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Here is an approach using the residue theorem. The residues of $\newcommand{\res}{\operatorname*{Res}}f(z)=\pi\cot\pi z\csc^2\omega\pi z$ at its poles are: $$\res_{z=0}f(z)=\frac{1-\omega}{3};\quad\res_{z=n}f(z)=\csc^2\omega\pi n;\quad\res_{z=n/\omega}f(z)=-\omega\csc^2\omega\pi n\quad(n\in\mathbb{Z}_{\neq 0})$$ (easy to obtain using power series; we keep in ...

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$\sum\limits_{k=1}^{\infty} k(a_{2k}+a_{2k+1})=\frac 1 2 \sum\limits_{k=1}^{\infty} (2ka_{2k}) + \frac 1 2\sum\limits_{k=1}^{\infty} (2k+1)a_{2k+1} -\frac 1 2 \sum\limits_{k=1}^{\infty} a_{2k+1}$ [as seen by splitting $\sum\limits_{k=1}^{\infty} (2k+1)a_{2k+1}$ into two parts]. Combining the first two terms here we get $\sum\limits_{k=1}^{\infty} k(... 0 Use the key formula $$\sum_{k=0}^p (-1)^k \binom{p}{k} \binom{k}{s} x^k = \binom{p}{s}(-x)^s (1-x)^{p-s}$$ Multiply both sides by$\sqrt{x}and integrate from 0 to 1. Use the Beta integral to get an answer in a ratio of gamma functions. Keep p as as a non-integer (the sum upper limit then runs to infinity) so that you don't get contradictory factors ... 1 One possible way is find the value of the geometric series $$G(x)=1+e^x+e^{2x}+\cdots+e^{nx}$$ using geometry. For example use the following type diagram with appropriate common ratio. Then compute $$G^{(9)}(0)=1^9+2^9+\cdots+n^9.$$ 2 We'll make use of $$\frac 1 {n + 1} = \int_0^1 x^{n}dx$$ and the well-known relationship between the Beta function with the Gamma function: $$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ $$\begin{split} \sum _{i=s}^p(-1)^i\binom{p}{i}\binom{i}{i-s}\frac{1}{2i+1} &= {p \choose s}\sum_{i=s}^p {p-s \choose i-s}\frac {(-1)^i} {2i+1}\\ &=(-1)^s{p \... 0 Recall that \binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c}, so$$\sum _{i=s}^p(-1)^i\binom{p}{i}\binom{i}{i-s}\frac{1}{2i+1}=\binom{p}{s}\sum _{i=s}^p(-1)^i\binom{p-s}{i-s}\frac{1}{2i+1}=(-1)^s\binom{p}{s}\sum _{i=0}^{p-s}(-1)^i\binom{p-s}{i}\frac{1}{2(i+s)+1}=\frac{(-1)^s}{2s+1}\binom{p}{s}\sum _{i=0}^{p-s}(-1)^i\binom{p-s}{i}\frac{1}{\frac{2i}... 1 Unfortunately the sum of Stirling Numbers, either 1st and 2nd kind, does not have a "closed" form, shorter than performing the sum directly. However for the "diagonal" sum might be interesting an identity which can be derived by the expression through the Eulerian Numbers of 2nd kind \eqalign{ & \left[ \matrix{ x \cr x - n \cr}... 1 I don't know if this answers your question or not. The LHS is \frac{\pi}{2} which is well-known. Now you work on the RHS like this: \begin{eqnarray} \sum_{n=0}^\infty\frac{\sin(n\theta)}{n}&=&\Im\sum_{n=0}^\infty\frac{e^{n\theta}i}{n}\\ &=&-\Im\ln(1-e^{\theta i})\\ &=&-\Im\ln[(1-\cos(\theta))-i\sin\theta]\\ &=&-\Im\ln\left\{... 1 By C-S\sum_{cyc}\frac{7+2b}{1+a}=\sum_{cyc}\left(\frac{7+2b}{1+a}-\frac{7}{2}\right)+\frac{21}{2}=\frac{1}{2}\sum_{cyc}\frac{4b+7(1-a)}{1+a}+\frac{21}{2}==\frac{1}{2}\sum_{cyc}\frac{11b+7c}{1+a}+\frac{21}{2}=\frac{1}{2}\sum_{cyc}\frac{(11b+7c)^2}{(11b+7c)(1+a)}+\frac{21}{2}\geq\geq\frac{1}{2}\frac{324(a+b+c)^2}{\sum\limits_{cyc}(11b+7c)(1+a)}+\... 