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Let , $a_n=\frac{4^n}{(2n)!}$. Then , $$\frac{a_{n+1}}{a_n}=\frac{4}{(2n+2)(2n+1)}\to 0 \text{ as } n\to \infty.$$ So, $$\lim_{n\to \infty}a_n=0.$$

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Here is one way. Call the term $a_n$. Then $$a_n=\frac4{1\times2}\frac4{3\times4}\frac4{5\times6}\cdots\frac4{(2n-1)(2n)}\ .$$ Now it is easy to see that every term except the first is less than $\frac12$, so $$0<a_n<\frac2{2^{n-1}}\ .$$ The RHS clearly has limit $0$, so by the pinching theorem (squeeze theorem, sandwich theorem) ...

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I don't see an application of l'hopital's rule here but using Stirling approximation you can deduce that the limit is zero. If you don't want to use any advanced results like Stirling, you can always go for an elementary proof like the following $$\frac{4^n}{(2n)!} \leq \frac{4^n}{(n+1)(n+2)...(2n)} \leq \frac{4^n}{(n+1)^n}$$ for $n \geq 4$ we have ...

0

There is no "simple number theory" proof anywhere and it would be nice if some "elementary" proof was presented. However there is an elementary proof here http://www.jstor.org/stable/1969454?seq=1#page_scan_tab_contents but is extremely complicated....

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This is not a novel result, but rather an alternate derivation of @Start wearing purple's result. I tried to keep everything simple and explained, which resulted in a slightly verbose solution. So if you are not interested in details, you may follow only the tagged equations until Step 3. Step 1 (Reduction of Integral). We begin with the following ...

5

Try writing out a few terms: $\left(\frac{4}{1}-\frac{4}{3}\right)+\left( \frac{4}{5}-\frac{4}{7} \right)+\left( \frac{4}{9}-\frac{4}{11} \right)+\cdots$. Then realize that you can rewrite the sum as $$4\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}$$ which is four times the series expansion of $\arctan$ evaluated at $1$. Since $\arctan 1= \frac{\pi}{4}$, we ...

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hint: $\dfrac{8}{(4n-1)(4n-3)} = 4\left(\dfrac{1}{4n-3}-\dfrac{1}{4n-1}\right)$

