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Prove some inequalities about (1+1/n)^n for positive integers n

  1. For n \in \mathbb{Z}_{>0} prove,

        \[ \left( 1 + \frac{1}{n} \right)^n =  1 + \sum_{k=1}^n \left( \frac{1}{k!} \cdot \prod_{r=0}^{k-1} \left(1 - \frac{r}{n}\right) \right). \]

  2. For n > 1, prove

        \[ 2 < \left( 1 + \frac{1}{n} \right)^n < 1 + \sum_{k=1}^n \frac{1}{k!} < 3. \]


  1. Proof. Recall the Binomial theorem,

        \[ (a+b)^n = \sum_{k=0}^n \binom{n}{k} \left( \frac{1}{n} \right)^k. \]

    So, letting a = \frac{1}{n} and b = 1 we have,

        \begin{align*}  \left(1 + \frac{1}{n}\right)^n &= \sum_{k=0}^n \binom{n}{k} \left(\frac{1}{n} \right)^k \\ &= 1 + \sum_{k=1}^n \binom{n}{k} \left( \frac{1}{n} \right)^k \\ &= 1 + \sum_{k=1}^n \frac{n!}{k! (n-k)!} \left( \frac{1}{n} \right)^k &(\text{Def. of } \binom{n}{k})\\ &= 1 + \sum_{k=1}^n \frac{1}{k!} \left( \left( \frac{\prod_{r=1}^n r}{\prod_{r=1}^{n-k} r} \right) \cdot \left( \frac{1}{n} \right)^k \right) &(\text{Def. of } n!)\\ &= 1 + \sum_{k=1}^n \frac{1}{k!} \left(\prod_{r=n-k+1}^n r \right) \left( \frac{1}{n} \right)^k &(\text{cancelling})\\ &= 1 + \sum_{k=1}^n \frac{1}{k!} \left(\prod_{r=0}^{k-1} (n-r) \right) \left( \frac{1}{n} \right)^k &(\text{reindexing prod.})\\ &= 1 + \sum_{k=1}^n \frac{1}{k!} \left( \prod_{r=0}^{k-1} (n-r) \right) \left( \prod_{r=0}^{k-1} \frac{1}{n} \right) &(\text{rewritting } (1/n)^k \text{ as prod})\\ &= 1 + \sum_{k=1}^n \frac{1}{k!} \left( \prod_{r=0}^{k-1} \left( 1 - \frac{r}{n} \right) \right). \qquad \blacksquare \end{align*}

  2. Proof. First, we prove the left inequality,

        \[ 2 < \left( 1 + \frac{1}{n} \right)^n. \]

    If n > 1, then n \geq 2, so using the Binomial theorem we have,

        \[ \left( 1 + \frac{1}{n} \right)^n = \sum_{k=0}^n \binom{n}{k} \left( \frac{1}{n} \right)^k \ \implies \ \left( 1 + \frac{1}{n} \right)^n = 1 + n\left(\frac{1}{n} \right) + \sum_{k=2}^n \binom{n}{k} \left( \frac{1}{n} \right)^k > 2. \]

    Where we know the inequality is strict since there is at least one term (which is necessarily positive) in \sum_{k=2}^n \binom{n}{k} \left( \frac{1}{n} \right)^k since n \geq 2.
    Next, we prove the middle inequality,

        \[ \left( 1 + \frac{1}{n} \right)^n < 1 + \sum_{k=1}^n \frac{1}{k!}. \]

    By part (a) we know,

        \[ \left( 1+ \frac{1}{n} \right)^n = 1 + \sum_{k=1}^n \frac{1}{k!} \left( \prod_{r=0}^{k-1} \left( 1 - \frac{r}{n} \right) \right). \]

    Further, for n > 1 we have \left(1 - \frac{r}{n}\right) < 1 for all r = 0, \ldots, n-1. Thus, we know that,

        \[ \prod_{r=0}^{k-1} \left(1 - \frac{r}{n} \right) < \prod_{r=0}^{k-1} 1 = 1. \]

    Hence,

        \[ \frac{1}{k!} \left( \prod_{r=0}^{k-1} \left(1 - \frac{r}{n} \right) \right) < \frac{1}{k!} \]

    for all k. Therefore, we have established the second inequality,

        \[ \left(1 + \frac{1}{n} \right)^n = 1+ \sum_{k=1}^n \frac{1}{k!} \left( \prod_{r=0}^{k-1} \left( 1 - \frac{r}{n} \right) \right) < 1 + \sum_{k=1}^n \frac{1}{k!}. \]

    Finally, we prove the right inequality,

        \[ 1 + \sum_{k=1}^n \frac{1}{k!} < 3. \]

    Here we expand the first few terms and use a previous result,

        \begin{align*}  1 + \sum_{k=1}^n \frac{1}{k!} &= 1 + 1 + \frac{1}{2} + \frac{1}{6} + \sum_{k=4}^n \frac{1}{k!} \\  &< \frac{8}{3} + \sum_{k=4}^n \frac{1}{2^k} & (2^k < k! \implies \frac{1}{2^k} > \frac{1}{k!})\\  &= \frac{8}{3} + \frac{1}{16} \left( \sum_{k=0}^{n-4} \frac{1}{2^k} \right) & (\text{reindexing and pulling out } \frac{1}{2^4}) \\  &= \frac{8}{3} + \frac{1}{16} \left( 2 - \frac{1}{2^{n-4}}\right) \\  &= \frac{8}{3} + \frac{1}{8} - \frac{1}{2^n} \\  &< 3 \end{align*}

    for all n > 1. In the second to last line we used this result on the kth powers of a real number x (in this case x = \frac{1}{2}). This completes the proof for all of the inequalities requested. \qquad \blacksquare

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