Give a precise meaning to evaluate the following:   $$\large{\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dotsb}}}}}$$   Since I think it has a recursive structure (does it?), I reduce the equation to   $$ p=\sqrt{1+p} $$ $$ p^2=1+p $$ $$ p^2-p-1=0 $$ $$ p=\frac{1\pm\sqrt{5}}{2} $$  Did I do this right?
3167
(Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12)  For $c \gt 0$, consider the quadratic equation $x^2 - x - c = 0, x &gt; 0$.  Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and then, if $n$ is an index for which $x_n$ has been defined, defining  $$x_{n+1} = \sqrt{c+x_n}$$  Prove that the sequence $\{x_n\}$ converges monotonically to the solution of the above equation.  Note: The answers below might assume $x_1 \gt 0$, but they still work, as we have $x_3 \gt 0$.    This is being repurposed in an effort to cut down on duplicates, see here: .  and here: .
36774
Well, the original task was to figure out what the following expression evaluates to for any $n$.  $$\begin{pmatrix} 1 &amp; 1 \\ 0 &amp; 1\end{pmatrix}^{\large n}$$  By trying out different values of $n$, I found the pattern: $$\begin{pmatrix} 1 &amp; 1 \\ 0 &amp; 1\end{pmatrix}^{\large n} = \begin{pmatrix} 1 &amp; n \\ 0 &amp; 1\end{pmatrix}$$  But I have yet to figure out how to prove it algebraically.   Suggestions?
99792
Let $a\ge1$ and define the sequence $(x_n)$ recursively by: $$x_1 = \sqrt{a}$$ $$x_{n+1}= \sqrt{a+x_n}$$  Here's what I did:  Plugging in some values makes it seem as if the sequence is increasing. I first solve the equation $t=\sqrt{a+t}$, equivalently $t^2 -t -a =0$, to get an idea what's going on. The roots are $$t_{1,2}= \frac{1 \pm \sqrt{1+4a} }{2}.$$ Therefore, between the roots, that is for $$t\in \left( \frac{1 - \sqrt{1+4a} }{2}, \frac{1 + \sqrt{1+4a} }{2}\right)$$  we have $t^2-t-a \le 0 \iff t \le \sqrt{a+t}$. Here's my attempt at showing inductively that $$\forall n\in \mathbb{N}, \ x_n \le \frac{1 + \sqrt{1+4a} }{2}$$  For $n=1$, the statement is true (simple arithmetic). Suppose it's true for $n$. Consider  $$x_{n+1}= \sqrt{a+x_n}\le \sqrt{a+\frac{1 + \sqrt{1+4a} }{2}}=\frac{1 + \sqrt{1+4a} }{2},$$  because $\frac{1 + \sqrt{1+4a} }{2}$ satisfies $x=\sqrt{x+a}$.  So this establishes the sequence $(x_n)$ is bounded. Now, by induction again, I try to show it's monotonically increasing. The base case $x_1\le x_2$ is obvious. Assume (strong induction) $x_{n-1}\le x_n$ for all $n&gt;1$. But since we have $$x_1\le x_n \le \frac{1 + \sqrt{1+4a} }{2}, $$   then $x_n$ lies in the appropriate interval, so $x_n \le \sqrt{a+x_n}=x_{n+1}.$  Is this proof correct? Can the condition on $a$ be relaxed to $a&gt;0$? Maybe even small negative $a$? And could someone provide me with another proof of convergence of $x_n$? I absolutely hate this one!
87460
I want to show rigorously that this is 2. I'm sure there's a faster way than by trying to see if $\sqrt{5} = a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}$ (linear combination of the basis elements for $\mathbb{Q}(\sqrt{3}, \sqrt{2})$). I'm just not sure what.
53392
$$\lim_{n\rightarrow \infty}\left[{\sqrt{4n^2+n}-2n}\right]=\frac{1}{4}$$ I am trying to use the definition of the limit but have no idea how to simplify the expression with radical!  ---edit---  so by the definition, $\forall n &gt; N \rightarrow |a_n - a | &lt; \epsilon, \text{where}\, \ a_n=\sqrt{4n^2+n}-2n\ \, \text{and}\,\  a = \frac{1}{4}$  so after multiply the conjugate and negate $\frac{1}{4}$, I get $\frac{2n-\sqrt{4n^2+n}}{4(\sqrt{4n^2+n}+2n)}$  Since there is still radical in the numerator, I think I have multiply the conjugate again...right?? And then find some formula that is greater!!??? confuse me this real analysis!!
32456
I've got an equation that contains   $$x^p - 1$$   $x$ is any positive number (such as 2) and $p$ is a small positive number close to 0 (such as 0.001).  For some reason (that I may have known in High School), this can be approximated by   $$\ln(x) \times p$$  Can anyone explain why this approximation works?  Does it have something to do with approximating a natural log with a series?
102009
If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then  $$x^{2000}+\frac{1}{x^{2000}}=?$$  My try:  $$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$  Continuation ?
118005
I read  and found that they define summation via recursion, so I decided to formalize the proof that that this definition is actually valid. I've two questions:   Does my proof contain any error? Are there other simple ways to define summation?   Thank you so much!     Suppose that $(a_1,\cdots,a_n)$ is a finite sequence in $\mathbb N$. Show that there is a sequence $(s_1,\cdots,s_n)$ such that $s_1=a_1$ and $s_{i+1}=s_i+a_{i+1}$ for all $1\leq i&lt;n$.   My attempt:  We define mapping $f$ as follows: $f: \mathbb N\times\mathbb N\to\mathbb N\times\mathbb N: (i,a)\mapsto\begin{cases} (i+1,a+a_{i+1})&amp;\text{if }i&lt;n\\ (i+1,a)&amp;\text{if }i\geq n \end{cases}$  By recursion theorem, there is a unique sequence $(p_i\mid i\in\mathbb N)$ such that $p_0=(1,a_1)$ and $p_{i+1}=f(p_i)$. Let $\pi:\mathbb N\times\mathbb N\to\mathbb N$ be the projection to the second co-ordinate i.e. $\pi(i,a)=a$. Let $s_i=\pi(p_i)$ for all $1\leq i\leq n$, then $(s_i\mid 1\leq i\leq n)$ is the required sequence. It's clear from the definition of $s_i$ that $s_{i+1}=s_i+a_{i+1}$ for all $1\leq i&lt;n$.