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This is not complete answer, but this answer is illustrative of one way we could try to attempt to answer the first question posted. Consider $\lbrace a_k \rbrace ^{p_x} _{p_{x-2}}$. $\lbrace a_k \rbrace ^{p_x} _{p_{x-2}}$ has possible two forms: $\lbrace 2, ..., 2,3,2,p_x,2,p_{x-1},2,3,2,...,2 \rbrace$ or $\lbrace 2, ..., ... 4 Hint: The inequality$\sin(x) \le x$holds for all$x \ge 0$. The second assertion is trivial. 0 I think what you mean is this. Using Faulhaber's formula, $$\sum_{i=1}^N i^{59} = \dfrac{N^{60}}{60} + \dfrac{N^{59}}{2} + \dfrac{59}{12} N^{58} + O(N^{57})$$ and thus for$N = 10^n$with$n$large, the highest-order digits of the sum are$1 6 \ldots 6 7 1 6 \ldots 671$with$n-36's in each group. Similarly $$\sum_{i=1}^N i^5 = \dfrac{N^6}{6} + ... 2 Using the representation$$\psi(x) = \int_0^{\infty} dt \left (\frac{e^{-t}}{t} - \frac{e^{-x t}}{1-e^{-t}} \right ) $$and changing the order of summation and integration, we get the following integral representation of the sum:$$\int_0^{\infty} dt \frac{e^{-t/2}}{\left ( 1+e^{-t/2} \right )\left ( 1+e^{-3 t/8} \right )} $$Subbing u=e^{-t/8}, we get ... 4 The series definition of the digamma function gives:$$\begin{eqnarray*} -S&=&\sum_{n\geq 1}\sum_{m\geq 1}\frac{32(-1)^n}{(8m+3n-7)(8m+3n-3)}\\&=&8\sum_{n\geq 1}\sum_{m\geq 1}(-1)^n\left(\frac{1}{8m+3n-7}-\frac{1}{8m+3n-3}\right)\\&=&8\int_{0}^{1}\sum_{n\geq 1}\sum_{m\geq ... 1 \begin{align} P_n&=b[P_{n-1}-a]\\\\ \underbrace{P_n-\frac b{b-1}\cdot a}_{Q_n}&=b\underbrace{\left[P_{n-1}-\frac b{b-1}\cdot a\right]}_{Q_{n-1}}\\\\ Q_n&=bQ_{n-1}=b^2 Q_{n-2}=\cdots=b^nQ_0\\\\ P_n-\frac b{b-1}\cdot a&=b^n \left[c-\frac b{b-1}\cdot a\right]\\\\ P_n&=b^nc-\frac {b(b^n-1)}{b-1}\cdot a \end{align} 3 HINT : $$P_n=bP_{n-1}-ab$$ Dividing both sides byb^n$gives $$\frac{P_n}{b^n}=\frac{P_{n-1}}{b^{n-1}}-\frac{a}{b^{n-1}}.$$ Now setting$Q_n=\frac{P_n}{b^n}$gives $$Q_n=Q_{n-1}-\frac{a}{b^{n-1}}$$ which should be easy to deal with. 0 $$\sum\limits_{k=1}^\infty \left(\frac{k\sin k}{2k+1}\right)^k$$ First note that $$\sum\limits_{k=1}^\infty\left|\frac{k\sin k}{2k+1}\right|^k\leq\sum\limits_{k=1}^\infty \left|\frac{k}{2k+1}\right|^k$$ Now lets use the root test on the series on the right hand side ... 0 Note that$|\sin k|\leqslant 1$for all$k$. So by the root test $$\limsup_{k\to\infty}\frac{k\sin k}{2k+1} \leqslant \limsup_{k\to\infty} \frac k{2k+1} = \frac12<1,$$ and hence the series converges. 0$\lim_{x \to \infty} \left( \frac{k \sin k}{2k+1} \right)$does diverge, but it always remains in the interval$[-\frac{1}{2}, \frac{1}{2}]$. Therefore, as we raise it to increasing powers, we do indeed get something which tends to zero. That means it's plausible for the sum to converge. Notice that $$\frac{k}{2k+1} \leq \frac{1}{2}$$ so the summand is ... 1 We have: $$\begin{eqnarray*} \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{(k+1)^{s-1}} &=& \int_{0}^{+\infty}\frac{x^{s-2}}{(s-2)!}\sum_{k=0}^{n}\binom{n}{k}(-1)^k e^{-(k+1)x}\,dx\\&=&\int_{0}^{+\infty}\frac{x^{s-2}}{(s-2)!}e^{-x}(1-e^{-x})^n\,dx\end{eqnarray*}$$ hence: $$\begin{eqnarray*} \sum_{n\geq ... 