here $d(n)$ counts the number of positive divisors of $n$.  I've tried 2 things:   Using Bell series. But then again it just showed me that the bell series of the square of a function is not the square of the bell series of the function. Expand it all and use combinatorics. If: $$n = \prod_{i=1}^r p_i^{\alpha_i} $$   then if $d \vert n$:  $$d = \prod_{i=1}^rp_i^{\beta_i}, 0\leq \beta_i \leq \alpha_i$$  so  $$\sum_{t \vert n}d(t) = \sum \prod_{i=1}^r (\beta_i + 1)$$ for all possible combitions of $(\beta_1,\beta_2,...,\beta_r)$ but I couldn't develop it very much
120789
Prove that if $n \in \mathbb{N}$, then $$\sum_{d|n}{(d(n))^3}=(\sum_{d|n}{d(n)})^2$$ where $d(n)$ is the divisor function. I have know only information is $d(n)=\sum_{d|n}{1}$.
37120
[Sorry, but I couldn't come up with a better title...]  I'm trying to prove that $$\prod_{i=1}^n (1+a_i) \leq 1 + 2\sum_{i=1}^n a_i$$ if all $a_i$ are non-negative reals with $\sum_{i=1}^n a_i \leq 1$ and it seems I'm stuck.  (And, no, this is not homework!)
37253
Describe all positive integers,n such that $3^{n+1}$divides $2^{3^n}+1$. I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,$3^{n+1}$ divides $2^{3^n}+1$. Kindly clarify this doubt and if it's the former part, please verify my solution-n=1.
22623
I'd like to have my proof verified and if possible, to see other solutions that are interesting.  Proof: Suppose $n(n+1)$ is a square. Then we write $$n(n+1) = \prod_{p} p^{c(p)}$$ where $c(p) = a(p) + b(p)$ are such that \begin{align*} n &amp;= \prod_p p^{a(p)} \\ n+1 &amp;= \prod_p p^{b(p)} \end{align*} Now, by our hypothesis, $c(p)$ is even for all primes $p$. As $(n,n+1)=1$ for all $n$, it must be that $a(p)$ and $b(p)$ are even for all primes $p$ and moreover, $a(p) = 0$ whenever $b(p)&gt;0$ and reversely.  This indicates that both $n$ and $n+1$ are squares. This is impossible as there are no consecutive squares in the natural numbers.
14944
In the proof of some proposition, it appears that the following statement should hold:     $$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\beta} =2^{2n}. $$   However, using the definition of combinatorials does not help:  $$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\beta} =  \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} \frac{(n+1)!}{(n+1-r)!r!} \frac{n!}{(n-\beta)! \beta!}.$$  I suspect that this has something to do with the binomial expansion. Any ideas?
8987
If $\lambda_1,\dots,\lambda_n$ are distinct positive real numbers, then  $$\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1.$$ This identity follows from a probability calculation that you can find at the  top of page 311 in the 10th edition of Introduction to Probability Models by Sheldon Ross.   Is there a slick or obvious explanation for this identity?   This  question is sort of similar to ; clearly algebra is not my strong suit!
36221
Let $d(n)$ be the number divisors of a natural number $n$. Now let $m$ be a natural number. Find least natural number $n$ such that $d(n)=m$.  Thank you
3137
If in the series  $1-\dfrac {1} {2}+\dfrac {1} {3}-\dfrac {1} {4}+\ldots $ the order of the terms be altered, so that the ratio of the number of positive terms to the number of negative terms in the first $n$ terms is ultimately $a^{2}$, show that the sum of the series will become $\log \left( 2a\right) $.  Solution attempt. Let $p$ be the number of positive terms in the first $n$ terms of the reordered series so based on the question we are allowed $a^{2}=\dfrac {p} {n-p}$. Now solving for p we get $p=\dfrac {a^{2}n} {\left( 1+a^{2}\right) }$ and $n-p=\dfrac {n } {\left( 1+a^{2}\right) }$.  We also observe that in the original series only the odd terms are positive and only the even terms are negative.  Let's define $S_{odd}=1+\dfrac {1} {3}+\dfrac {1} {5}+.\ldots +\dfrac {1} {2n-1}$ and $S_{even}=\dfrac {-1} {2}\dfrac {-1} {4}\ldots -\dfrac {1} {2n}$  $S_{Reordered}=S_{odd_{p}}+S_{even_{n-p}}$  $S_{Reordered}=\sum _{t=1}^{t=P}\dfrac {1} {2t-1}+\sum _{t=1}^{t=(n-P)}\dfrac {-1} {2t}$ $S_{Reordered}=\sum _{t=1}^{t=\dfrac {a^{2}n} {\left( 1+a^{2}\right) }}\dfrac {1} {2t-1}+\sum _{t=1}^{t=\dfrac {n } {\left( 1+a^{2}\right) }}\dfrac {-1} {2t}$    In order to extend $S_{Reordered}$ from n terms to an infinite length. I guess i should take the limit of the $S_{Reordered}$ as $n\rightarrow \infty $ which would make the upper values of $t$ (Not sure of technical term here) to be $\infty$ giving me   $S_{Reordered}=\sum _{t=1}^{t=\infty}\dfrac {1} {2t-1}+\sum _{t=1}^{t=\infty}\dfrac {-1} {2t}$   and i have lost the $a$ from the expression. I guess i am stuck i need to perform some step to capture a before taking the limit. Any help would be much appreciated.
