Prove that $\prod_{i=1}^n (1+x_i/n) \sim \exp (\sum_{i=1}^n x_i/n)$ as $n\rightarrow\infty$?

Question

Let $x_1,x_2,\dots$ be an infinite sequence of real numbers. Assume that they are bounded, $|x_i| \le C < \infty$ for all $i$ for some $C$.

Is it true that, for any such sequence

$$\lim_{n \rightarrow \infty} \prod_{i = 1}^n \left( 1 + \frac{x_i}{n} \right) = \lim_{n \rightarrow \infty} \exp \left(\frac{1}{n} \sum_{i = 1}^n x_i \right)?\qquad\qquad (1)$$

So far I only have the heuristic argument

$$\prod_{i = 1}^n \left[ \left( 1 + \frac{x_i}{n} \right)^n \right]^{1 / n} \rightarrow \prod_{i = 1}^n \mathrm e^{x_i / n} = \exp \left( \frac{1}{n} \sum_{i= 1}^n x_i \right)$$

but I can't be sure that it is correct.

Update: The answer to my original question, that the lims in (1) are equal, is NO. In fact the limits might not exist. However their ratio goes to 1, so the two expressions are asymptotically equivalent. I've updated the title to reflect the actual true statement that was proved in the answers here to make this easier to find.

Answer 1

(I). $\lim_{n\to \infty}n^{-1}\sum_{j=1}^n x_j$ need not exist, even if $(x_j)_j$ is a bounded sequence.

(2). Let $A_n=\prod_{j=1}^n(1+x_j/n) .$ Let $B_n=\exp (n^{-1}\sum_{j=1}^n x_j).$ Let $n>C.$

We have $\forall j\leq n\,( |x_j|<n)$ so $$\log A_n=\sum_{j=1}^n\log (1+x_j/n)=$$ $$=\sum_{j=1}^n\sum_{s=1}^{\infty }(-1)^{s-1}(x_j/n)^s/s=$$ $$=(\sum_{j=1}^nx_j/n) + R_n$$ $$\text { where } \quad |R_n|=|\sum_{j=1}^n\sum_{s=2}^{\infty}(-1)^{s-1}(x_j/n)^s/s|\leq$$ $$\leq \sum_{j=1}^n \sum_{s=2}^{\infty}(C/n)^s/s=n\sum_{s=2}^{\infty}(C/n)^s/s=$$ $$=(C^2/n)\sum_{t=0}^{\infty} (C/n)^t/(t+2)\leq$$ $$\leq (C^2/n)\sum_{t=0}^{\infty}(C/n)^t=(C^2/n)\cdot 1/(1-C/n).$$ So $\lim_{n\to \infty}R_n=0.$

But we also have $R_n=\log A_n-\log B_n=\log(A_n/B_n).$ So $\lim_{n\to \infty}\log (A_n/B_n)=0.$ So $\lim_{n\to \infty}(A_n/B_n)=1.$

So if the Cesaro mean $\lim_{n\to \infty}n^{-1}\sum_{j=1}^n x_j$ exists then the LHS and RHS in your Q are equal. If the Cesaro mean does not exist, then the ratio of the LHS to the RHS still converges to $1.$

Answer 2

Given $$ S(n) = \prod\limits_{i = 1}^n {\left( {1 + {{x_{\,i} } \over n}} \right)} \quad \left| {\;\left| {x_{\,i} } \right| \le C \le \left\lceil C \right\rceil = D} \right. $$ let's take $D<<n$ so that we can put $n=mD\quad | \; 2 \le m \in \mathbb Z$.
Thereafter taking the logarithm we get $$ \eqalign{ & \ln S(n) = \sum\limits_{i = 1}^n {\ln \left( {1 + {{x_{\,i} } \over n}} \right)} = \sum\limits_{i = 1}^{mD} {\ln \left( {1 + {{x_{\,i} } \over {mD}}} \right)} = \cr & = \sum\limits_{i = 1}^{mD} {\ln \left( {1 + {{\left( {x_{\,i} /D} \right)} \over m}} \right)} \cr} $$

