I am looking for the best way to format fractions in the upper limit of summations.  In particular, the fraction is the argument of the floor function.  In the following example, the floor function and fraction appear disproportionately large.  \begin{equation}   \sum_{k=1}^{\left \lfloor{\frac{n+1}{2}}\right \rfloor}\frac{1}{2k-1} \end{equation}   I'm sure there is a better way to achieve this result...
122199
When I write  \lfloor\dfrac{1}{2}\rfloor   the floors come out too short to cover the fraction. How can I lengthen the floor symbols?
37354
$$ \sum_{i=1}^N\left\lfloor\frac{N}{i}\right\rfloor $$ Is there a closed form expression to the above sum? (Mathematica doesn't give me anything)
118599
I am working on an assignment and have to write a proof for the following:  $$\lfloor 4x \rfloor =  \lfloor x \rfloor + \lfloor x+\frac{1}4 \rfloor + \lfloor x+\frac{1}2 \rfloor + \lfloor x+\frac{3}4 \rfloor$$  I have to do by proof by case I know the cases to be as follows  $$0 \le x-\lfloor x \rfloor \lt \frac 1 4$$ $$\frac 1 4 \le x-\lfloor x \rfloor \lt \frac 1 2$$ $$\frac 1 2 \le x-\lfloor x \rfloor \lt \frac 3 4$$ $$\frac 3 4 \le x-\lfloor x \rfloor \lt 1$$  The problem is I don't understand how I can express this.  Case 0 In this case I know that it would be $4n$ here  Case 1 In this case I know that it would be $4n+1$  Case 2 Here it would be $4n+2$  Case 3 Now the last case would end up being  $4n+3$  Can someone help me express this?
31020
I am trying to size the numerator and denominator of the below equation so that they are equal in size.  I have not had much luck suggestions are appreciated.  \begin{equation} \sum_{i=1}^{n} \frac{\frac{cf_{n}}{(1+i)^n}} {\displaystyle\sum_{i=1}^{n}\frac{cf_n}{(1+i)^n}} , n_{i} \end{equation}
44239
How to solve $$\sum_{i=1}^n \lfloor e\cdot i \rfloor $$    For a given $n$.    For example, if $n=3$, then the answer is $15$, and it's doable by hand. But for larger $n$ (Such as $10^{1000}$) it gets complicated . Is there a way to calculate this summation?
67206
I am writing a short discussion of Farey numbers and was wondering if there are any examples of when the mediant function is ever actually equal to the sum of the two fractions in the usual sense? (Not to produce Farey numbers, I just thought it might be an amusing way to introduce Farey addition).   Explicitly: Are there any examples of fractions $\frac{a}{b}$ and $\frac{c}{d}$ where  $$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$$ for positive integers?   Currently I have found the example  $$\frac{1}{1}+\frac{1}{i} =\frac{1+1}{1+i}$$ if we remove the integers requirement but it would be nice to find a case that didn't involve complex numbers though!
58261
Need help computing: $\displaystyle\lim _{n\to\infty}\sum _{k=1}^{n}\frac{\sqrt{k}}{{n}^{\frac{3}{2}}}$  Now my intuition is that using Stolz-Cesaro  $\displaystyle\lim _{n\to\infty}\sum _{k=1}^n\frac{\sqrt{k}}{n^{\frac{3}{2}}}=\lim _{ n\to\infty}\frac{1}{n}\sum _{k=1}^n\sqrt{\frac{k}{n}}=1$  Is it correct or not?
86503
I am trying to compute $$\sum_{k=1}^{n-1} \frac{1}{k(n-k)}$$ using a term by term partial fraction decomposition and also with generating functions, but I'm stuck.
37404
In my assignment I have to prove that the following function doesn't have a limit:  $$f(x)=\frac{x}{x-\lfloor{\sin x}\rfloor}$$ when $${x \to 0}$$  In other words, I have to prove that for every $\delta$ such that $0&lt;|x-{x_0}|&lt;\delta,$ $|f(x)-L|&gt;\epsilon$.  In fact, I don't even know how to start. Do I choose my $\delta$? I know that the sin function has certain values that's supposed to help me. How do I use it?  I have to prove this by definition, so I can't use limit arithmetic for example  Thanks.
