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The book contains $p$ pages and has therefore pagenumbers from $1$ to $2p$. The sum of all the pagenumbers is then given by $$\sum_{i=1}^{2p}i =p \Big( 2p + 1 \Big).$$ The father holds the page with page number $n$ in his hand, so we need to solve $$81,707 = p \Big( 2p + 1 \Big) - n.$$ As $81,707 \le p \Big( 2p + 1 \Big)$, we obtain $$p \ge 202,$$ ...

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Let $p,p+1$ be the pagenumbers of the pages he is holding. Assume the book has $n$ pages. Then: $$81707=\sum_{i=1}^ni-(p+p+1)=\frac{n(n+1)}{2}-2p-1.$$ It is clear that $\frac{n(n+1)}{2}$ must be an even number, say $2m.$ So $$81707=2m-2p-1=2(m-p)-1\implies m-p=40854.$$ That is $$\frac{n(n+1)}{4}=p+40854,$$ from where ...

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Yes, there is a closed form for this famous sequence: $$u_n = \frac{1}{n+1}\binom{2n}{n}.$$

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Sure. Note that $\lim_{n \to \infty} \frac{P(n+1)}{P(n)} = 1$ for any (non-zero) polynomial $P$, whereas $\limsup_{n \to \infty}\frac{S_{n+1}}{S_n} = 2$.

