My attempt:  $1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{1.3.5}{3.6.9}+\frac{1.3.5.7}{3.6.9.12}....$  =$1+\frac{1}{6}$.$2\choose 1$+$\frac{1}{6^2}$.$4\choose 2$+$\frac{1}{6^3}$.$8\choose 4$+.......+$\frac{1}{6^{n-1}}$.$2n-2\choose n-1$+...........  But I am perplexed as to what should be my next step?  Options:  (A)$\sqrt{2}$  (B)$\sqrt{3}$  (C)$\sqrt{\dfrac{3}{2}}$  (D)$\sqrt{\dfrac{1}{3}}$  One edit $t_n=\dfrac{1.3.5.7.....(2n-3)}{3^{n-1}(1.2.3....(n-1))}$  $=\dfrac{(2n-2)!}{(6^{n-1}(n-1)!(n-1)!)}$  $=\dfrac{1}{6^{n-1}}$.$2n-2\choose n-1$  Is this going to help me?
6914
How to find infinite sum How to find infinite sum $$1+\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\dfrac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $$   I can see that 3 cancels out after 1/3, but what next? I can't go further.
68737
$$\frac1{(1-x)^3} = \sum_{n=0}^\infty \binom{n+2}2x^n$$  I am not sure how to begin. $$\begin{aligned}\frac1{1-x}&amp;= \sum_{n=0}^{\infty} x^n\\\frac1{(1-x)^2}&amp;=\sum_{n=0}^{\infty} nx^{n-1}\end{aligned}$$ Then you would get: $$\frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} nx^{2n - 1}$$ Which is nowhere near the same...?
38858
Prove that:     $$1-\frac 12+\frac 13-\frac 14+...+\frac {1}{199}- \frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}$$   I know only this method:   $\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+....=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-...$  But, unfortunately, I could not a hint.
15535
Problem: Show that if $x_n\to 5$ then $\dfrac{x_n +1}{\sqrt{x_n -1}} \to 3$.  So I know $|x-5|&lt;\epsilon$ and I need to show that $\left|\dfrac{x_n +1}{\sqrt{x_n -1}} -3\right| &lt; \epsilon$.  I tried applying the triangle inequality (and a bit of algebra) to get $$\left|\frac{x_n +1}{\sqrt{x_n -1}} -3\right| \leq |x_n -5| \left |\frac{1}{\sqrt{x_n -1}}\right| + \left|\frac{6}{\sqrt{x_n-1}} -3\right|.$$  From this I know that $|x_n -5| &lt;\epsilon$ and can show that $\left |\dfrac{1}{\sqrt{x_n -1}}\right|&lt;1$. So my only problem is the second part of the sum. I have no idea what to do with it to get everything to be less than $\epsilon$.
96105
I have the following question:          Find the value of:     $$ \frac {1} {\sqrt {4} + \sqrt {5}} +  \frac {1} {\sqrt {5} + \sqrt {6}} + ... +  \frac {1} {\sqrt {80} + \sqrt {81}}$$      The book already provide the answer for this question but it didn't include the ways. Hint or anything that will help is appreciated
36572
I would like to show that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$ holds for all natural numbers. I got stuck here:  $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}.$  I would appreciate your help.
42568
How to compute $\int \frac{-3}{2x^2-x-1}\ dx$?  I tried completing the square   and ended with :  $\frac{-3}{(2(x-1/4)^2-9/8)}$  Then I did U-sub for $u = x-1/4$  and got  $-3 \cdot 8 \cdot \int  1/(16u^2)-9 du$  What do I do from here?
114298
I'm currently a high school junior enrolling in AP Calculus, I found this website that's full of "math geeks" and I hope you can give me some clues on how to solve this problem.  I'm pretty desperate for this since I'm only about $0.4%$ to an A- and I can't really afford a B now...  The problem is to simplify:  $$\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}}$$  What I did, was using basic "limits" taught in class and I figured out that the denominator would just keep going like this and approaches $1$, so this whole thing equals $1$, but I think it's not that easy...
105485
The common difference is d  Multiplying and dividing by d  $$\frac 1d [\frac{3-1}{\sqrt 3 + \sqrt 1} + \frac{5-3}{\sqrt 5 + \sqrt 3}....]$$ $$=\frac 1d [\sqrt 3 - \sqrt 1 + \sqrt 5 - \sqrt 3........\sqrt {2n-1} - \sqrt {2n-2}]$$ $$\frac 1d [\sqrt {2n-1} - \sqrt 1]$$  Since n term will be $$a_n=1+(n-1)2$$ $$=2n-1$$  Final answer  $$\frac 12 [\sqrt{2n-1}-1]$$  This isn’t the right answer however. What am I doing wrong?
42887
$$ \left\lfloor\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +\dots+ \frac{1}{\sqrt{100}}\right\rfloor =\,? $$   I rationalized the denominator and then I think I should somehow group the numbers, but i don't know how.  Thanks in advance!
30027
A series is given as follows   $$\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots$$  Can you give me hints to get started finding its value? Thanks.
35034
The following expression:  $$\frac{\sqrt{4+h}-2}{h}$$  should be simplified to:  $$\frac{1}{\sqrt{4+h}+2}$$  (even if I don't agree that this second is more simple than the first).  The problem is that I have no idea of the first step to simplify that.. any help?
35641
I have the following problem:  Prove that if $k &gt; 1$, then $$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{4k}{(k^2 - 1)^2}$$ Hence simplify $$\sum\limits_{k = 2}^n \dfrac{k}{(k^2 - 1)^2}$$  My Work: $$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k - 1)^2 (k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k^2 - 2k + 1)(k^2 + 2k + 1)}$$ I could have continued and multiplied the denominator out, but, at this point, I'm thinking that I'm actually expected to have proceeded differently, rather than multiplying the denominator out fully? And I don't have solutions for this problem, so I can't check anything. I would appreciate it if someone could please take the time to explain this problem.
93378
$$\frac{1}{a} + \frac{1}{b} = \frac{1}{12}$$  I'm trying to find all integer solutions to this equation. I've played with this equation algebraically to see if I can figure out the restrictions on $a$ and $b$. For example, I know at least one of them must be divisible by $12$ because  $$   a + b = \frac{ab}{12}  $$  I'm wondering if there are more general approaches.
41206
$$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$  How to find sum of this series?  I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that  I am not able to solve this.
70693
How to compute $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}$$ &amp; $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}$$?  I understand that  $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}=(1^{1/2})(2^{1/4})(3^{1/8})\cdots$$  and  $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}=(1+(2+(3+(\cdots))^{1/8})^{1/4})^{1/2} $$  How to Proceed further?
89894
Evaluate : $$1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot 5\cdot 8}{6\cdot12\cdot 18}+\ldots$$  thought: I have tried to manipulate the series so that the sum upto n terms , can be simplified. But I can't figure out any profitable result.