I found this formula in a textbook in which the proof to the formula was not given  Ramanujam's formula $$\sqrt{1 +n\sqrt{1 +(n+1)\sqrt{1 + (n+2)\sqrt{1 + (n+3)\sqrt{1 +....\infty}}}}} = n+1$$ Its a great equation andhow do you prove this. its a bit difficult for me and tried different methods to solve this. Iam an undergraduate and I want you to elaborate the method of solving if its complex.
44967
I think I have sort of a proof of the following nested radical expression due to Ramanujan for $x\ge 0$.     $$\large x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}$$ for $ x\ge -1$   I just want to know if my proof is okay or there is a flaw, and if there is one I request to give some suggestions to eliminate them. Thank you. The proof is the following:  Proof: Let us define $$ a_n(x)=\underbrace{\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}}_{n \ \mbox{terms}}$$ for $x\ge 0$ so that $$ a_1(x)=\sqrt{1+x},\ a_2(x)=\sqrt{1+x\sqrt{1+(x+1)}},\ a_3(x)=\cdots$$    and so on. Since $x\ge 0$ each one of the $a_n$ is defined (I am taking only the positive square root). Also, we note that $$a_{n+1}^2(x)=1+xa_{n}(x+1)$$  Now, we note that $\{a_n(x)\}$ is an increasing sequence and that $$a_n(x)&lt;x+1$$ $\forall n\ge 1$, this is easy to prove by induction as below:  For $n=1$, $a_1(x)=\sqrt{1+x}&lt;1+x$ since $x\ge 0\Rightarrow 1+x\ge 1$. SO it is true for $n=1$. Similarly, the truth can be proved for $n&gt;1$.  Then $a_n(x)$ converges to $$l(x)=\sup_{n}a_n(x)\le x+1$$ Now I make the following claim:  Claim: $l(x)=x+1\quad \forall i\ge 0$  Proof: Fix $x$. Let $l(x)&lt;x+1$. Then, $l(x)=x+1-\epsilon$ for some $\epsilon&gt;0$. Now, I claim that there must be a $n$ such that $$x+1-a_n(x)&lt;\epsilon$$, and if that is true then $$a_n(x)&gt;x+1-\epsilon=l(x)$$ which is a contradiction since $$l(x)=\sup_{n}a_n(x)$$ and then it implies that $$l(x)=x+1$$ To prove my claim it requires $$x+1-a_n(x) &lt;  \epsilon$$ Now, \begin{align} x+1-a_n(x) = &amp; x+1-\sqrt{1+xa_{n-1}(x+1)} \\ \ =&amp; \frac{(x+1)^2-({1+xa_{n-1}(x+1)})}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\ \ =&amp; x\frac{(x+1)+1-a_{n-1}(x+1)}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\ \ &lt;&amp; \frac{x}{x+2}((x+1)+1-a_{n-1}(x+1))\\ \ &lt;&amp; \frac{x}{x+2}\cdot\frac{x+1}{x+3}((x+2)+1-a_{n-2}(x+2))\\ \ &lt;&amp; \frac{x}{x+n-1}\cdot \frac{x+1}{x+n} ((x+n-1)+1-a_{1}(x+n-1))\\ \ &lt;&amp; \frac{x}{x+n-1}\cdot \frac{x+1}{x+n}(x+n-\sqrt{x+n})\\ \ &lt;&amp; \frac{x(x+1)}{x+n-1} \end{align}  Now, if one is able to find $n$ such that $$\frac{x(x+1)}{x+n-1}&lt;\epsilon \Rightarrow x&lt; \frac{-(1-\epsilon)+\sqrt{(1-\epsilon)^2+4\epsilon(n-1)}}{2}$$ then we're done.   Now from the upper bound it seems that there always exists some $n$ that satisfies this requirement. Hence the claim is proved.
42675
I'm trying to find a formula for the following sequence:  $\{\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}},...\}$  I thought of solving it recursively and I got this formula:  $a_{n}=\sqrt{3*a_{n-1}}$  $a_{0}=1$  Is there a better and non-recursive formula for the given sequence?
28325
Possible Duplicate:                I know that the sum of the squares of the first n natural numbers is $\frac{n(n + 1)(2n + 1)}{6}$. I know how to prove it inductively. But how, presuming I have no idea about this formula, should I determine it? The sequence $a(n)=1^2+2^2+...+n^2$ is neither geometric nor arithmetic. The difference between the consecutive terms is 4, 9, 16 and so on, which doesn't help. Could someone please help me and explain how should I get to the well known formula assuming I didn't know it and was on some desert island?
122311
$$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$$  I tried to say we can erase the $1$ from the equation, as it's a constant. But I don't know how to do the rest without running into this mistake: $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-n=\frac{\sqrt[3]{\frac{n^3}{n^3}+\frac{n^2}{n^3}}-\frac{n}{n}}{\frac{1}{n}}=\frac{1-1}{0}$$
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.
76052
I am not mathematician by any means so this question might be rather stupid. I came across this Wikipedia article on  and found this bewildering formula,   $$1 + 2 + 3 + \dots  = - \frac1{12}$$  The article also says that "Ramanujan summation of a divergent series is not a sum in the traditional sense". I am wondering then what this summation actually implies?  Is there any interpretation of this confounding result in the physical world that someone not so math-savvy can understand?