1 we have to prove $$7\sum \frac{1}{1+a}+2\sum_{cyc}\frac{b^2}{b+ab}\ge 69/4$$ but byAM\ge HM$$$7\sum \frac{1}{1+a}\ge 63/4$$ also using$ab+bc+ca\le {(a+b+c)}^2/3=1/3$$$2\sum_{cyc} \frac{b^2}{b+ab}\ge 2\frac{(a+b+c)^2}{ab+bc+ca+a+b+c}\ge 6/4$$ The conclusion is now obvious 2 From Wolfram 3F2 list the formula $${}_{3}F_{2}(1, 1, c; 2, e; 1) = \frac{e-1}{c-1} \, ( \psi(e-1) - \psi(e - c) )$$ can be obtained for$c \neq 1$and$\psi(x)begin the digamma function. Now, \begin{align} \sum_{k=1}^{\infty} \frac{\Gamma(k+a)}{k \, \Gamma(k+a+b)} &= \sum_{k=0}^{\infty} \frac{\Gamma(k+1) \, \Gamma(k+a+1)}{\Gamma(k+2) \, \Gamma(k+a+b+... 2 Titu's lemma gives us Since(p+q+r) ^2 \geq 3 (pq+qr+rs)$, so$3 \geq pq+qr+rs $and thus $$\sum \frac{q}{ 3 + p } \geq \frac{9}{9 + pq+qr+rs} \geq \frac{3}{4}.$$ The original problem could be approach in as similar manner. Then, since$ \frac{1}{3} \geq ab+bc+ca, thus $$\sum \frac{7+2b}{1+a} \geq \frac{23^2}{30 + 2ab+2bc+2ca} \geq \frac{69}{4}.$$ 1 The observed pattern is fine. We have \begin{align*} S(n)&=4\cdot 9^{n-1}+4\left(n\frac{9^{n-1}-1}{9-1}-\sum_{k=0}^{n-2}k\,9^k\right)\\ &=\frac{1}{2}9^{n-1}\left(n+8\right)-\frac{1}{2}n-4\sum_{k=1}^{n-2}k\,9^{k}\tag{1} \end{align*} In (1) we have collected terms and we start the indexk$with$1$, since the term with$k=0$is zero. In order to get a ... 1 Combinatorial proof The number$n \choose k$tells us in how many ways we can choose$k$elements out of$n$elements. Now split those$n$elements into two parts with$m$and$n-m$elements. To choose$k$elements out of the$n$elements we can choose$l \in \{0, \ldots, k\}$elements out of the part with$m$elements and$k-l$elements out of the part with ... 2 @cosmo5's tip lets us write the sum as$\sum_{n\ge1}\int_0^1x^{3n-2}(1-2x+x^2)dx$, which by the dominated convergence theorem is$$\int_0^1\frac{x(1-x)^2}{1-x^3}dx=\int_0^1\left(\frac{1+2x}{1+x+x^2}-1\right)dx=\left[\ln(1+x+x^2)-x\right]_0^1=\ln3-1.$$ 1 Hint : $$\frac{1}{3n-1}+\frac{1}{3n}+\frac{1}{3n+1}-\color{blue}{\frac{3}{3n}}$$ $$\frac{1}{3n-1}+\frac{1}{3n+1}-\frac{2}{3n}$$ $$\frac{1}{3n-1}+\frac{1}{3n+1}-\frac{1}{3n}-\frac{1}{3n}$$ $$\Big(\frac{1}{3n+1}-\frac{1}{3n}\Big)-\Big(\frac{1}{3n}-\frac{1}{3n-1}\Big)$$ 4 In my eyes, this is most naturally linked to Eisenstein series. For$\tau\in\mathbb{C}$with$\Im\tau>0$, and an integer$k>1$, these are defined by $$G_{2k}(\tau)=\sum_{m,n\in\mathbb{Z}}'(m+n\tau)^{-2k}:=\sum_{(m,n)\in\mathbb{Z}^2\setminus\{(0,0)\}}(m+n\tau)^{-2k}$$ (i.e.,$\sum'$means$\sum$over$(m,n)$excluding$m=n=0$), and satisfy the well-... 0 let there is a square grid of size$n$x$nwith two particles situated at the opposite corners so we find the no. of ways they can meet themselves if both of them have the same speed and they can move only up and down as they both have to cover equal distance they must meet on the diagonal so number of ways of meeting is $$\sum_{r=0}^{r=n} {\binom{n}{r}}^... 