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I'm right now reviewing some Calc. 1 again and came across an Inequality which I got stuck proving. The inequality is:

$\left| { (1+\frac { x }{ n } ) }^{ n }- (1+x)-\sum _{ j=2 }^{ m }{ { \frac { { x }^{ j } }{ j! } } } \right| \le \quad \sum _{ j=2 }^{ m }{ \frac { { \left| x \right| }^{ j } }{ j! } } (1-({ 1-\frac { j-1 }{ n } ) }^{ j-1 })+\sum _{ j=m+1 }^{ \infty }{ \frac { { \left| x \right| }^{ j } }{ j! } } $ and
$m\le n$.

I'm pretty sure the inequality holds since we have on the RHS almost the whole series representation of $e^x$ and on the LHS only a finite piece of it. I tried to prove it directly which turned out to be pretty difficult. Then I tried to find a "squeeze" so that $A\le C\le B$ holds but I got stuck with the multiplication on the RHS. Could someone help me out please? (Calc I methods would be appreciated)

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  • $\begingroup$ On the r.h.s., you have an opening ( without the corresponding ) $\endgroup$ – Bernard Nov 28 '17 at 19:58
  • $\begingroup$ Do you have any ( ) after the minus sign in the LHS of the inequality? I don't understand why you put a \quad after the first minus sign on the left. $\endgroup$ – Jack Nov 28 '17 at 20:13
  • $\begingroup$ im sorry now its corect $\endgroup$ – MasterPI Nov 28 '17 at 20:25
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$$\begin{align}\left(1 + \frac xn\right)^n - (1+x) &= \left(\sum_{j=0}^n {n \choose j}\frac {x^j}{n^j}\right) - (1+x) \\&= \sum_{j=2}^n {n \choose j}\frac {x^j}{n^j} \\&= \sum_{j=2}^n \frac{n!}{(n-j)!n^j}\frac{x^j}{j!} \end{align}$$ Hence $$\begin{align}\left| \left(1+\frac xn\right)^n-(1+x)-\sum_{j=2}^m\frac { x^j}{j!} \right| &= \left|\sum_{j=2}^m \left(\frac{n!}{(n-j)!n^j}-1\right)\frac{x^j}{j!} + \sum_{j=m+1}^n \frac{n!}{(n-j)!n^j}\frac{x^j}{j!}\right|\\&\le \sum_{j=2}^m\left|\frac{n!}{(n-j)!n^j}-1\right|\frac{\left|x^j\right|}{j!}+\sum_{j=m+1}^n \frac{n!}{(n-j)!n^j}\frac{\left|x^j\right|}{j!}\end{align}$$

Now, $$\frac{n!}{(n-j)!n^j} = \frac {\overbrace{n(n-1)(n-2)\dots(n-j+1)}^{j\text{ factors}}}{n^j} = 1\left(1-\frac 1n\right)\left(1-\frac 2n\right)\dots\left(1-\frac{j-1}n\right)$$

Since $j > 1$, we get $$\frac{n!}{(n-j)!n^j} < 1$$ But also since $\left(1-\frac{k}n\right) \ge \left(1-\frac{j-1}n\right)$ for $0 \le k \le j- 1 < n$, $$\frac{n!}{(n-j)!n^j} =\left(1-\frac 1n\right)\left(1-\frac 2n\right)\dots\left(1-\frac{j-1}n\right) \ge \left(1-\frac{j-1}n\right)^{j-1}$$ and so $$0 < 1 - \frac{n!}{(n-j)!n^j} \le 1 - \left(1-\frac{j-1}n\right)^{j-1}$$ Therefore $$\sum_{j=2}^m\left|\frac{n!}{(n-j)!n^j}-1\right|\frac{\left|x^j\right|}{j!} = \sum_{j=2}^m\frac{\left|x^j\right|}{j!}\left(1-\frac{n!}{(n-j)!n^j}\right)\le \sum_{j=2}^m\frac{\left|x^j\right|}{j!}\left(1 - \left(1-\frac{j-1}n\right)^{j-1}\right)$$ and $$\sum_{j=m+1}^n \frac{n!}{(n-j)!n^j}\frac{\left|x^j\right|}{j!} \le \sum_{j=m+1}^n \frac{\left|x^j\right|}{j!} \le \sum_{j=m+1}^\infty \frac{\left|x^j\right|}{j!}$$ and your inequality follows.

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  • $\begingroup$ nice work, very helpful and nice trick at the beginning! $\endgroup$ – MasterPI Nov 29 '17 at 8:33

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