I've been working on a proof of the irrationality of $e$. I was thinking that this would move in a similar routine to proving that the square root of a prime number is irrational, but I think that routine might be a little overdone. I want to try and prove it in a manner that deals with the expansion of $e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + ...  = \sum_{i=0}^{\infty} \frac{1}{n!}.$ If I assume this sum is rational, will I run into some contradictory properties related to the sum of an infinite series?
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Prove that e is an irrational number.  Recall that $\,\mathrm{e}=\displaystyle\sum_{n=0}^\infty\frac{1}{n!},\,\,$ and assume $\,\mathrm{e}\,$ is rational, then  $$\sum\limits_{k=0}^\infty \frac{1}{k!} = \frac{a}{b},\quad \text{for some positive integers}\,\,\, a,b.$$  so  $$b\sum\limits_{k=0}^\infty \frac{1}{k!} =a$$  or  $$ b\left(1+1+\frac{1}{2} + \frac{1}{6} +\cdots \right)= a. $$  Where can I go from here?
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I have read a few proofs that $\sqrt{2}$ is irrational.  I have never, however, been able to really grasp what they were talking about.  Is there a simplified proof that $\sqrt{2}$ is irrational?
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In university last semester I was asked to prove that $\sin1$ (1 radian that is) is irrational,  and ended up simply using the Taylor Series Expansion. This method provides a very quick solution, but I am curious as to whether anyone has a method for proving this without making use of the Taylor Series Expansion. I feel as though doing so must be possible using some number theory, but am low on ideas as to an alternative approach to the question.  Note:  If anyone is interested in my solution using Taylor Series Expansion (although it is not the focus of my question), here it is :      From    $$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} -  \dots$$   We see that    $$ \alpha = \sin 1 = 1 - \frac{1^3}{3!} + \frac{1^5}{5!} -  \dots $$   Given integers $a$ and $b$, if $\alpha = \frac{a}{b}$ then it follows that $b!\alpha \in \mathbb{Z}$, and $b!\alpha = C + D$ where $C \in \mathbb{Z}$ and we have :   $$ D = \begin{cases} \pm(\frac{1}{b+1} - \frac{1}{(b+1)(b+2)(b+3)} + \dots) \text{    if $b$ is even}\\ \pm(\frac{1}{(b+1)(b+2)} - \frac{1}{(b+1)\dots(b+4)} + \dots ) \text{ if $b$ is odd}\\ \end{cases} $$   In each case we can see that $0 &lt; D &lt; 1$, giving us a contradiction.    Thus, we have that $\sin 1 $ is irrational.    $$ \blacksquare $$
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I have to calculate sum of series $\sum \frac{n^2}{n!}$. I know that $\sum \frac{1}{n!}=e$ but I dont know how can I use that fact here..
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A look at the first few zeros  $$14.134725,21.022040,25.010858,30.424876,32.935062,37.586178,\dots$$ is in accord with     Numerical evidence suggests that all values of $t$ (the imaginary part of a root of $\zeta$) corresponding to nontrivial zeros are irrational (e.g., Havil 2003, p. 195; Derbyshire 2004, p. 384).    (numbers and quote taken from ). What are the attempts to prove that all values of $t$ are irrational? Would it mean something to the distribution of primes, if one, some or plenty of rational roots $\frac{1}{2}+i\frac{q}{r}$ exist?
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My teacher posed an infinite series question to me today and I'm not quite sure how to start to go about it. $$\sum_{r=1}^{\infty} \dfrac{2^{-r}}{r^{2}}$$ Any hints would be much appreciated.
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So my math professor in college gave us this problem:  $$U_n=\sum_{k=1}^{k=n}\frac{1}{k!}$$  $$V_n=U_n+ \frac{1}{n\cdot n!}$$  The problem has two questions:  1- Prove that the two series are adjacent (easy)  2- Let $e$ be their limit, prove that $e$ is irrational.  I have no idea how to start the second part, since I don't have a professor that does the problems the first professor gives us, so I have a lack in reasoning skills, I tried proof by contradiction, but no result.  also I'm new to the site so I don't know how to use the symbols so sorry for that.
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Prove that between any two different real numbers there is a rational number and an irrational number.  My attempt at a proof: (Throughout this proof it is assumed without any loss of generality that $x &lt; y$)  1.) $x$ is irrational and $y$ is irrational.  In this case the arithmetic mean $\frac{x+y}{2}$ is always an irrational number between $x$ and $y$. Also, if $y$ is irrational then it has a non-terminating decimal expansion and so we can always find a positive integer $n$ such that $x &lt; y - 10^{-n}y &lt; y$. The middle of the three numbers here will have a terminating decimal expansion, and hence it is rational. It also clearly lies between $x$ and $y$. These examples of rational and irrational numbers should also work in the cases:  2.) $x$ is irrational and $y$ is rational.  3.) $x$ is rational and $y$ is irrational.  There is one last case to consider.  4.) $x$ is rational and $y$ is rational.  In this case the arithmetic mean of $x$ and $y$ is rational. An irrational number for example could be the ratio $\frac{ex + \pi y}{e + \pi}$  I'm not sure if this proof is a good method or if it is even valid, so some feedback would be appreciated.