0 (not an answer, only a comment) A Pari/GP-routine gives with some quick checks a likely different constant. fmt(200,12) m=10000; err = sum(n=1,m,n*log(n)) - (m*(m+1)*log(m+1)/2 - m*(m+2)/4 + log(m+1)/12 ) m=20000; err = sum(n=1,m,n*log(n)) - (m*(m+1)*log(m+1)/2 - m*(m+2)/4 + log(m+1)/12 ) m=30000; err = sum(n=1,m,n*log(n)) - ... 0 The binomial coefficient hints at using an exponential generating function:$$ \widehat{T}(z) = \sum_{n \ge 0} t(n) \frac{z^n}{n!} $$Shift indices and reorganize:$$ 2^n t(n + 1) = 2^n + \sum_{0 \le i \le n} \binom{n}{i} t(i) $$Multiply by z^n/n!, sum over n \ge 0, and recognize the resulting sums:$$ \sum_{n \ge 0} t(n + 1) \frac{(2 z)^n}{n!} = ... 0 $$\sum_{n\geq2}\frac{\zeta\left(n\right)\left(n-1\right)}{4n-1}\geq\frac{1}{4}\sum_{n\geq2}\frac{\zeta\left(n\right)}{n}\geq\frac{1}{4}\sum_{n\geq2}\frac{1}{n}.$$ 2 One may observe that $$\zeta(n)=1+\frac1{2^n}+\frac1{3^n}+\cdots,\quad n>1,$$ gives $$\lim_{n \to \infty}\zeta(n)=1,$$ then $$\lim_{n \to \infty}{\left(\frac{(n-1)\zeta(n)}{4n-1}\right)}=\frac14 \times 1\neq0.$$ Your series, as written, is divergent. 2$\sum_iX_i=n\bar X, so \begin{align*} \sum_i(X_i-\bar X)^2&=\sum_iX_i^2-2\bar X\sum_iX_i+\sum_i\bar X^2\\ &=\sum_iX_i^2-2\bar X(n\bar X)+n\bar X^2\\ &=\sum_iX_i^2-n\bar X^2\;. \end{align*} Moreover, \begin{align*} \sum_i(X_i-\bar X)(Y_i-\bar Y)&=\sum_iX_iY_i-\bar X\sum_iY_i-\bar Y\sum_iX_i+\sum_i\bar X\bar Y\\ ... 2 Any differentiable function is continuous. So 1 is clearly "yes". 2 is "no", there are lots of continuous functions which are not even differentiable at any point, the most famous examples of which are due to Weierstrass. For 3, I might recommend reviewing the definition of continuity; intuitively, a sawtooth wave can be drawn without picking one's pencil up ... 0 \We can also say f_{n_1 \cdots n_r}(x) = \prod_{i=1}^r f_{n_i}(x) = \odot_{i=1}^r f_{n_i}(x). Also for boolean or we have f_{n_1}(x) \oplus \dots \oplus f_{n_r}(x) = \lfloor \max\{g_{n_1}(x), \dots, g_{n_r}(x)\}\rfloor 6 Here is how one can compute \mathcal S_m for arbitrary m. Your formula (3) can be rewritten as\mathcal{S}_m=-m\int_0^1\frac{\ln\frac{1+z}{2}}{z^m-1}z^{m-1}dz=-m\sum_{k=0}^{m-1}\alpha_{km}\int_0^1\frac{\ln\frac{1+z}{2}}{z-e^{2\pi i k/m}}dz,$$where$$\alpha_{km}=\lim_{\quad z\to\; \exp{\frac{2\pi i k}m}}\frac{z^{m-1}\left(z-e^{2\pi i ... 1 Or still simpler, with equivalences, since it is a series with positive terms: $$\frac1{\sqrt{n(n+1)}}\sim_\infty\frac 1n,$$ which diverges, hence the original series diverges. 5 It just used the comparison test, as the summand $$\frac{1}{\sqrt{n(n + 1)}} \geq \frac{1}{\sqrt{(n + 1)^2}} = \frac{1}{n + 1}$$ and $$\sum_{n = 1}^\infty \frac{1}{n + 1} = \sum_{n = 2}^\infty \frac{1}{n}$$ diverges. 3 The limit you want to have calculated is the following $$\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k(k+1)}.$$ Note that\tfrac{1}{k(k+1)}=\tfrac{1}{k}-\tfrac{1}{k+1}$, so this is a telescoping series, meaning that consecutive terms cancel eachother out. It is then easy to see that for all$n$we have ... 3 We can write sum as$\displaystyle \lim_{n\rightarrow \infty}\sum_{r=1}^{n}\left[\frac{1}{r\cdot (r+1)}\right] = \lim_{n\rightarrow \infty}\sum_{r=1}^{n}\left[\frac{1}{r}-\frac{1}{r+1}\right]$Now Using Telescopin Sum(or Simply open Sigma) We get Sum$\displaystyle = \lim_{n\rightarrow \infty}\left(1-\frac{1}{n+1}\right) = 1-0 =1$1 This won't get you far. Try$\frac{1}{n(n + 1)} = \frac{n + 1 - n}{n(n+1)} = \frac{1}{n} - \frac{1}{n + 1}$instead. 