The Lagrange formulation of the Taylor remainder gives $$ \ln \left( {1 + z} \right)\quad \left| {\,\left| z \right|} \right. < 1/2 = z - {1 \over {2\left( {1 + \zeta } \right)^{\,2} }}z^{\,2} \quad \left| {\, - 1/2 < \zeta } \right. < 1/2 $$ so that $$ \left| {\, - {1 \over {2\left( {1 + \zeta } \right)^{\,2} }}\;} \right|z^{\,2} < 2z^{\,2} $$

Therefore we can write $$ \eqalign{ & \ln S(n) = \sum\limits_{i = 1}^{mD} {\ln \left( {1 + {{\left( {x_{\,i} /D} \right)} \over m}} \right)} = \sum\limits_{i = 1}^{mD} {\left( {{{\left( {x_{\,i} /D} \right)} \over m} + O\left( {{1 \over {m^{\,2} }}} \right)} \right)} = \cr & = \sum\limits_{i = 1}^{mD} {{{\left( {x_{\,i} /D} \right)} \over m} + O\left( {{1 \over m}} \right)} = \sum\limits_{i = 1}^n {{{x_{\,i} } \over n} + O\left( {{1 \over {n/D}}} \right)} \cr} $$ and $$ \eqalign{ & \mathop {\lim }\limits_{n\, \to \;\infty } \ln S(n) = \mathop {\lim }\limits_{n\, \to \;\infty } \left( {\sum\limits_{i = 1}^n {{{x_{\,i} } \over n} + O\left( {{1 \over {n/D}}} \right)} } \right) = \cr & = \mathop {\lim }\limits_{n\, \to \;\infty } \left( {{1 \over n}\sum\limits_{i = 1}^n {x_{\,i} } } \right) \cr} $$

Thus, whether the two limits exist or not, their ratio is in any case $1$.

Answer 3

It follows quite simply from the one of the definitions of $e^x$ (saying "one of" the definitions a bit loosely to indicate that any one of them could be taken to be the definition) - $$e^x\equiv\lim_{n\rightarrow\infty}\left(1+{x\over n}\right)^n .$$

The product of $n$ factors is a generalisation of the raising to the power of $n$ in the given (by me) limit definition, and that the argument of your exponential term is the arithmetic mean over all $x$ is the generalisation in that place corresponding to it.

One way we have of proving it in the simple case (let's call it) when it's just a single value of $x$, is to show that the binomial expansion morphs into the Taylor series for $e^x$ with $(n)_k$ (Pochhamner notation desending) becoming ever more nearly $n^k$ for a given value of $k$ (index along the series - the binomial or the Taylor) as $n$ increases. This will work also for the compound case (let's call it - infinite product of $n$ factors), if we can show that the sum of the products of the $x_i$ taken $k$ at a time becomes more nearly like $${\left(\sum_{j=1}^n x_j\right)^k\over k!}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\operatorname{¶}$$ for a given $k$ as $n$ increases in the same manner as, or a similar manner to that in which, $(n)_k$ becomes more nearly like $n^k$ ... or even in a very different manner, as long as it becomes more nearly like it: but that is not really to be expected. The difference consists in this last sum having $k$-fold products in it that are not fully hererogeneous - whereas in $$\prod_{j=1}^n\left(1+{x_j \over n}\right)\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\operatorname{§}$$ all the $k$-fold products occuring in a sum of them are fully heterogeneous. The total number of $k$-fold products including the non-fully-heterogeneous ones in ¶ is $n^k$, so it becomes the more nearly completely dominated by the fully-heterogeneous ones in the same proportion as $(n)_k$ becomes the more like $n^k$. This is an exact analogy of the similutude - ever increasing with $n$ - of the $(n)_k$ in the binomial expansion to the $n^k$ in the Taylor series in the simple case - and differs in that rather than being a pure number, it is the number of terms in a sum.

I cannot prove at this present time that there is absolutely no way a pathological set of $x_j$ could be devised such that the fully heterogeneous terms in the sum fail to dominate the non-fully heterogeneous ones despite their ever-increasing preponderance - constituting a proportion $${(n)_k\over n^k}$$ of the total termage. But it certainly would be an extremely pathological set, the devising of which would demand surpassing ingenuity.

Ah! but I've just noticed that you said "assume they are bounded"! That prettymuch precludes anysuch "pathological set", I should think.