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Compute the limit:

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

Did
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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? – preferred_anon Aug 17 '12 at 10:03
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    @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 – preferred_anon Aug 17 '12 at 10:19
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    $1= \lim_{n \rightarrow \infty} 1 = \lim_{n \rightarrow \infty} (\sum_{i= 1}^{n} \frac{1}{n} )= \lim_{n \rightarrow \infty} \frac{1}{n} +\lim_{n \rightarrow \infty} \frac{1}{n} \ldots \lim_{n \rightarrow \infty} \frac{1}{n} =0+0+\ldots+0 = 0$ – clark Aug 17 '12 at 11:00

2 Answers2

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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 $$ \ell=\int_0^1f(x)\mathrm dx=\int_0^1\left(\sum_{n\geqslant1}(-1)^{n+1}\frac{x^{n-1}}n\right)\mathrm dx=\sum_{n\geqslant1}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}. $$

Did
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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 $p$.

  • I don't really understand why this solution gets downvoted. I think it's by far my best post in math.se. If there's a wrong step can you please comment it so that I learn. –  Jan 25 '16 at 15:22