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I would appreciate it if somebody could help me with the following problem:

Q: How to proof (using by mathematical induction)($n=2,3,4,\cdots$)

$$\left(1+\frac{1}{n}\right)^n<1+\frac{1}{1!}+\frac{1}{2!}+\cdots+ \frac{1}{n!}$$

Young
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    Have you attempted a proof yourself, and if so, where did you get stuck? If you don't know where to start, you could take a look at the Wikipedia article on induction. – Peter Woolfitt Jul 01 '14 at 05:34
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    Also note that the statement isn't true for $n=1$, so perhaps you want $\le$ instead of $<$, or say $n>1$, $n\in\mathbb{N}$ – Peter Woolfitt Jul 01 '14 at 05:38
  • Also note as n approaches infinity, left converges to e and right converges to e, but if this statement where true then it would say e<e which is not true so I don't think this statement is true – Kamster Jul 01 '14 at 05:39
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    @user159813 That is not contradictory: under taking limit only weak inequalities are preserved, though in this case the result fails when $n=1$ apparently – user160738 Jul 01 '14 at 05:40
  • @user159813 $\frac 1 n \overset{n → ∞} \longrightarrow 0$, but $\frac 1 n > 0$, so $0 > 0$? – k.stm Jul 01 '14 at 05:40
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    @k.stm and user160738 thanks for pointing that out I didn't think of that – Kamster Jul 01 '14 at 05:42
  • Sorry $n=2,3,4,\cdots$ – Young Jul 01 '14 at 05:42
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    There are better ways than using induction – Hagen von Eitzen Jul 01 '14 at 06:24
  • I agreed with Hagen von Eitzen. For example Use that for $n\geq 1$ and $0\leq k\leq n$ we have $\displaystyle \binom{n}{k}\frac{1}{n^k}=\frac{n(n-1)\cdots(n-k+1)}{n^k}\frac{1}{k!}\leq \frac{1}{k!}$ (and if $n\geq 2$, we have strict inequality for $k\geq 2$) and use the binomial theorem for $\displaystyle(1+\frac{1}{n})^n$. Of course this is a direct proof, not by induction. – Kelenner Jul 01 '14 at 06:44

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We invoke strong induction:

$\textbf{I:}$ We wish to show the statement is true for $n=2$. This can be done with a simple calculation of this inequality: $\left(1 + \frac{1}{2}\right)^2 < \left(1 + \frac{1}{1!} + \frac{1}{2!}\right)$.

$\textbf{II:}$ Suppose the above statement is true for $n \in [2,k-1]$, we wish to show the statement is true for $n=k$. From out hypothesis we know that $\left(1+\frac{1}{k-1}\right)^{k-1} < \left(1 + \frac{1}{1!} + \cdots + \frac{1}{(1-k)!}\right)$. Multiplying both sides by $1+\frac{1}{k-1}$, we have the string of inequalities $$ \left(1+\frac{1}{k}\right)^k < \left(1+\frac{1}{k-1}\right)^{k} < \left(1+\frac{1}{1!} + \cdots + \frac{1}{(k-1)!}\right) \times \left(1+\frac{1}{k-1}\right). $$ We can expand the right side of this inequality to obtain $$ 1+\frac{1}{1!}+\cdots+\frac{1}{(k-1)!} + \frac{1}{k-1} + \cdots + \frac{1}{(k-1)!(k-1)} = \sum_{i=1}^{k-1}\frac{k}{i!(k-1)}. $$ Finally, we can utilize the usual convergence tests to show that $$ \sum_{i=1}^{k-1}\frac{k}{i!(k-1)} < 1+ \sum_{1}^{k-1} \frac{1}{i!} + \frac{1}{k!}. $$

$\textbf{III:}$ Thus by the principle of strong induction, $\left(1+\frac{1}{n}\right)^n < \left(1 + \frac{1}{1!} + \cdots + \frac{1}{n!}\right)$ for all $n = 2, 3, \ldots \hspace{25pt} \square$

Kevin Sheng
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