The first term is an integer, and every fraction in the sum is actually an integer because n ≤ b for each term. \sum_{n=0}^{\infty} \frac{1}{k^n} = \frac{1}{1-\frac{1}{k}} The third property is: Now let r=p/q and eʳ be rational numbers. First, to prove that x is strictly positive, we insert the above series representation of e into the definition of x and obtain.
Note that
They are: Let us start with the simpler but less general first proof (see Simmons).
Writing this as $\alpha=(c_1,c_2,\ldots,c_k)_!$ ones has the familiar property of finite decimal representations (without integer part), that the order of two such numbers is given by lexicographic comparison of their representations $(c_1,c_2,\ldots,c_k)$. : Addison-Wesley. It's also an important number in physics, where it shows up in the equations for waves, such as light waves, sound waves, and quantum waves. For example first number is #4/3#, second number is #-7351/990# and third is #95742/700#. In the first sum, $k\leq b$ so $k!|b!$, so each term is an integer, so the sum is an integer. }$$, Satisfies the more-or-less easily verifiable identity, $$_0F_1(k-1;z) - {}_0F_1(k; z) = \frac{z}{k(k-1)}{}_0F_1(k+1;z)$$, Iterating this, one ends up with the continued fraction, $$\frac{{}_0F_1(k+1;z)}{k{}_0F_1(k;z)} = \frac1{k+\cfrac{z}{k+1+\cfrac{z}{k+2+\cfrac{z}{k+3+\cdots}}}}$$, Now note that $_0F_1(3/2;x^2/4) = \cosh(x)$ and $x \,{}_0F_1(3/2;x^2/4) = \sinh(x)$, hence applying the above one has the pretty well-known continued fraction, $$\tanh(x) = \cfrac{x/2}{\frac{1}{2} + \cfrac{\frac{x^2}{4}}{\frac{3}{2} + \cfrac{\frac{x^2}{4}}{\frac{5}{2} + \cfrac{\frac{x^2}{4}}{\frac{7}{2} + \cdots}}}} = \cfrac{x}{1 + \cfrac{x^2}{3 + \cfrac{x^2}{5 + \cfrac{x^2}{7 + \cdots}}}}$$. Among the set of irrational numbers, two famous constants are e and π.
{\displaystyle p,q} So we certainly have $$.
\sum_{k=0}^\infty \frac{b!}{k! Use MathJax to format equations. 11 ) $A$ e $B$ are clearly integer by definition, so if we prove that $C$ is not integer we have an absurdity starting by the hypothesis of rationality of $e$ and the proof is ended.
q + \frac{1}{(q+1) !} \displaystyle \left. We have
If one would want to turn this into a formal argument, one could argue as follows. 2 ) Summing first 13 terms of $\sum \frac{1}{n! + \ldots \right) q! e How we test gear. \frac{1}{q+1} + \frac{1}{(q+1) ^2} + \ldots = \sum_{n=1}^{\infty} \frac{1}{(q+1) ^n} = \sum_{n=0}^{\infty} \frac{1}{(q+1) ^n} - 1 \displaystyle \left( 2 + \frac{1}{2!} Why is $\sum_{k=1}^\infty \frac{1}{(b+1)^k}=\frac{1}{b}$? Hence $e$ is not rational either. isn't harder than proving the irrationality of $\sqrt{2}$, just a bit more laborious (I fear that with $\pi$ things go in a very different way...). $$
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