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$\lim _{n\to \infty }\left(\left(1+\frac{1}{1}\right)\cdot \left(1+\frac{1}{2}\right)^2\cdot ...\cdot \left(1+\frac{1}{n}\right)^n\right)^{\frac{1}{n}}$

Hi. I'm trying to solve this limit without using special theorems, is that possible? Here's what I tried. Thanks.

I transformed the limit to: $\lim _{n\to \infty \:}\left(\prod _{k=1}^n\left(\left(1+\frac{1}{k}\right)^k\right)^{\frac{1}{n}}\right)$

Now, the problem is, I don't know how to do limits with k. It's either a finite number and most of the products equal to 1 because of that, but in case when k=n I get a 1^infinity^0 situation.

Because of that I rewrote the limit as: $\lim \:_{n\to \:\infty \:}\left(e^{\prod \:\:_{k=1}^n\left(\frac{1}{n}\cdot \:\:ln\left(\left(1+\frac{1}{k}\right)^k\right)\right)}\right)$

This also has two situations. Most of the products are equal to 1 because k is a finite number and so is the stuff in ln() and 1/n is 0. e^0 = 1.

But when k=n->infinity I get a 1^infinity situation and that's an indeterminate form as far as I know.

How would you solve this in a simple way similar to this, with the product, without using any special theorems?

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  • $\begingroup$ You can use the fact that $(1+\frac{1}{n})^n$ is increasing to get an upper bound of $e$. $\endgroup$ – Rene Schipperus Aug 13 '14 at 14:25
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We have $\log((\Pi_{k=1}^n (1+1/k)^k)^{1/n}) = \frac{1}{n}\sum_{k=1}^n \log((1+1/k)^k)$. Now $\lim_{k\to\infty}(1+1/k)^k = e$, and it follows that $\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \log((1+1/k)^k) = 1$. So $\lim_{n\to\infty} (\Pi_{k=1}^n (1+1/k)^k)^{1/n} = e$.

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Use the fact that $\left(1+\frac{1}{n}\right)^{n}$ is increasing and $\left(1+\frac{1}{n}\right)^{n+1}$ is decreasing.

This implies that

$$\left ( \left(1+\frac{1}{1}\right)^{1} \cdots \left(1+\frac{1}{n}\right)^{n}\right)^{\frac{1}{n}} \leq \left(1+\frac{1}{n}\right)^{n}$$

and that $$\left(1+\frac{1}{n}\right)^{n+1} \leq \left ( \left(1+\frac{1}{1}\right)^{2} \cdots \left(1+\frac{1}{n}\right)^{n+1}\right)^{\frac{1}{n}}$$

It remains only to see that both series have the same sum and this amounts to $(n+1)^{\frac{1}{n}}\rightarrow 1$

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Hint : $$\lim_{n\to \infty} x_n=\ell\implies \lim_{n\to \infty}\frac{x_1+x_2+\cdots+x_n}{n}=\ell$$ Now using it twice squeeze this expression, using : $$\frac{n}{\dfrac{1}{x_1}+\dfrac{1}{x_2}\cdots +\dfrac{1}{x_n}}\le \displaystyle \sqrt[n]{x_1x_2\cdots x_n}\le \dfrac{x_1+x_2+\cdots+x_n}{n}$$ From the previous hint you get LHS and RHS converge to $\ell$. Here $x_i=\left(1+\frac{1}{i}\right)^i$ hence $\ell=e$. Try to prove the hint, it is easy and you are done. Though Rene Schipperus has given a better solution, but this is how I proved that a sequence and its geometric mean of first terms converge to the same point. So I couldn't resist giving it. :)

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[after writing $f(n) = e^{\log f(n)}$] use Taylor series: $$ P = \log \prod_{k=1}^{n}\Big(1+\frac{1}{k}\Big)^k = \sum_{k=1}^{n}k \log \Big(1+\frac{1}{k} \Big) \sim \sum_{k=1}^nk \cdot \frac{1}{k} = n $$ hence $$ \lim_{n \to \infty}e^{\frac{P}{n}} = e $$

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