Prove that the series $a_n = $ $2^n \over n!$ converges to 0.
I am trying to find a variable that will be bigger than $2^n\over n$ , with no luck. Any help, besides using binoms? Thanks in advance!!!
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Write the terms out. The $n$th term is bounded above by $4/n$. – David Mitra Nov 14 '15 at 13:03
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Note that $$ \frac{2^{n}}{n!} = 2\frac{2^{n-2}}{3\cdot 4 \cdots n} \leq 2\cdot \frac{2}{n} = \frac{4}{n} $$ for all suitable $n$.
If you want $\varepsilon$-analysis: given any $\varepsilon > 0$, we have $n > 4/\varepsilon$ only if $4/n < \varepsilon$.
Yes
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$\frac{2^{n}}{n!} = 2\frac{2^{n-2}}{3\cdot 4 \cdots n} \leq \frac{4}{n}$. @YonatanIzutskiver – Yes Nov 14 '15 at 13:19
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Since there is already a suitable answer, I will offer another method which can be used for a large class of these sorts of problems.
Note that the Taylor series of the exponential function is given by $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ naturally this converges for all $x \in \mathbb{R}$ and its value is $e^x$.
For all convergent series it follows that the terms of the series must converge to $0$ so $\forall x \in \mathbb{R}$ $$ \lim_{n \to \infty} \left( \frac{x^n}{n!} \right) = 0$$ Plug in $x = 2$ and the result follows.
Kayle of the Creeks
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Sorry, i'm not familiar with Taylor's theorem yet, it's a few lectures ahead of me :) – Yonatan Izutskiver Nov 14 '15 at 13:48
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Of course! Come back to this comment once you have covered more theory about series and convergence, I am sure you will appreciate this method then! – Kayle of the Creeks Nov 14 '15 at 14:04