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I am trying to find an upper limit for the product $(1+1/2)(1+1/3)\dots(1+1/n)$. I tried using AM-GM inequality as follows: \begin{align} (1+1/2)(1+1/3)\dots(1+1/n) \leq \left(\frac{1}{n-1}\left( -1 + n + 1/2 + 1/3 + 1/4 + \dots + 1/n\right)\right)^{n-1} \end{align}

From Wikipedia, $1/2 + 1/3 + 1/4 + \dots + 1/n \leq \log n$.

Therefore, I get

\begin{align} (1+1/2)(1+1/3)\dots(1+1/n) \leq \left(\frac{1}{n-1}\left( -1 + n + 1/2 + 1/3 + 1/4 + \dots + 1/n\right)\right)^{n-1} \leq \left( 1+ \frac{\log n}{n-1} \right)^{n-1} \end{align}

As $n \to \infty$, I find an upper limit is $n$.

I was wondering if there are more tighter upper limits. Any comments/thoughts are welcome.

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    $\begingroup$ isn't the product equal to $(3/2)(4/3)(5/4)\cdots ((n+1)/n)=(n+1)/2$ $\endgroup$
    – Gareth Ma
    Jul 11, 2021 at 18:26
  • $\begingroup$ Oh.. my bad! Thanks! $\endgroup$ Jul 11, 2021 at 18:27

1 Answer 1

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The product is equal to

$$\prod_{i=2}^n \left(1+\frac{1}{i}\right)=\prod_{i=2}^n \frac{i+1}{i}=\frac{n+1}{2}$$

:D

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