33499
We're given a sequence defined by the recursive relation: $$a_n = \sqrt{a_{n-1}a_{n-2}}$$  $a_1$ and $a_2$ are positive constants. We have to show the following:   The sequences $\{ b_n \} = \{ a_{2n-1} \}$ and $\{ c_n \} = \{ a_{2n} \} $ are monotonic, and if one is increasing, the other is decreasing. The limit of the sequence $\{ a_n \} $ is $ \left(a_1a_2^2\right)^{\frac13} $     Now, I have proved the first part. Besides that, I have also proved a few other things:  If $a_1 &gt; a_2$, then:  a. $ \{ b_n \} $ decreases, and $ \{ c_n \} $ increases.  b. $ c_n &lt; b_n $  If $a_1 &lt; a_2$, we just flip $\{ b_n \}$ and $\{c_n\}$  Besides, I have also shown that both the sequences : $\{b_n\}$ and $\{ c_n\}$ have the same limit. What I don't know, is how to evaluate the limit.
88282
Find $\displaystyle \lim_{n\to\infty}2^n\underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}_{n \textrm{ square roots}}$.   By geometry method, I know that this is $\pi$. But is there algebraic method to find this ?   Thank you.
119711
I'm trying to evaluate the limit     $$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$   I used  to find the result, which gives  $$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$  Then, plugging the value $1$ for $x$, you get $\frac{1}{3}$.  I don't see how did they reach that conclusion. This is how I tried to tackle it:  $$\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} = \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} \cdot \frac{\sqrt[3]{x}+1}{\sqrt[3]{x}+1},$$  which then yields  $$\frac{x-1}{(2\sqrt{x}-2)(\sqrt[3]{x}+1)},$$  and that becomes  $$\frac{x-1}{2\cdot(\sqrt{x}-1)\cdot(\sqrt[3]x+1)}.$$  That's  $$\frac{x-1}{2\cdot(\sqrt[6]{x}+\sqrt{x}-\sqrt[3]{x}-1)},$$  and this will still evaluate to $\frac{0}{0}$.  How did they solve this, exactly?
80157
Question:     show that: the beautiful ${\tt sqrt}$-identity:      $$ \left({2 \over \sqrt{\vphantom{\Large A}\, 4\ -\ 3\,\sqrt[4]{\,5\,}\ +\ 2\,\sqrt[4]{\,25\,}\ - \,\sqrt[4]{\,125\,}\,}\,}\ -\ 1\right)^{4} =5 $$   Can you someone have methods to prove this by hand? (Maybe this problem have many methods?because this result is integer. It's a surprise to me.) Thank you   Because I found this $$4\ -\ 3\sqrt[4]{\,5\,}\ +\ 2\sqrt[4]{\,25\,}\ -\ \sqrt[4]{\,125\,}$$ is not square numbers.
72935
I have a sequence: $1,3,5,8,10,14,16,20,23,27,\dots$  I know that the recursive relation is:  $$p[i] := p[i-1] + \text{number of factors of $i$}, \quad \text{with $p[1]=1$.}$$  How do I solve this recursive relation in $O(n^{0.5})$ time? I have to write a program for it, and $n$ can be as large as $10^9$. So, some appropriate precomputation and prestorage is allowed, provided that they comply (memory and time wise) with the range of $n$.
91740
Rationalize   $\dfrac{1}{\sqrt[3]{p^2}+\sqrt[3]{pq}+\sqrt[3]{q^2}}.$  How would I go about doing this without wading through lots of algebra?  Is there a trick similar to how you can multiply by $\dfrac{\sqrt a-\sqrt b}{\sqrt a-\sqrt b}$ with square roots?  Thanks in advance!
26583
I found this problem on a well-known problem solving website. It is apparently from Ramanujan.     With   $$\LARGE{a = \frac{1}{1 + \frac{e^{-2\pi}}{1 + \frac{e^{-4\pi}}{1 + \ddots}}}},$$   what is $a$?   The basic methods I know for questions of this type were not helpful to me.  I do not require a full solution, but feel free to post one if you would like. Hints or lecture notes on how to solve problems of this nature would be appreciated.   Keep in mind that I am at an undergraduate level in mathematics.
113257
prove that $[\sqrt{x}+\sqrt{x+1}]=[\sqrt{4x+2}], x  \in \Bbb N$.  Could someone help me to solve. I try many ways but I can't solve it ( I tried : Let $x=n^2 +k$ and don't know what to do afterwards).
5287
Let $x_{n} = \sqrt{1 + \sqrt{2 + \sqrt{3 + ...+\sqrt{n}}}}$  a) Show that $x_{n} &lt; x_{n+1}$  b) Show that  $x_{n+1}^{2} \leq 1+ \sqrt{2} x_{n}$  Hint : Square $x_{n+1}$ and factor a 2 out of the square root  c) Hence Show that  $x_{n}$ is bounded above by 2. Deduce that $\lim\limits_{n\to \infty} x_{n}$ exists.  Any help? I don't know where to start.