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0

Maybe it's not what you want, but it uses integral method... We compute the integral $$\int_0^{e^{-1}}1 dx$$ using the partition $\{e^{-n}, n\geq 1\}$, since the integrand is constant we have $$\int_0^{e^{-1}}1 dx = \sum_{n=1}^{+\infty}\left(e^{-n} - e^{-(n+1)}\right)$$ i.e. $$e^{-1} = \sum_{n=1}^{+\infty}\left(e^{-n} - e^{-(n+1)}\right) = ... 1 We aim to show s_n= \frac{1}{n} \sum_{k=1}^n a_k-a\to 0 \iff \frac{1}{n} \sum_{k=1}^n (a_k-a)\to 0 Hence it suffices to show the case when s=0, that is a_n\to 0 \implies s_n\to 0 That is \forall \epsilon > 0 \ \exists N_1 \in \mathbb{N} \ \text{such that} \ \forall n > N_1 \vert a_n\vert < \epsilon Let M=\sum_{i=0}^N |a_i|, \exists N_2 ... 1 a) We can't entirely neglect the part from 1 to N because it's part of the mean. But for each \varepsilon the part from 1 to N contributes a constant to the numerator and a constant to the denominator. You can show, then, that after sufficiently many terms (how many?), the mean is within 2\varepsilon of the limit. Then you are done because ... 1 If \sum a_{n}^{2} converges, then \sum \left(a_{n}^{2} + \frac{1}{n^{2}}\right) also converges. But$$0 \leq \left(a_{n} - \frac{1}{n}\right)^{2} = a_{n}^{2} + \frac{1}{n^{2}} - 2\frac{a_{n}}{n},$$so$$\frac{a_{n}}{n} \leq \frac{1}{2}\left(a_{n}^{2} + \frac{1}{n^{2}}\right).$$Thus, \sum \frac{a_{n}}{n} converges. 4 Just let a_n be any enumeration of the rationals. 4 Cauchy-Schwarz says$$ \sum_{n=1}^\infty\frac{|a_n|}{n}\le\left(\sum_{n=1}^\infty a_n^2\right)^{1/2}\left(\sum_{n=1}^\infty\frac1{n^2}\right)^{1/2} so the series is absolutely convergent. 1 If it's not easy to see the connection between C(x)=\sum_{n=0}^{\infty}c_nx^n with A(x)=\sum_{j=0}^{\infty}a_jx^j and B(x)=\sum_{k=0}^{\infty}c_kx^k we could look at the coefficients c_n for small n and check if we detect some pattern or regularity. Assuming c_n is given by \begin{align*} c_n=\sum_{{j+4k=n}\atop{j,k\geq 0}}a_jb_k\qquad ... 0 The direct way to compute S is to plug in the value x=\frac12 into the identities, valid for every |x|\lt1,\sum_{n=0}^\infty x^n=\frac1{1-x},\qquad\sum_{n=0}^\infty nx^{n-1}=\frac1{(1-x)^2}.$$2$$\sum_{n=1}^\infty\frac{2n-1}{2^n}=\sum_{n=1}^\infty\frac n{2^{n-1}}-\sum_{n=1}^\infty\frac1{2^n}=\frac1{\left(\frac12\right)^2}-\frac{\frac12}{1-\frac12}=4-1=3$$The above follows from$$|x|<1\implies\frac1{1-x}=\sum_{n=0}^\infty x^n\;,\;\;\left(\frac1{1-x}\right)'=\sum_{n=1}^\infty nx^{n-1}$$and the splitting of the first sum in the first line ... 4 Hint From an algebraic point of view, consider$$S=\sum_{n=1}^{\infty}(2n-1)x^n$$and you can write$$S=2x\sum_{n=1}^{\infty}n x^{n-1}-\sum_{n=1}^{\infty} x^{n}$$I am sure that you can take from here. When you finish, replace x by \frac12. 1 \forall x\in X, f_n(x)\to f(x), we have |f_n|^p(x)\to|f|^p(x), since g(x)=|x|, h(x)=x^p are continuous functions. Hence \liminf_{n \rightarrow \infty} |f_n|^p =\lim_{n \rightarrow \infty} |f_n|^p = = |f|^p. 0 OK, the progression appears to be from "high" to "low" in terms of 1's occuring in the bitstring. The lowest occurrences (of 1's) appear first. So all you need to do is count the number of 1's in your bit string. Then figure out how many came before it, then you can subtract from the highest value of the "substring". Here is an example:$$ n = 4, ...

