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Thread: Induction again

  1. #1
    No one in Particular VonNemo19's Avatar
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    Induction again

    So, here I'm asking for a check..

    For each natural number $n$ with $n\geq3$, $\left(1+\frac1{n}\right)^n<n$.

    Proof:

    Basis $n=3$: true.
    Assume for some $k$ with $k\geq3$, that $\left(1+\frac1{k}\right)^k<k$. Multiplying by $\left(1+\frac1{k}\right)$ yields $\left(1+\frac1{k}\right)^{k+1}<k+1$. But since $k+1>k\implies{1/(k+1)<1/k}\implies{}\left[1+\frac1{k+1}\right]^{k+1}<\left(1+\frac1{k}\right)^{k+1}<k+1$ as desired.
    the set $T$ of natural numbers which make the above statement true is such that $3\in{}T$, and if $k\in{}T,~k\geq3$, then $k+1\in{}T$.
    Therefore, by the principle of induction, for every natural number $n\geq3$, $\left(1+\frac1{n}\right)^n<n$.

    The part I worry about is when I show that k<k+1... Is this ok?
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  2. #2
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    Re: Induction again

    I pretty much agree with your proof. Your question reduces to: For any positive integer k, the function $f(x)=x^k$ is strictly increasing for $x>0$. The easy reason that this is true is that the derivative of f is positive. Alternatively, it's easy to give an inductive proof.

    My only criticism is that the base case n = 3 is not really proved. I would add some some argument.
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  3. #3
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    Re: Induction again

    Quote Originally Posted by VonNemo19 View Post
    So, here I'm asking for a check..
    For each natural number $n$ with $n\geq3$, $\left(1+\frac1{n}\right)^n<n$.
    Let
    $\begin{align*}e_n=\left(1+\frac{1}{n}\right)^n&=1 +\frac{n}{1!}\cdot\frac{1}{n}+\frac{n(n-1)}{2!}\cdot\frac{1}{n^2}+\cdots+\frac{1}{n^n}\;\; \;\text{binomial theorem}\\&<1+\frac{1}{1!}+\frac{1}{2!}\cdots+ \frac {1} {n!}\\&\le 1+\left(1+\frac{1}{2}+\frac{1}{2^2}\cdots+ \frac{1}{2^{n-1}} \right)\\&<1+2=3 \end{align*}$

    That has nothing to do with induction. It is a standard theorem in advanced calculus.
    Moreover, more importantly, each term of the $e\text{-sequence}$ is $<3<n$
    Last edited by Plato; Mar 11th 2017 at 03:59 PM.
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