# difficult sequence

Prove that the sequence:

$$a_n= \frac{1}{n}\left(e\cdot\sqrt e \cdots\sqrt[3]e\cdot\sqrt[n]e\right)$$ is decreasing to a finite limit. After having shown that the sequence: $$b_n=\left(\sum_{k=1}^n\frac{1}{k}\right)-\log n$$ converges to a positive real number $b,$ say who is the limit of $a_n$

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Note that $b_n=\log a_n$. The limit of the $b$s is the definition of the Euler-Mascheroni constant. –  Henning Makholm Jan 10 '12 at 0:05

\eqalign{ & {a_n} = \frac{1}{n}\prod\limits_{k = 1}^n {{e^{\frac{1}{k}}}} \cr & \log {a_n} = - \log n + \sum\limits_{k = 1}^n {\frac{1}{k}} \cr & \log {a_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} - \log n \cr & \mathop {\lim }\limits_{n \to \infty } \log {a_n} = \gamma \cr}

You can prove $0 < \gamma < 1$ since we can replace $\log n$ by $\log (n+1)$ and put

\eqalign{ & {b_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} - \sum\limits_{k = 1}^n {\log \frac{{k + 1}}{k}} \cr & {b_n} = \sum\limits_{k = 1}^n {\left( {\frac{1}{k} - \log \frac{{k + 1}}{k}} \right)} \cr}

Since we know $$1 - \frac{1}{x} \leqslant \log x \leqslant x - 1$$

We have

$$\frac{1}{{k + 1}} \leq \log \left( {1 + \frac{1}{k}} \right) \leq \frac{1}{k}$$

We can prove both bounds with this. \eqalign{ & \frac{1}{k} - \frac{1}{{k + 1}} \geqslant \frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right) \geqslant 0 \cr & \sum\limits_{k = 1}^n {\frac{1}{k} - \frac{1}{{k + 1}}} \geqslant \sum\limits_{k = 1}^n {\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right)} \geqslant 0 \cr & 1 \geqslant \sum\limits_{k = 1}^n {\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right)} \geqslant 0 \cr}

The first inequality also proves each term is positive or zero (though the last is discarded by a simple look at the image), i.e

$$\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right) \geqslant 0$$

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First, notice that

$$\log(a_n) = \log\frac{1}{n} + \log(e\cdot\sqrt{e}\cdots\sqrt[n]{e}) = \sum_{k = 1}^n\log \sqrt[k]{e} - \log n = \sum_{k=1}^n\frac{1}{k} - \log n = b_n$$

Assume that $\lim_{n\rightarrow\infty}b_n = b\in\mathbb{R}$. Then by the above equation,

$$\lim_{n\rightarrow\infty}a_n = \lim_{n\rightarrow\infty}e^{b_n} = e^b$$

Hence it is enough to show that $b_n$ is decreasing to a finite limit. Now, notice that

$$\int_1^{n+1}\frac{1}{x}dx \leq \sum_1^n \frac{1}{k}\leq 1 + \int_1^n\frac{1}{x}dx$$

In particular,

$$\int_1^{n + 1}\frac{1}{x}dx - \log n \leq b_n \leq 1 + \int_1^n\frac{1}{x}dx - \log n$$

so that

$$\log\frac{n+1}{n} \leq b_n \leq 1$$

So $\{b_n\}$ is bounded and positive. It remains to show that $b_n$ converges. Try to prove monotonicity of $b_n$.

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Actually, $(b_n)$ is decreasing. You will have a hard time proving that $\ln (n) - \ln (n+1)$ is positive, and $b_{n+1} - b_n = 1/(n+1) - \int_n^{n+1} 1/x dx < 0$. –  D. Thomine Jan 10 '12 at 0:38
Yes, decreasing, sorry! –  William Jan 10 '12 at 13:33