1 If |\{0, 1\}| = 2 then the sum in the question equals 2. It is similar to \sum_{k=1}^n 1 = n. 1 Proof by induction: I leave it up to to show that the relation is valid for n=1, it should be trivial. Induction step:$$S_n=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}}{4\sinh^2{u}}S_{n+1}=S_n+(n+1)\sinh2(n+1)u=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}}{4\sinh^2{u}}+(n+1)\sinh(2n+2)uS_{n+1}=\frac{n\sinh{(2n+2)u-(n+1)\sinh{2nu}}+4(n+1)\sinh^2u\sinh(2n+2)... 2 Core idea: show that $$\sum_{m,n,p}\frac1{2^m3^n5^p}=\sum_m\frac1{2^m}\cdot\sum_{n}\frac1{3^n}\cdot\sum_p\frac1{5^p}$$ 4 $$\sinh(x\pm y) = \sinh x \cosh y \pm \cosh x \sinh y$$ Hence \begin{align}&\quad(\sinh u)(2n+1)\cosh((2n+1)u)-\sinh((2n+1)u)\cosh u \\~\\&= (n+1)(\sinh u\cosh ((2n+1)u) - \sinh((2n+1)u)\cosh u) \\&\;\;+n(\sinh u\cosh ((2n+1)u) + \sinh((2n+1)u)\cosh u) \\~\\&=(n+1)\sinh(u-(2n+1)u)+n\sinh(u+(2n+1)u) \\~\\&=(n+1)\sinh(-2nu)+n\sinh((2n+2)u) \... 1 The recurrent equation is \begin{align} a_n-\dfrac{1}{4}a_{n-1}=\dfrac{1}{4}\left(\dfrac{2}{3}\right)^{n-1}, n=1,2,\ldots. \end{align} Solve the homogeneous equation, $$a_n-\dfrac{1}{4}a_{n-1}=0.$$ The characteristic equation is $$r-\dfrac{1}{4}=0$$ which gives $$r=\dfrac{1}{4}.$$ The solution of homogeneous equation is $$a_n^{(c)}=C\left(\dfrac{1}{4}\right)^... 0 The question is whether the following change of the order of integration and summation is valid:$$ \sum_{k=0}^\infty\int_0^\infty f_k(x)\,dx = \int_0^\infty \sum_{k=0}^\infty f_k(x)\,dx, $$where$$ f_k(x) = \frac{1}{k!}\sin(2(k-y)x)\,e^{-x}. $$Here y is treated as a fixed real number. If the series in question were merely a finite sum, then we could ... 1 The telescoping summation helps:$$a_n=\frac{1}{4}a_{n-1}+\frac{1}{4}\left(\frac{2}{3}\right)^{n-1},\frac{1}{4}a_{n-1}=\frac{1}{4^2}a_{n-2}+\frac{1}{4^2}\left(\frac{2}{3}\right)^{n-2},\frac{1}{4^2}a_{n-2}=\frac{1}{4^3}a_{n-3}+\frac{1}{4^3}\left(\frac{2}{3}\right)^{n-3},\cdot\cdot\cdot\frac{1}{4^{n-2}}a_2=\frac{1}{4^{n-1}}a_1+\... 2 Note that $$4^na_n-4^{n-1}a_{n-1}=\left(\dfrac{8}{3}\right)^{n-1}$$ now telescope. Add: Let me compete the computation to get a closed form. After taking the summation $$4^na_n-a_0=\sum_{k=1}^n\left(\dfrac{8}{3}\right)^{k-1}=\dfrac{1-\left(\dfrac{8}{3}\right)^{n}}{1-\left(\dfrac{8}{3}\right)}$$ and hence $$4^na_n=a_0+\dfrac{3}{5}\left(\left(\dfrac{8}{3}\... 1 Because$$\sum_{i=0}^{n-1}\left(1+\frac{1}{n}+\frac{1}{4n^2}\right)^i\left(1+\frac{1}{n}\right)^{n-1-i}=\sum_{i=0}^{n-1}\frac{\left(1+\frac{1}{2n}\right)^{2i}}{\left(1+\frac{1}{n}\right)^i}\left(1+\frac{1}{n}\right)^{n-1}==\sum_{i=0}^{n-1}\left(1+\frac{1}{4n^2+4n}\right)^i\left(1+\frac{1}{n}\right)^{n-1}>\sum_{i=0}^{n-1}1\left(1+\frac{1}{n}\right)^{n-... 1 $$\sum_{i=0}^{n-1}\left(1+\frac{1}{n}+\frac{1}{4n^2}\right)^i\left(1+\frac{1}{n}\right)^{n-1-i}\ge\sum_{i=0}^{n-1}\left(1+\frac{1}{n}\right)^i\left(1+\frac{1}{n}\right)^{n-1-i}=n\left(1+\frac{1}{n}\right)^{n-1}$$ 8 Leta_n$be defined by $$a_n = (1+\sqrt{2})^n + (1-\sqrt{2})^n.$$ Then it is straightforward to verify that $$a_0 = a_1 = 2, \qquad a_{n+2} = 2a_{n+1} + a_n.