73822
Prove that, for every natural n, $\sum_{d|n} \frac{\mu(d)}{d}=\prod_{p|n}(1-\frac{1}{p})$ and conclude that $\phi(n)=n \sum_{d|n} \frac{\mu(d)}{d}$.  Proof of $\sum_{d|n} \frac{\mu(d)}{d}=\prod_{p|n}(1-\frac{1}{p})$:  I am havent trouble doing the proof of this statement,  Proof of $\phi(n)=n \sum_{d|n} \frac{\mu(d)}{d}$:  By definition, the Euler phi-function is given by $\phi(n)=\sum_{m \leq n,(m,n)=1}1.$ Eliminating the coprimality condition $\gcd(m,n)=1$ yields  $\phi(n)=\sum_{m \leq n} \sum_{d| \gcd(m,n)} \mu(n)=\sum_{d|n} \mu(n) \sum_{m \leq n, d|m}1 = \sum_{d|n} \mu(d) \frac{n}{d}$
27508
I've seen a question in my math book with this explanation above it: "If the product of n positive real numbers is 1, then the sum of these numbers must be more than n". I was wondering if this is correct, and if it is, could someone give an explanation or a proof for it?
1448
Apparently $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$.  How? What's the proof? Or maybe it is self apparent just looking at the above?   PS: This problem is known as "The sum of the first $n$ positive integers".
10928
Given an integer $n$, we are asked to investigate about the existence of integer divisors of $n^2+1$ of the form $4k+3$. Can you provide some insights about it?
73735
If $D_{n}$ is the Dihedral group of degree $n$ (order $2n$) where n is an odd number, can I then conclude that the numbers of elements of order 2 in $D_{n}$ is equal $n?$  I suppose there are the following types of elements in $D_{n}$: b, $a^{i}$, $a^{i}b$ for some $0\leq i &lt; n$ .
35659
The question is:  Develop a formula for $\delta_2(n)$, the sum of the squares of the positive divisors of $n$.
54001
I am trying to show $$\sum_{d \leq x} \mu(d)\left\lfloor \frac{x}{d} \right\rfloor = 1 \;\;\;\; \forall \; x \in \mathbb{R}, \; x \geq 1 $$ I know that the sum over the divisors $d$ of $n$ is zero if $n \neq 1$. So we can rule out integers that are divisors of $x$ (if $x &gt; 1$). I am not sure if I am on the right track to prove this. Any hints would be helpful. This is homework.
12931
$$(\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+(\sqrt 5-\sqrt 4)…$$  I have found partial sums equal to natural numbers. The first 3 addends sum to 1. The first 8 sum to 2. The first 15 sum to 3. When the minuend in an addend is the square root of a perfect square, the partial sum is a natural number. So I believe this series to be divergent.  Am I right? Have I used correct terminology? How would this be expressed using sigma notation? Is there a proof that this series diverges?
108730
I have found two formulas. I want to connect them!  The number of ways in which a given positive integer $n≥3$ can be expressed as a sum of three positive integers $x,y,z$ (i.e. $x+y+z=n$) , subject to $x≤y≤z$ is the integer closest to $\frac{n^2}{12}$. If $x+y+z=n$ where $x,y,z$ are postive integers then the number of ways is ${n-1 \choose 2}$ .  The difference between these two formula is the condition $x≤y≤z$. Can we connect these two ?? Please help me! I mainly wants to know how one implies the other.