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Proving that $$\sum_{n=0}^{\infty }\frac{1}{(2n)!!}=\sqrt{e}$$ Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.
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I am studying the proof that $\sqrt 2$ is an irrational number. Now I understand most of the proof, but I lack an understanding of the main idea  which is:     We assume $\frac{m^2}{n^2} = 2$. Then both $m$ and $n$ can't be even.   I do not understand, why can't both $m$ and $n$ be even?
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As a follow-up to a , I'd like to know if  $$\sum_{n=1}^\infty\frac{1}{2 ^ n}$$  is convergent? If yes, then what does it converge to?  Just as a sidenote - the whole issue started with reading about Zeno's Turtle Paradox which made me curious about the underlying mathematics.
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Suppose $\sum_{n=1}^\infty \frac{1}{n} = S$ where $S$ is finite.  Then  $$S =\sum_{n=1}^\infty \frac{1}{n}= \sum_{n=1}^\infty \frac{1}{2n-1} + \frac{1}{2n} &gt; \sum_{n=1}^\infty \frac{1}{2n} + \frac{1}{2n} = S$$  which is a contradiction.  Is this valid?
91022
I had a little back and forth with my logic professor earlier today about proving a number is irrational. I proposed that 1 + an irrational number is always irrational, thus if I could prove that 1 + irrational number is irrational, then it stood to reason that was also proving that the number in question was irrational.  Eg. $\sqrt2 + 1$ can be expressed as a continuous fraction, and through looking at the fraction, it can be assumed $\sqrt2 + 1$ is irrational. I suggested that because of this, $\sqrt2$ is also irrational.  My professor said this is not always true, but I can't think of an example that suggests this.  If $x+1$ is irrational, is $x$ always irrational?   Actually, a better question is: if $x$ is irrational, is $x+n$ irrational, provided $n$ is a rational number?
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We have the sum:  $\sum\nolimits_{n=0}^\infty \frac{n}{n!}$  Why is this sum a representation of $e$ when the sum:  $\sum\nolimits_{n=0}^\infty \frac{1}{n!}$  is already a representation of $e$?  For me it seams that the first row is greater then the second.
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Could someone provide, please, a proof of the theorem below?  "Being $x$ and $b$ integers greater than $1$, which can not be represented as powers of the same basis (positive integer) and integer exponent, then the logarithm of $x$, in base $b$, is an irrational number."  Well... Assuming that the logarithm is a rational number $p/q$ ($p$ and $q$ are relatively prime integers), I know I can write $x^q = b^p$, but I can not conclude from this fact that $b$ and $x$ are powers of a same integer (and with integer exponent).
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I need to study this sum: $\sum_{n=1}^{\infty} \frac{n!\, i^n}{n^n}$. Taking: $$\lim_{n\rightarrow \infty} \; \left|\left(\frac{n!\: i^n}{n^n}\right)^{\frac{1}{2}}\right|$$ $$\rightarrow \lim_{n\rightarrow \infty}  \left|\left(\frac{n!}{n^n}\right)^{\frac{1}{2}}\right|$$  Using Stirling approximation considering the limit:  $$\ln n! \approx n \ln\:n-n$$ then $n! \approx (\frac{n}{e})^n$ (is this correct?):  $$\rightarrow \lim_{n\rightarrow \infty} \left|\frac{n}{n\, e}\right| = \frac{1}{e}$$ This doesn't make too much sense because I know the series diverges.  I think I'm missing a $\sqrt{2\pi n}$ in Stirling approximation but I don't understand why that pops out taking the $e^{(\;)}$ from the first expression.
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I'm reading R. Courant &amp; H. Robbins'  for fun. I'm on page $60$ and $61$ of the second addition. There are three exercises on proving numbers irrational spanning these pages, the last is as follows.     Exercise $3$: Prove that $\phi=\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational. Try to make up similar and more general examples.   My Attempt:     Lemma: The number $\sqrt{2}+\sqrt{3}$ is irrational. (This is part of Exercise 2.)   Proof: Suppose $\sqrt{2}+\sqrt{3}=r$ is rational. Then $$\begin{align} 2&amp;=(r-\sqrt{3})^2 \\ &amp;=r^2-2\sqrt{3}+3 \end{align}$$ is rational, so that $$\sqrt{3}=\frac{r^2+1}{2r}$$ is rational, a contradiction. $\square$  Let $\psi=\sqrt{2}+\sqrt{3}$. Then, considering $\phi$, $$\begin{align} 5&amp;=(\phi-\psi)^2 \\ &amp;=\phi^2-\psi\phi+5+2\sqrt{6}. \end{align}$$  I don't know what else to do from here. My plan is/was to use the Lemma above as the focus for a contradiction, showing $\psi$ is rational somehow.  Please help :)  Thoughts:  The "try to make up similar and more general examples" bit is a little vague.    The question is not answered  as far as I can tell.