59750
I want some advice about how to format it better if I have to write a long text below a $\sum$ or a $\prod$ sign. Basically I want to explain the restriction over which the sum or the product is being taken.   For example how can I improve the looks of these equations,  $N_i (\rho) = \sum _ {B \in \rho s.t n_{\delta_j}(B) = 0    \forall j \neq i} n_{\delta_i} (B)$  $\sum _{ \rho s.t N_i(\rho) = a_i and M_i(\rho) = b_i}    \prod _{B \in \rho} \langle \prod_{i=1}^{n} \delta_i ^{n_{\delta_i}(B)}    y_i ^{n_{y_i}(B)} \rangle_c$
33516
Does anyone know the general strategy for summing a series of the form:  $$\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2},$$  where $a$ is a positive integer?  Any hints or ideas would be great!
115704
Prove using combinatorics   $$\sum\limits_{k=0}^n 2^k \binom{n}{k} \binom{n-k}{\lfloor{\frac{n-k}{2}}\rfloor}=\binom{2n+1}{n}$$  The right side seems easy for the right side we have floor function it seems we have to count the number of subsets that $A_i \notin A_j$ but when I try to prove I can't show the right side.Any hints?
69137
Does $\sum\frac{(-1)^{\lfloor \sqrt n\rfloor}}{n}$ converge ?  I know there is a technical method that uses an adapted Abel's rule  $(*)$ ; I would like to know if you see another method (which is understable with basic knowledge) to prove it converges. Thank you.  Remark: Actually, it can be proved that for all $z \in \mathbb C$ such that $|z| = 1$, $\sum\frac{(-1)^{\lfloor \sqrt n\rfloor}}{ n}z^n$ converges.  $(*)$ Consider two series $\sum a_n$ and $\sum b_n$. Define $B_m = \sum_{n=0}^mb_n$   If the sequence $(a_n)$ tends to $0$ and the sequence $(B_n)$ is bonded, and if the series $\sum (a_{n+1} - a_n)$ converges absolutely, then $\sum a_nb_n$ converges.  You have to adapt this theorem to use it here.
98997
We fix $y \ge 0$. What is the following limit?  $$\text{lim}_{n \rightarrow \infty}\frac{n!}{n^{\lfloor y\sqrt{n} \rfloor}(n-\lfloor y \sqrt{n}\rfloor)!}$$  My attempt so far is as follows:  By Stirling's formula, $n! \approx e^{-n}n^{n+1/2} \sqrt{2\pi}$, so that we can compute:  $$\text{lim}_{n \rightarrow \infty}\frac{e^{-n}n^{n+1/2} \sqrt{2\pi}}{n^{\lfloor y\sqrt{n} \rfloor}e^{-n+\lfloor y\sqrt{n} \rfloor}(n-\lfloor y\sqrt{n} \rfloor)^{n-\lfloor y\sqrt{n} \rfloor+1/2} \sqrt{2\pi}} \\   = \text{lim}_{n \rightarrow \infty}e^{-\lfloor y\sqrt{n} \rfloor} \left( \frac{n}{n-\lfloor y\sqrt{n} \rfloor} \right)^{n-\lfloor y\sqrt{n} \rfloor+1/2}$$  upon cancellation and rearrangement of terms.  This is endlessly confusing for me to compute at this point.  Does anyone know how to deal with the limit from here?  I'm asking this question as part of another:  for reference.
95733
Calculate $$\sum \limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$$  I use software to complete the  series is  $\frac{2}{27} \left(18+\sqrt{3} \pi \right)$  I have no idea about it.  :|
114099
Given the Series $$\sum_{k=1}^\infty \frac{1}{k(k+2)}$$  How exactly would I find out the limit is $\frac34$ as suggested by Wolframalpha? I already found out I can prove it actually converges by performing the comparison test and seeing that the underlying sequence isn't a null-sequence. But unfortunately I am absolutely clueless on how to prove that it converges to $\frac34$.  Regards,  Dennis
74423
If $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$$ Then $\lfloor A \rfloor\;\;,$ Where $\lfloor x \rfloor $ Represent floor fiunction of $x$   My Try:: Using $\bf{A.M\geq H.M\;,}$ We get  $$\frac{1980+1891+1982+....+2012}{33}&gt;\frac{33}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}$$  So $$\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}&lt;\frac{1980+1981+....+2012}{(33)^2}=\frac{1996}{33}\approx 60.5&lt;61$$  Now how can i prove that the above expression $A$ is $&gt;60$  Help me, Thanks