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Compute the limit:

$$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{\frac{1}{2}}\cdots\left(1+\frac{n}{n}\right)^{\frac{1}{n}}$$

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I realise the question has been answered, but can someone please explain to me why one can't simply take the logarithm of the product, show that each term in that sum goes to zero, and conclude that the limit is 1? – Daniel Littlewood Aug 17 '12 at 10:03
@DanLitt A posteriori: because the limit is not 1. A priori: because no theorem says that the limit of some sums is the sum of the limits when the number of terms in the sums is not bounded. – Did Aug 17 '12 at 10:14
Thank you, I needed the a priori statement – Daniel Littlewood Aug 17 '12 at 10:19
$1= \lim_{n \rightarrow \infty} 1 = \lim_{n \rightarrow \infty} (\sum_{i= 1}^{n} \frac{1}{n} )= \lim_{n \rightarrow \infty} \frac{1}{n} +\lim_{n \rightarrow \infty} \frac{1}{n} \ldots \lim_{n \rightarrow \infty} \frac{1}{n} =0+0+\ldots+0 = 0$ – clark Aug 17 '12 at 11:00
up vote 25 down vote accepted

Note at the onset that $1+\frac{k}n\leqslant\mathrm e^{k/n}$ for every $k$ hence the $n$th product $P_n$ is such that $P_n\leqslant\mathrm e$, in particular, the sequence $(P_n)_{n\geqslant1}$ is bounded.

To show that $(P_n)_{n\geqslant1}$ actually converges and to identify its limit, note that, for every $n$, $$ \log(P_n)=\frac1n\sum\limits_{k=1}^nf\left(\frac{k}n\right), \qquad\text{with}\quad f(x)=\frac{\log(1+x)}x. $$ The function $f$ is continuous on $[0,1]$ (define $f(0)=1$) hence its Riemann sums converge to its integral and $P_n\to\mathrm e^\ell$ with $$ \ell=\int_0^1f(x)\mathrm dx=\int_0^1\left(\sum_{n\geqslant1}(-1)^{n+1}\frac{x^{n-1}}n\right)\mathrm dx=\sum_{n\geqslant1}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}. $$

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Thanks for your nice answer. The integral way seems to be very useful here. – user 1618033 Aug 17 '12 at 9:46
in fact, on the second line you already have a Riemann sum and that is $-Li_{2}(-1)=\frac{\pi^2}{12}$. Oh, I missed this nice point :( – user 1618033 Aug 17 '12 at 9:52
+1: nice! (and nontrivial) – Fabian Aug 17 '12 at 12:21
I got $\frac 4 e$ for the answer and Wolfram Alpha agrees. Shouldn't $f$ be $f(x) = \ln (1+x)$? – Ennar Jul 23 '15 at 15:35
@Ennar No. You misread $(1+k/n)^{1/\color{red}{n}}$ for $(1+k/n)^{1/\color{red}{k}}$. – Did Jul 23 '15 at 15:39

We want to calculate $$\lim_{n \to \infty} \prod_{1 \leqslant k \leqslant n} \left(1 + \frac {k} {n}\right)^{\frac {1} {k}}.$$ Denote it by $\ell$. Taking logarithms we have $$\begin {eqnarray} \log \ell & = & \lim_{n \to \infty} \sum_{1 \leqslant k \leqslant n} \frac {1} {k} \log \left(1 + \frac {k} {n}\right) \nonumber \\ & = & \lim_{n \to \infty} \sum_{1 \leqslant k \leqslant n} \frac {1} {k} \sum_{m \geqslant 1} (-1)^{m + 1} \frac {k^m} {m n^m} \nonumber \\ & = & \lim_{n \to \infty} \sum_{1 \leqslant k \leqslant n} \sum_{m \geqslant 1} (-1)^{m + 1} \frac {k^{m - 1}} {m n^m} \nonumber \\ & = & \lim_{n \to \infty} \sum_{m \geqslant 1} \sum_{1 \leqslant k \leqslant n} (-1)^{m + 1} \frac {k^{m - 1}} {m n^m} \nonumber \\ & = & \lim_{n \to \infty} \sum_{m \geqslant 1} \left(\frac {(-1)^{m + 1}} {m^2} + O \left(\frac {1} {n}\right) \right) \nonumber \\ & = & \sum_{m \geqslant 1} \frac {(-1)^{m + 1}} {m^2} \nonumber \\ & = & \frac {\pi^2} {12}. \end {eqnarray}$$ Hence, $\ell = \exp \left(\frac {\pi^2} {12}\right)$.

Note that in step 2 we changed $\log \left(1 + \frac {k} {n}\right)$ with its Taylor expansion, and in step 5 we used the fact that $\sum_{1 \leqslant k \leqslant n} k^p = \frac {n^{p + 1}} {p + 1} + O (n^p)$ for any non-negative integer $p$.

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I don't really understand why this solution gets downvoted. I think it's by far my best post in If there's a wrong step can you please comment it so that I learn. – Nima Bavari Jan 25 at 15:22

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