2 This is not the full solution, it's too long for a comment, and gives some intuition of @user153012's observation about the appearance of$\operatorname{Li_2(\frac12)}$and$\ln^2 2. We can see that $$S_m=\sum_{n,k=1}^{\infty}\frac{(-1)^n}{k(mk+n)}=\sum_{n,k=1}^{\infty}\frac{(-1)^n}{k}\int_0^1 x^{mk+n-1}dx=\int_0^1 \frac{\ln(1-x^m)}{1+x}$$ Now we can divide ... 3 And what about this one? $$\mathcal{S}_6 \stackrel{?}{=} \ln^2(2) + \ln(2)\ln(3) -\frac{5\pi^2}{36} .$$ Furthermore, you probably know that $$\operatorname{Li}_2\left(\frac{\sqrt5 - 1}{2}\right) = \frac{\pi^2}{10} - \ln^2\left(\varphi\right) = \frac{\pi^2}{10} - \ln^2(2) -\ln^2\left(1+\sqrt{5}\right) + 2\ln(2)\ln\left(1+\sqrt{5}\right),$$ ... 4 An "easier way" as requested. If you know (or can guess) the answer, then you can show that $$\frac 1{(2k-1)(2k+1)}=\frac k{2k+1}-\frac {k-1}{2k-1}$$ which allows you complete the summation by telescoping immediately without the need for decomposing into partial fractions in the traditional way. 7 A general approach which works quite well is partial fraction decomposition: $$\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}\left(\frac{1}{(2k-1)}-\frac{1}{(2k+1)}\right)$$ After this, you can use the technique of telescoping. In your example, this yields: \begin{align} & \sum_{k=1}^{n}{\frac{1}{(2k-1)(2k+1)}} = ... 2 Your definition of denizen is a very baroque, confused way to describe an extremely simple concept: you're just asking for the longest interval that can be covered with arithmetic progressions, one for each prime up top$. When the same number is covered by multiple progressions, you went to great pains to force the denizen to record only the least ... 1 For a helix you did already, e positions of each block is given by: $$(x,y,z) = ( a \cos \theta, a \sin \theta, c \cdot \theta, )$$ where$c$is the pitch through which the spiral advances for 1 turn. Choose your$c$depending on how fast you want to climb,$a$on how big your helix base should be, and also choose a small enough$\theta $increment. ... 1 $$x=14\left(\overbrace{3+1}+\overbrace{\frac15+\frac1{15}}+\overbrace{\frac1{75}+\cdots}\cdots\right)\\ =14\left[(3+1)\left(1+\frac 1{15}+\frac1{15^2}+\cdots\right)\right]\\ =14\cdot 4\cdot \frac 1{1-\frac1{15}}=?$$ 0 Finding the transposition distance is an NP-hard problem. This is proved in this paper: Laurent Bulteau, Guillaume Fertin, and Irena Rusu Sorting by Transpositions is Difficult SIAM J. Discrete Math., 26(3), 1148–1180 (2012). DOI: 10.1137/110851390, arXiv:1011.1157. NP-hard means that most probably only approximation algorithms can be efficient ... 2 Take odd$f_n$s with $$f_n(x) = \begin{cases} nx, & \text{if 0\le x < 1/n} \\ 2-nx, & \text{if 1/n\le x \le 2/n.} \\ 0, & \text{if x > 2/n} \end{cases}$$$f_n(x)\to 0$pointwisely, since for any$x$there is some$N_x$that$x\in[-2/n,2/n]^c$(namely$f_n(x)=0$) for any$n\ge N_x$. However the ... 0 You wrote$\psi$for$f$which is just a typo. BUT the sum of the LHS is not valid if you don't assume that $$\lim_{x \to 0}f(x)=0 .