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for $m>1$$|\frac{1}{3m+1}+\dots+\frac{1}{6m-2}| > m\times\frac{1}{6m-2} \nrightarrow 0$$ 4 Let there be$n$pages, and suppose your father is on page$k$. Then you want $$\sum_{i=k+1}^ni=81707$$ Do you know how to calculate this sum? 3 If$x\leqslant1$, then$a_n\geqslant1$for every$n$hence the series$\sum\limits_na_n$diverges. If$x\gt1$, then the expansion$x^{1/n}=\exp((\log x)/n)=1+(\log x)/n+o(1/n)$yields $$\log(a_{n+1}/a_n)=\log(2-x^{1/n})\sim-(\log x)/n,$$ hence$\log a_n\sim-(\log x)\cdot\log n$. Thus: If$x\gt\mathrm e$, one can pick$1\lt y\lt\log x$then ... 3 We have$n+1+r$people lined up in a row, and want to choose$n+1$of them. That can be done in$\binom{n+1+r}{n+1}$ways, which is$\binom{n+1+r}{r}$, for we are rejecting$r$people. Let us count the number of choices another way. Suppose the first person in the line that is chosen is person$1$. Then there are$\binom{n+r}{n}ways to choose the others. ... 2 There is an easier way. Notice that: $$\sum_{r = 1}^n a + (r-1)d = \sum_{r = 1}^n a + \sum_{r = 1}^n rd - \sum_{r = 1}^n d = na + d\left (\sum_{r = 1}^n r\right) - nd$$ Now, just prove by induction that $$\sum_{r = 1}^n r = \frac{n(n+1)}{2}$$ which is much easier, and manipulate the previous expression to get what you need. 2 Deriving from integral approach is easier: \begin{align} \int(1+x^n)^{-\frac{1}{m}}~dx &=\int_0^x(1+t^n)^{-\frac{1}{m}}~dt+C\\ &=\int_0^{x^n}(1+t)^{-\frac{1}{m}}~d(t^\frac{1}{n})+C\\ &=\dfrac{1}{n}\int_0^{x^n}t^{\frac{1}{n}-1}(1+t)^{-\frac{1}{m}}~dt+C\\ &=\dfrac{1}{n}\int_0^1(x^nt)^{\frac{1}{n}-1}(1+x^nt)^{-\frac{1}{m}}~d(x^nt)+C\\ ... 2 We must have0<a_n<1$for all$n$. Also note that for$k=2$, we must have for all$n$the inequelity$\displaystyle a_n^2\leq \frac{1}{2}$hence$\displaystyle a_n\leq \frac{1}{\sqrt{2}}$. We have then for large$k$: $$1-a_n\geq 1-\frac{1}{\sqrt{2}}>(\frac{1}{\sqrt{2}})^{k^2}\geq a_n^{k^2}$$ and thus$1-a_n-a_n^{k^2}> 0$for all$n$and large ... 2 Hint: The supremum is a function of$x$, which you can define piecewise. Consider what the supremum is in the two cases$x \ge 1$and$x < 1$. 2 A sequence of functions can be thought of as a mapping$g:\mathbb{N}\to\mathbb{R}^{\mathbb{R}}$defined by$g(n)=f_{n}$where$\mathbb{R}^{\mathbb{R}}$is the space of functions that map$\mathbb{R}$into$\mathbb{R}$. The supremum will be the map defined by$f(x)=\sup_{n\in\mathbb{N}}f_{n}(x)$. This may not be a function in$\mathbb{R}^{\mathbb{R}}$since ... 2 If I have not mistaken something, $$[x^{2n+1}]\frac{1}{(1-x)(1-x^2)(1-x^3)} = \frac{(n+1)(n+3)}{3},$$ when$n\equiv 0,2\pmod{3}$, and $$[x^{2n+1}]\frac{1}{(1-x)(1-x^2)(1-x^3)} = \frac{(n+1)(n+3)+1}{3}$$ when$n\equiv 1\pmod{3}$, hence: ... 1 Note that the supremum is an operator on a set, in other words when we consider sequences we actually look at $$\sup \{ a_n \mid n\in \mathbb{N} \}$$ And similarly for functions. The usual way I have seen for defining$\sup_{n \in \mathbb{N} } f_n$where each$f_n$is a function is to define a new function$f$Point wise like so $$f(x) = \sup_n f_n(x)$$ ... 1 Your sequence$S_n$has the property that for$n$large, $$S_n > \frac{\sqrt{2}^n}{2}.$$ Suppose that$f$is a polynomial of degree$p$with$f(n) = S_n$. Then you'd have $$c n^p > \sqrt{2}^n$$ for some constant$c$(the leading coefficient of$f$, a factor of 2, ...) and all sufficiently large$n$. Taking logs, that becomes $$\log(c) + p \log(n) ... 1 Note that$$a_n=\sum_{k=1}^n\frac{1}{3k-2}>\sum_{k=1}^n\frac{1}{3k}=\frac{1}{3}\sum_{k=1}^n\frac{1}{k}.$$Now, since \displaystyle\sum_{k=1}^n\frac{1}{k}\to\infty, when n\to\infty, we have$$a_n\to\infty,\ \mbox{when}\ n\to\infty.$$1 I was in the end able to derive the correct expression as...$$\frac{(1/n)_k}{(1+1/n)_k}=\frac{\frac1n(\frac1n+1)(\frac1n+2)\cdots(\frac1n+k-1)}{(1+\frac1n)(1+\frac1n+1)(1+\frac1n+2)\cdots(\frac1n+1+k-2)(\frac1n+1+k-1)} =\frac{1}{nk+1}$$1 Hint: \frac{a^n}{2^nn^2} = \left(\frac{a}{2}\right)^nn^2 = \frac{b^n}{n^2} with b > 1 1 HINT: \displaystyle\frac{a^n}{2^nn^2}=\frac{a^n}{2^n}\cdot\frac{1}{n^2}=\frac{(1+\epsilon)^n}{n^2} 1 The rth term, u_r=a+(r-1)d with u_1=a be the first term & d be the common difference If \displaystyle S_n= \sum_{r=1}^nu_n=\frac n2[2a+(n-1)d] \displaystyle\implies S_{n+1}=\displaystyle \sum_{r=1}^{n+1}u_n=S_n+u_{n+1}=\frac n2[2a+(n-1)d]+a+nd \displaystyle\implies S_{n+1}=(n+1)a+d\cdot\frac n2(n-1+2)=\frac{n+1}2[2a+(n+1-1)d] 1$$103$$your book has n=404 pages, so the sum of all pages is n(n+1)/2=81810, since the sum must be 81707, the page missing must be 103 if your book had 405 pages, the sum would be 82215, and the page missing would have to be 508 --impossible if your book had 403 pages, the sum would be 81406, not enough edit: The solution before is assuming 1 number per ... 1 A more general approach to solving for the nth term of such sequences uses matrix multiplication. Suppose the even terms are nonzero constant r\neq 1 times the preceding odd terms, while the odd terms are constant d plus the preceding even terms. We have:$$ \begin{pmatrix} 1 & 0 & d \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} ... 1 The generating function of Catalan's numbers$y_n=\frac 1{n+1}\binom{2n}n$is $$\sum_{n\geqslant 0 }\frac1 {n+1}\binom{2n}nx^n=\frac{1-\sqrt{1-4x}}{2x}$$ Multplying by$x\$ and differentitating gives that $$\sum_{n\geqslant 0}\binom {2n}nx^n=\frac{d}{dx}\frac{1-\sqrt{1-4x}}{2}=\frac{1}{\sqrt{1-4x}}$$ One can also try to prove this directly, by noting that ...

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