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Does anyone have any idea how to prove that  $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}$$
98463
I want to express $$\sum_{n=-\infty}^\infty \dfrac{1}{(z+n)^2+a^2}$$ in closed form. What comes to mind is the formula $$\pi\cot\pi z = \dfrac{1}{z}+\sum_{n\ne 0}\left(\dfrac{1}{z-n}+\dfrac1n\right)=\dfrac{1}{z}+\sum_{n=1}^\infty\dfrac{2z}{z^2-n^2}$$  and also $$\dfrac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty \dfrac{1}{(z-n)^2}.$$ But neither of these gives the term $(z+n)^2+a^2$ that we want. Perhaps we can adjust somehow?
86635
I try to solve the following sum:  $$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$$  I'm very curious about the possible approaching ways that lead us to solve it. I'm not experienced with these sums, and any hint, suggestion is very welcome. Thanks.
46297
Find $$M:=\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}$$   There's a solution  that uses complex numbers which I didn't understand and I was wondering if the following is also a correct method.   My proposed solution   $$\begin{align} &amp;\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}\\ =&amp;\sum_{n=1}^{\infty}\tan^{-1}\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\ =&amp;\sum_{n=1}^{\infty}(\tan^{-1}(1+n)+\tan^{-1}(1-n))\\ =&amp;\sum_{n=1}^{\infty}(\tan^{-1}(n+1)-\tan^{-1}(n-1)) \end{align} $$  And this implies  $$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\frac{3\pi}{4}$$
94022
Possible Duplicate:     I've tried to work on this equation for about 4-5 Hours.  I'm trying to show that the following equation is true with induction: $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}&gt;2(\sqrt{n+1}-1)$  after about 4-5 hours and numerous attempts I couldn't really get it to work out.  Any ideas ?
46416
Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical $$ \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}} $$ Taking a cue from Ramanujan's solution method, I defined a function $f(x)$ such that $$ f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} $$ We can see that $$\begin{align} f(0) &amp;= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\ &amp;= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}} \end{align}$$ And we begin solving by $$\begin{align} f(x) &amp;= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\ f(x)^2 &amp;= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\ &amp;= 2^x + f(x + 1) \\ f(x + 1) &amp;= f(x)^2 - 2^x \end{align}$$ At this point I find myself stuck, as I have little experience with recurrence relations.  How would this recurrence relation be solved? Would the method extend easily to $$\begin{align} f_n(x) &amp;= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\ f_n(x)^2 &amp;= n^x + f_n(x + 1)~\text ? \end{align}$$
119726
A1. $(150,9,1,0,0,0,0,0,1,1,6,33)$ Let $$T_0=2,\quad T_1=3,\quad T_2=6,$$ and for $n\ge3$, $$T_n=(n+4)T_{n-1}-4nT_{n-2}+(4n-8)T_{n-3}.$$ The first few terms are $$2,3,6,14,40,152,784,5168,40576,363392.$$ Find, with proof, a formula for $T_n$ of the form $T_n=A_n+B_n$, where $(A_n)$ and $(B_n)$ are well-known sequences.  I found that: $$T_n - n! = 2^n$$ and verified that $T_0 = 2^0 + 1 = 2$ Hence, $$T_n = 2^n + n!$$ I need to prove that: $$T_{n+1} = 2^{n+1} + (n+1)!$$ Which I can't do. I went back and stated: $$T_{n+1} = nT_{n} + T_n + 4T_n - 4nT_{n-1} - 4T_{n-1} + 4T_{n-2} + T_{n-2} - 8T_{n-2}$$ What to do next?
25721
$$\sum_2^\infty e^{3-2n}$$  The formulas for these things are so ambiguous I really have no clue on how to use them.  $$\frac {cr^M}{1-r}$$  $$\frac {1e^2}{1-\frac{1}{e}}$$  Is that a wrong application of the formula and why?
43804
It is an interesting task to try finding the limit of nested square root expressions.      $$\lim_{n \to \infty}\left( 1 + \sqrt{2 + \sqrt{3+ ... + \sqrt {n + \sqrt{n+1}}}}\right)$$   How to solve this one?
105500
$\sum_{n=1}^{\infty}{\frac{1}{\sqrt{1}+\sqrt{2}+...+\sqrt{n}}}$ $\\$  I tried a comparison test, Leibniz, Cauchy, Cauchy-integral and D'Alembert criteria but nothing gives me the result. Any hints would be helpful!
77004
Possible Duplicate:          I once read that some mathematicians provided a very length proof of $1+1=2$.  Can you think of some way to extend mathematical rigor to present a long proof of that equation? I'm not asking for a proof, but rather for some outline what one would consider to make the derivation as long as possible.  EDIT: It seems the proof I heard about is a standard reference given here multiple times :) I stated that the proof itself is less useful than an outline for me as I know only "physics level maths". Can someone provide a short outline what's going on in the proof? Some outline I can look up section by section in Wikipedia to at least get a feel of what could be needed to make such a proof?