$$ But the RHS sum is wrong. We have $$\ln ((3/4)^nx)= \ln x +n \ln (3/4)$$ so $$\sum _{n=0}^{n=m} (B \ln ((3/4)^n.x)-C=Bm \ln x +B((m^2+m)/2) \ln (3/4) -mC$$ which goes to$\infty$or to$- \infty$as$m \to \infty$... 2 Instead of "series" the mathematical term in this case is a permutation. Mathematically speaking you want to determine the following: For two permutations$\pi$,$\tau$you want to find the minimal number$n$of transpositions$\sigma_i$so that $$\pi = \sigma_1 \circ \ldots \circ \sigma_n \circ \tau.$$ Equivalently one might ask for the minimal$n$for ... 3 The Comparison Test says that, if$0 \le a_n \le b_n$, then If$\sum\limits_{n=1}^\infty b_n$converges, then$\sum\limits_{n=1}^\infty a_n$also converges. If$\sum\limits_{n=1}^\infty a_n$diverges, then$\sum\limits_{n=1}^\infty b_n$also diverges. 0 Hint. As Carl Heckman noticed, you have probably mixed up sequences and series. For series$\sum a_n$and$\sum b_n$, such that $$0\leq a_n\leq b_n,$$ the following holds true: If the series$\sum a_n$diverges then the series$\sum b_n$diverges If the series$\sum b_n$converges then the series$\sum a_n$converges 0 The result you're trying to prove is false. Let$a_n = (-1)^n + 2$, and$b_n = 5$. Clearly$b_n \to 5$, but$a_n$oscillates between$3$and$1$without converging. You've basically mimicked that example, but with some functions (strictly, you should evaluate your functions at each natural number in order to get a true sequence). There is a related result, ... 0 It's well after this post has been published and I know only a little about the topic, but if the question is only where to start the program that should be pretty simple: First note the formula to solve the problem in two variables: Find the minimal N such that $$ax + by \neq N$$ Theorem: N = ab - a - b = (a - 1)*(b - 1) - 1 WLOG assume a < b. ... 4 The series converges absolutely and uniformly, independently of$t$. We can bound the numerator above by$3k\pi^2 + 16 + 8\pi$, which is in general less than$C k$for a constant$C$and entirely independent of$t$. Then the sum looks like $$\sum_k \frac{C}{k^2},$$ which converges absolutely. This gives absolute and uniform convergence by the Weierstrass M ... 3 Let$A$be a matrix consisting of all zeros and$B$be the identity matrix. Then by multiplying each row in$B$by$1/n$for any$n>1$infinitely many times, we can transform it into$A$. However, it is clearly impossible to transform$B$into$A$in finitely many steps. 0 This is a crucial bit of reasoning about limits. Note that a function$f\left(n\right)$can approach a constant value a$n \to \infty$. As you mention, the infinite sum $$1/2 + 1/4 + 1/8 + 1/16 \dots$$ converges to unity, of which there is a nice geometric proof. This happens if and only if the gradient (which describes how fast this function is growing) ... 1 I think your proof is correct. By the way, we can prove it directly as the following :$\$\begin{align}k\sum_{i=1}^{j}b_i-j\sum_{i=1}^{k}b_i&=(j+k-j)\sum_{i=1}^{j}b_i-j\left(\sum_{i=1}^{j}b_i+\sum_{i=j+1}^{k}b_i\right)\\&=(k-j)\sum_{i=1}^{j}b_i-j\sum_{i=j+1}^{k}b_i\\&\gt (k-j)jb_j-j(k-(j+1)+1)b_{j+1}\\&=(k-j)j(b_j-b_{j+1})\\&\gt ...

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