ \ln \left(1+\frac{1}{n}\right)^n<n \ln <A+\frac{p}{n^{m-1}}<n\left(\frac{1}{n}+\frac{1}{n^2}+\cdots+\frac{p}{n^{m-1}}\right)= 1+\frac{1}{n}+\cdots+\frac{p}{n^{m-2}} \\ \because \lim _{n \rightarrow \infty} \ln \left(1+\frac{1}{n}\right)^n=1=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}+\cdots+\frac{p}{n^{n+2}}\right)\\ \therefore \quad \lim _{n \rightarrow \infty} n \ln \left(A+\frac{p}{n^{m-1}}\right)=1 $$ Hence for any constant $p$, we get $$ \lim _{n \rightarrow \infty}\left(A+\frac{p}{n^{m-1}}\right)^n=e $$

In particular, $$\lim _{n \rightarrow \infty}\left(A+\frac{1}{n^{m-1}}\right)^n=\lim _{n \rightarrow \infty}\left(A+\frac{1}{2 n^{m-1}}\right)^n=e $$ Hence we can conclude that

$$\boxed{\lim _{n \rightarrow \infty} \prod_{k=1}^n\left(1+\frac{1}{n}+\frac{1}{n^2}+\cdots +\frac{1}{n^{m-1}} +\frac{k}{n^m}\right)=e\,}$$ where $m\ge 3$.

Answer 5

There is a general technique based on the following result which can be found in many texts on Probability theory (see Exercise 1.1. on Durret, R., Probability Theory and Examples, 3d Edition, Thomson Brooks/Cole, 2005, pp.78).

Theorem Let $\{c_{n,m}:1\leq m\leq m_n\}\subset\mathbb{C}$. Suppose that

1. $\lim\limits_{n\rightarrow0}\sup_{1\leq m\leq m_n}|c_{n,m}|=0$,
2. $\lim\limits_{n\rightarrow\infty}\sum^{m_n}_{m=1}c_{n,m}=c\in\mathbb{C}$,
3. and $M:=\sup_n\sum^{m_n}_{m=1}|c_{n,m}|<\infty$. Then \begin{align}\prod^{m_n}_{m=1}(1+c_{n,m})=e^c \end{align}

A proof of this result is given at the end of this posting.

Here I applied this result to the products that appear in the OP:

• For $a_{n,m}=\frac{m}{n^2}$, $1\leq m\leq n$ we have $a_{n,m}>0$ and $$\max_{1\leq m\leq n}a_{n,m}=\frac{1}{n}\xrightarrow{n\rightarrow\infty}0,$$ $$\sum^n_{m=1}a_{n,m}=\frac{1}{2}\frac{n+1}{n}\xrightarrow{n\rightarrow\infty}\frac12$$ Conditions (1) and (2) of the Theorem hold, and since $a_{n, m}>0$, (3) follows from (2). Hence $$\prod^n_{m=1}\Big(1+\frac{m}{n^2}\Big)\xrightarrow{n\rightarrow\infty}e^{1/2}$$

• Define $b_{n,m}=\frac1n+\frac{m}{n^2}$, $1\leq m\leq n$. Then $b_{n,m}>0$, $$\max_{1\leq m\leq n}b_{n,m}=\frac2n\xrightarrow{n\rightarrow\infty}0,$$ and $$\sum^n_{m=1}b_{n,m}=1+\frac12\frac{n+1}{n}\xrightarrow{n\rightarrow\infty}\frac32$$ This shows that (1) and (2) of the Theorem hold and since $c_{n, m}>0$, (3) follows from (2). Consequently $$\lim_m\prod^n_{m=1}\Big(1+\frac1n+\frac{k}{n^2}\Big)=e^{3/2}$$

• The same technique applied to $$c_{n,k}=\frac{1}{n}+\frac{1}{n^2}+\ldots +\frac{1}{n^{m-1}}+\frac{k}{n^m}, \qquad 1\leq k\leq n$$ with $m>2$. This yields $$\max_{1\leq k\leq n}c_{n, k}=\frac{1}{n}+\frac{1}{n^2}+\ldots +\frac{1}{n^{m-2}}+\frac{2}{n^{m-1}}\xrightarrow{n\rightarrow\infty}0,$$ and $$\sum^n_{k=1}c_{n,k}=1+\frac1n+\ldots+\frac{1}{n^{m-2}}+\frac{1}{2}\frac{n+1}{n^{m-1}}\xrightarrow{n\rightarrow\infty}1 $$ Hence $$\lim_n\prod^n_{k=1}\Big(1+\frac1n+\ldots+\frac{1}{n^{m-1}}+\frac{k}{n^m}\Big)\xrightarrow{n\rightarrow\infty}e$$

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Proof of Theorem:

If $\log$ is the principal logarithm on $\mathbb{C}\setminus (-\infty,0]\times\{0\}$, then for $\lim\limits_{z\rightarrow0}\frac{\log(1+z)}{z}=1$ whenever $|z|<1$. Given $\varepsilon>0$, there is $\delta>0$ such that $0<|z|<\delta$ implies $$|\log(1+z)-z|<\varepsilon|z|. $$ Without loss of generality, we can assume that $\sup_m|c_{n,m}|<\delta$ for all $n$. Then \begin{align} \Big|\sum^{m_n}_{m=1}\big(\log(1+c_{n,m})-c_{n,m}\big)\Big|\leq \sum^{m_n}_{m=1}|\log(1+c_{n,m})-c_{n,m}|\leq \varepsilon\sum^{m_n}_{m=1}|c_{n,m}| < M\varepsilon \end{align} By letting $n\rightarrow\infty$ and then $\varepsilon\rightarrow0$ we obtain \begin{align} \lim_{n\rightarrow\infty}\sum^{m_n}_{m=1}\log(1+c_{n,m}) =\lim_{n\rightarrow\infty} \sum^{m_n}_{m=1}c_{n,m}=c \end{align} The conclusion follows from the continuity of the exponential function.