$$ In particular,$a_n$is integer for all$n \geq 0$. So, if$n \geq 1$, then $$(1+\sqrt{2})^{2n-1} = a_{2n-1} + (\sqrt{2}-1)^{2n-1}.$$ This and$0 < \sqrt{2}-1 < 1$together then show that ... 3 This is actually a simple geometric progression, in disguise. The first term in the sequence is$1+\sqrt2$and its fractional part is$\sqrt2-1$. But $$(\sqrt2+1)(\sqrt2-1)=1$$ Hence for any integer$k$, $$(\sqrt2+1)^k(\sqrt2-1)^k=1$$ and the fractional part of each$(\sqrt2+1)^k$is$(\sqrt2-1)^k$when$k$is odd. Now$(1\pm\sqrt2)^2=3\pm2\sqrt2$, so the ... 0 Yes, and the reason is simple - the addition rule. If you have: $$\int (du + dq)$$ We can break this integral up at the plus sign, givin: $$\int du + \int dq$$ Because these are equivalent, we can also reverse the process. A summation is just a whole lot of these squeezed together. So, therefore, you can bump the summation on either side of the integral ... 3 An alternative version. $$\int_0^x\sum_{n=1}^\infty \frac{(x^n)'}{(n-1)!}dx = \sum_{n=1}^\infty \frac{x^n}{(n-1)!} = x \sum_{n-1}^\infty \frac{x^{n-1}}{(n-1)!} = x \cdot e^x$$ Derivate the result and you get $$(x e^x)' = e^x+xe^x = (1+x)e^x$$ 3 Hint:$x^n=x\cdot x^{n-1}$the product rule says that$(x^n)'=(x\cdot x^{n-1})'=x^{n-1}+(n-1)x\cdot x^{n-2}.7 We have \begin{align} \sum_{n=1}^{\infty} \frac{nx^{n-1}}{(n-1)!} &= \sum_{n=1}^{\infty} \frac{(n-1+1)x^{n-1}}{(n-1)!}\\ &= \sum_{n=1}^{\infty} \frac{(n-1)x^{n-1}}{(n-1)!} + \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}\\ &= x\sum_{n=2}^{\infty} \frac{x^{n-2}}{(n-2)!} + e^x\\ &= e^x(x+1). \end{align} 0{\bf Hint:}\quad 0 \leq p^2 - 2pq+ q^2$. 0 Thanks to Kavi Rama Murthy, I understand how to deduce the Cauchy–Schwarz inequality ! Let$c_k = a_k/\sqrt{\sum|a_i|^2}$and$d_k = b_k/\sqrt{\sum|b_i|^2}$So we have: $$\sum_{k=1}^{n}|c_kd_k|\le\frac{1}{2}\left(\sum_{k=1}^{n}c_k^2+\sum_{k=1}^{n}d_k^2\right)$$ $$\left|\sum_{k=1}^{n}c_kd_k\right|\le\sum_{k=1}^{n}|c_kd_k|\le\frac{1}{2}\left(\sum_{k=1}^{n}c_k^... 0 The answer is \frac{1}{6}(n - 1)(4n^2 + 13n + 12) which is approximately \frac{2}{3}n^3 as noted. The summation term simplifies to 2k^2 - k(4n+ 5) + (2n^2 + 5n + 2) in descending powers of k. The two identities above can then be applied along with \sum_{k=1}^{n-1} 1 = n-1, to give the answer. In more detail, \sum_{k=1}^{n-1} 2k^2 - k(4n+ 5) + (2n^... 2 By the De Morgan's laws we obtain:$$\sum_{cyc}(A-B)+ABC=\sum_{cyc}(A(-B)+ABC)=\sum_{cyc}A(-B+BC)==\sum_{cyc}A(-B+B)(-B+C)=\sum_{cyc}A(-B+C)==\sum_{cyc}(A(-B)+AC)=\sum_{cyc}(A(-B)+AB)=\sum_{cyc}A(-B+B)=\sum_{cyc}A.$$I used the cyclic summation. For example,$$\sum_{cyc}(A\setminus B)=(A\setminus B)+(B\setminus C)+(C\setminus A)=(A\setminus B)\cup(... 1 EDIT: Oh I see you wanted if and only if statements. I guess need to say$x\in A\cup B\cup C \iff x\in A\vee x\in B\vee x\in C \iff \ldots. But the idea can be the same. Here is a similar proof for only two sets: $$\begin{split} (A\setminus B) \cup (B\setminus A) \cup (A\cap B) &= (A\cap B^c) \cup (B\cap A^c) \cup (A\cap B) \\ &= (A \cap (B \cup B^... 