A Problem Of Convergence

Question

I am newly learning about the convergence of sequences and I want to share a problem which I am unable to make through. The problem is as follows:-

Find a closed form for the limit $$f(m)=\lim_{n\to\infty} \left(\prod_{i=1}^n \frac{m+i}{n}\right)^{1/n}$$

I have learned a theorem in my book where it was written that if there is a sequence whose terms are all positive and the sequence converges to a finite limit then the Geometric mean of $n$ terms of that sequence will approach the same limit as that of the sequence itself, provided $n$ approaches infinity. I thought of using the above result somehow to figure out the answer of my question.

Though it is not provided here I assumed $m$ as a positive real number to increase the simplicity and to make use of the theorem. Is the result which I am thinking to make use of (unable to make use of though) will help to solve the problem? If yes then please provide me with a hint else help me with a solution of the problem

Answer 1

These kinds of limits can be reduced nicely with the root-to-ratio limit:

$$\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$$

In your case this becomes:

\begin{align}L&=\lim_{n\to\infty}\frac{m+n+1}{n+1}\prod_{k=1}^n\frac{m+k}{n+1}\frac n{m+k}\\&=\lim_{n\to\infty}\prod_{k=1}^n\frac n{n+1}\\&=\lim_{n\to\infty}\left(\frac n{n+1}\right)^n\\&=\lim_{n\to\infty}\left[\left(1-\frac1{n+1}\right)^{n+1}\right]^{n/(n+1)}\\&=e^{-1}\end{align}

Answer 2

Your hint is the way to go here. Let's do it that way. Your hint can be stated as

Lemma 1: Let $\{a_n\} $ be a sequence of positive terms such that $a_n\to L$ as $n\to\infty $. Then $$\sqrt[n] {a_1a_2\dots a_n} \to L$$ as $n\to \infty $.

The key here is to choose $a_n$ properly. Let us choose $$a_n=\left. \prod_{i=1}^{n+1}\frac{m+i}{n+1}\middle/\prod_{i=1}^{n}\frac{m+i}{n}\right.$$ so that $$a_n=\frac{m+n+1}{n+1}\cdot\frac{n^n}{(n+1)^n}$$ so that $a_n\to 1/e$.

Next note that $$a_1a_2\dots a_n=\frac{1}{m+1}\prod_{i=1}^{n+1}\frac{m+i} {n+1}$$ and by Lemma 1 the $n$-th root of above expression tends to $1/e$. Since $\sqrt [n] {m+1}\to 1$ it follows that $$\left(\prod_{i=1}^{n+1}\frac{m+i}{n+1}\right) ^{1/n}\to \frac{1}{e}$$ Raising the expression on left to power $n/(n+1)$ we can see that the above equation holds if the exponent $1/n$ is replaced with $1/(n+1)$. And then we can replace $n$ by $n-1$ everywhere to see that the desired limit is $1/e$.

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You should be able to show that lemma 1 above is equivalent to the root-to-ratio limit result used in another answer here.