According to  the continued fraction $[0, 1, 2, 3, 4...]$ is equal to $\frac{I_1(2)}{I_0(2)} \approx 0.697774657964...$, where $I_n$ is the Modified Bessel Function of the First Type.  This is rather surprising to me, I can't think of a good reason why Bessel Functions ought to be involved.  Why is this the case?
17100
Does this continued fraction converge?  $$\cfrac { 1 }{ 1+\cfrac { 1 }{ 2+\cfrac { 1 }{ 3+\cfrac { 1 }{ 4+\ddots  }  }  }  } $$  ($[0;1,2,3,4, \dots]$)  I tried approximating a few values but I couldn't make out whether it converges or diverges. Can anyone provide a proof whether this converges or diverges. If it converges, please add what it converges to.
28253
While I was looking at the values of the zeta function for the first natural numbers, I noticed that the sum of the values minus $1$, converge to $1$. Better put: $$\sum_{n=2}^{\infty} \left(\zeta(n)-1\right) = 1 $$ Furthermore, if you use only the even numbers for the zeta function, the sum will converge to $\frac{3}{4}$, or $$\sum_{n=1}^{\infty} \left(\zeta(2n)-1\right) = \frac{3}{4}$$  Leaving $$\sum_{n=2}^{\infty} \left(\zeta(2n-1)-1 \right)= \frac{1}{4}$$  This is probably common knowledge among mathematicians, but I couldn't find much about it on the internet. Is there a proof of this or perhaps even a simple explanation why this is so?
98951
There are so many available bases. Why is the strange number $e$ preferred over all else?  Of course one could integrate $\frac{1}x$ and see this. But is there more to the story?
110214
I need to fins the real vlaue of x for the continued fraction (Image attached)  I have tried partial coefficient method, but didn't get  the exact answer.  I there any way where we can identify the value of x algebrically?
114880
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.
37277
Every infinite continued fraction is irrational.  But can every number, in particular those that are not the root of a polynomial with rational coefficients, be expressed as a continued fraction?
122212
Let $\{a_n\}$ be a sequence of positive real numbers such that   $a_1 =1,\ \  a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.   Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...  (A) $(1,2]$, (B) $(2,3]$, (C) $(3,4]$, (D)$(4,5]$.   Solution attempt:   Firstly, we figure out what $\frac{a_{n+1}}{a_n}$ is going to look like. We get, from the recursive formula, $\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$ (remembering the fact that $a_n&gt;0$, the other root is rejected).   We know that, if $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}&gt;1$, then $\lim a_n \to \infty$.   Further, $(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}&gt;0$. (Again, the other root is rejected due to the same reason).   Hence, $(a_n)$ increases monotonically.   Therefore, the largest value of $\frac{a_{n+1}}{a_n}$ is approximately $1+\sqrt{1+\frac{1}{1}} \approx 2.15$  Now, the sum can be approximated as $\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$ (In actuality, $\mathbb{sum}&lt; 1.3$).  So, option $(A)$ is the correct answer.   Is the procedure correct?   I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.   Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?
24958
I found$\ \prod_{i=1}^\infty \frac{p_i}{p_i-1}\approx 11.0453$ on . Moreover, writing a paper, should one provide a proof or it is trivial? Thanks in advance.
33100
$1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ......$ Can anyone help me out how to solve this.   My try : I was thinking about the expansion of $tan^{-1}x$. But in that series  positive and negative will come alternatively.
34610
What is $1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cdots}}}$ ?  What is $1+\cfrac{2}{1+\cfrac{3}{1+\cdots}}$ ?  It does bear some resemblance to the continued fraction for $e$, which is $2+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\cdots}}}$.   Another thing I was wondering: can all transcendental numbers be expressed as infinite continued fractions containing only rational numbers? Of course for almost all transcendental numbers there does not exist any method to determine all the numerators and denominators.
11444
I'm having a hard time proving the following $$\int_0^{\infty} \frac{x^8 - 4x^6 + 9x^4 - 5x^2 + 1}{x^{12} - 10 x^{10} + 37x^8 - 42x^6 + 26x^4 - 8x^2 + 1} \, dx = \frac{\pi}{2}.$$  Mathematica has no problem evaluating it while I haven't the slightest idea how to approach it. Of course, I would like to prove it without the use of a computer. Is this an explicit form of a special function I fail to recognize?
14652
If I want to find the continued fraction of $\sqrt{n}$ how do I know which number to use for $a_0$? Is there a way to do it without using a calculator or anything like that? What's the general algorithm for computing it? I tried to read the wiki article but was overwhelmed and lost. I tried Googling but couldn't find a website that actually explained this question.  If anyone has a good site that answers these questions either, please let me know. Thanks!
42671
Say you have the infinitely continued fraction: $$x+\cfrac{1}{x^2+\cfrac{1}{x^3+\cfrac{1}{x^4+\ddots}}}$$ When $x=1$, you can see that it's $$1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ddots}}}$$ which converges upon the golden ratio $\phi = 1.61803398875...$. What if I were to plug in $x=2$? $$2+\cfrac{1}{4+\cfrac{1}{8+\cfrac{1}{16+\ddots}}}$$ Using a caluclator, it seems to converge upon the irrational number: $2.24248109286...$ Is there any way I can represent this irrational number algebraically?
90437
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find   $$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}.$$   I have tried adding the first two and the second two separately, then adding those sums but how do I get a numerical value as the answer?   Thanks
82582
$\lim_{n \to \infty}\left({^n\mathrm{C}_0}{^n\mathrm{C}_1}\dots{^n\mathrm{C}_n}\right)^{\frac{1}{n(n+1)}}$ is equal to: a) $e$ b) $2e$ c) $\sqrt e$ d) $e^2$  Though it looks really innocent at first sight, it certainly isn't. Attempt: It's $\infty^{\infty}$ form. I had tried taking the product raised to the power $\frac{1}{n(n+1)}$ as function $f(n)$. Then I took logarithm of both sides to see if things simplifying.  Even after factoring out the extra factorials it wasn't easy. Note that ${^n\mathrm{C}_x} = \binom{n}{x}$.
45472
Given a fraction:  $$\frac{a}{b}$$  I now add a number $n$ to both numerator and denominator in the following fashion:  $$\frac{a+n}{b+n}$$  The basic property is that the second fraction is suppose to closer to $1$ than the first one. My question is how can we prove that?  What I have tried:  I know $\frac{n}{n} = 1$ so now adding numbers $a$ and $b$ to it would actually "move it away" from $1$. But I cannot understand why $\frac{a}{b}$ is actually farther away from $1$ than $\frac{a+n}{b+n}$.  Why is that? What does it mean to add a number to both the numerator and denominator?
80740
Using 5000-digit precision in PARI/GP, I discovered that the fractional part of $(\sqrt{2}+\sqrt{3})^{2008}$ is extremely small, less than $10^{-999}$.  Is there a simple explanation for this fact ?  This looks like a Pisot number issue (similar questions have already been studied on MSE, see for example ), but it’s a more complicated situation.   Related :