2 \newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\... 2 \sum\limits_{k=0}^{54}20^k is a geometric series, so it's \dfrac{20^{55}-1}{19}. 20^{55}=2^{55}\times10^{55}, so it concludes with a long string of 0s, and 20^{55}-1 concludes with a long string of 9s. Picture dividing 20^{55}-1 by 19 using long division. Once the long string of 9s starts, there will be a particular remainder, which is a ... 0 Hint:$$ \eqalign{ & {1 \over {n\left( \matrix{ n - 1 \cr q + k \cr} \right)}} = {{\Gamma \left( {q + k + 1} \right)\Gamma \left( {n - q - k} \right)} \over {n\Gamma \left( n \right)}} = \cr & = {\rm B}\left( {q + k + 1,n - q - k} \right) = \int_{t = 0}^1 {t^{\,q + k} \left( {1 - t} \right)^{n - \,q - k - 1} dt} = \cr & = \... 0 I'd say it starts withk=2$and the inner summation is for$i_1 = 1$up to$n$then$i_2 = 2$up to$n$and so on for all combinations of these indices beginning at$1$and also satisfying the condition$i_1 < \cdots < i_k$. The expansion would begin$\Delta \tilde{W}(c_1, c_2) + \Delta \tilde{W}(c_1, c_3) + \cdots + \Delta \tilde{W}(c_1, c_n) + \...

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$$n\binom{2n}n=n\cdot\dfrac{(2n!)}{n! n!}=2\binom{2n-1}{n-1}$$ Now like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots?$. comparing the expansion of $(1+x)^m$ with $$2\sum_{n=1}^\infty\binom{2n-1}{n-1} (-z)^{n-1}$$ $$mx=2\binom31(-z)^{2-1}\text{ and }\dfrac{m(m-1)}2x^... 1 By the generalized binomial theorem,$$(1+4z)^{-1/2}=\sum_{n\ge0}\frac{(-1/2)_n}{n!}4^nz^n=\sum_{n\ge0}\binom{2n}{n}(-z)^n.$$Differentiating,$$-2(1+4z)^{-3/2}=\sum_{n\ge0}\binom{2n}{n}(-1)^nnz^{n-1}=\sum_{n\ge1}\binom{2n}{n}(-1)^nnz^{n-1}.$$Multiplying by z,$$-2z(1+4z)^{-3/2}=\sum_{n\ge1}\binom{2n}{n}n(-z)^n.$$1 More generally$$\sum_{k=0}^n\sum_{i=0}^ka_{ki}b_ic_{ki}=\sum_{k=0}^n\sum_{i=0}^kb_ia_{ki}c_{ki}=\sum_{0\le i\le k\le n}b_ia_{ki}c_{ki}=\sum_{i=0}^nb_i\sum_{k=i}^na_{ki}c_{ki}.$$1 Imagine the terms in a matrix indexed by k for rows and i for columns. The right side finds the total by summing along the rows instead of down the columns. 2 Intuitively best seen by recognizing the double sum as single sum$$\sum_{(i,k)\text{ with }0\le i\le k\le n} 

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Okay I think I found a solution to my question. It's actually harder than expected. Anyone who wants to verify it is gladly welcomed. First of all, notice that the property we want to found is invariant by translation (i.e we can consider $z_i=y_i-Z$ for any $Z \in \mathbb{R}^d$). Therefore we can consider that $i=1$ and $y_1=0$ without loosing generality. ...

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Let $f(x) = \sum_{n=0}^{\infty} a_n x^n$ be a series. If it converges on an open interval $(-a,a)$, then $f$ is smooth and is equal to its Taylor series \begin{align} f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \end{align} By uniqueness of the coefficients of the Taylor series, one has \begin{align} \forall{n} \geqslant 0,~ n!a_n = f^{(n)}(0) \end{...

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