1

I'm probably going a little overboard here, but I remember some of the simplifications were confusing to me the first time I learned this. You need to use the ratio test: $$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}} \right| = L$$ Where: $$L > 1 \implies \text{Divergent} \\ L < 1 \implies \text{Convergent} \\ L = 1 \implies ... 2 Hint$$0< \frac {(1)(2)(3)\ldots(n)}{(n)(n)(n)\ldots(n)} < \frac{1}{n}$$0 A very useful formula is Stirling approximation of n! as Milly commented. One of the simplest form is given by$$n!\approx n^n \sqrt{2 \pi n} e^{-n}\frac{n!}{n^n} \approx \sqrt{2 \pi n} e^{-n}$$I am sure that you can take from here. 1 Put$$a_n=\frac{n!}{n^n}\implies\frac{a_{n+1}}{a_n}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e<1\implies \sum_{n=1}^\infty\frac{n!}{n^n}\;\;\text{converges}\implies\lim_{n\to\infty}\frac{n!}{n^n}=0$$4 Hint: If n is even, then half of the fractions \dfrac{1}{n}, \dfrac{2}{n}, \ldots, \dfrac{n}{n} are less than or equal to \dfrac{1}{2} and the other half are less than or equal to 1. Therefore, 0 \le \dfrac{n!}{n^n} \le \left(\dfrac{1}{2}\right)^{n/2} \cdot 1^{n/2} = \dfrac{1}{2^{n/2}}. You can do a similar thing if n is odd. 0 1.) In your given string AB and BC are repeating substrings, not patterns. A pattern in a sequence would be a repeating substring that occurs every n characters, where n \in \{0, 1, 2, \ldots, n\}. In your given string AB appears after 5 characters, then after 13 characters, then after 0 characters (3 times); which is not a pattern. However if ... 1 The first one is a mistake: either it was intended to be 3,7,16,32,57,\ldots, corresponding to the description s_1=3, s_n=s_{n-1}+n^2 for n>1, or there is simply not enough information to allow one to guess how it is intended to differ from that sequence. For the second one you can try ‘unwinding’ it:$$\begin{align*} s_n&=3s_{n-1}+2\\ ...

4

One way to approach the problem is to begin by looking at the subsequence of terms with indices of the form $10n+1$: $$\begin{array}{rccc} 10n+1:&1&11&21&31&41\\ s_{10n+1}:&1&21&51&91&141\\ s_{10n+1}-(10n+1):&0&10&30&60&100 \end{array}$$ If you divide that last row of numbers by $10$, it might be ...

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Judging from what you have posted, it looks like you have $$f(10)=10,\ f(20)=30,\ f(30)=60,\ f(40)=100$$ and so on. A possible formula for this would be $$f(10q)=10(1+2+\cdots+q)=5q(q+1)\ .$$ The numbers reading backwards from $10$ to $1$, from $20$ to $11$ and so on appear to just decrease by $1$ each time. So these can be included by writing ...

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group terms in pairs, 1st and 2nd, then 3rd and 4th, etc. Note $1/1^s>1/2^s$, then $1/3^s>1/4^s$, so $1/1^s-1/2^s>0$, also $1/3^s-1/4^s>0$, so $1/1^s-1/2^s + 1/3^s-1/4^s+ ... > 0$.

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Here's a proof that $f$ only vanishes at $x = 0$ (you can use a similar method to get some asymptotic results as well). Write $f(x)/x$ as \begin{align*} {f(x)\over x} &= \sum_{n\geq 1} {\operatorname{sinc}{(x/n^2)}\over n^2} \end{align*} Since $f(x)/x$ is even, we need only treat the case $x\geq 0$. Split the sum into the regions where $x/n^2$ is ...

0

Let $(x_n)$ and $(a_n)$ be sequences of real numbers such that $x_n\to\pm\infty$ and $a_n\to L$ for some $L\in\mathbb{R}$. Then $\displaystyle\frac{a_n}{x_n}\to 0$ since $\displaystyle\frac{a_n}{x_n}=\big(a_n\big)\left(\frac{1}{x_n}\right)$ where $\big(a_n\big)$ is bounded and $\displaystyle\frac{1}{x_n}\to 0$, so applying this result with $a_n=x_n y_n$ ...

1

$$\sum_{k=1}^n\frac{1}{k(k+1)} = (1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots+(\frac{1}{n}-\frac{1}{n+1})=\frac{n}{n+1}$$ So $$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \lim_{n\to \infty} \sum_{k=1}^n\frac{1}{k(k+1)}= \lim_{n\to \infty} \frac{n}{n+1}=1.$$

1

Here's a slightly informal way to get the asymptotic expansion of this function. Split the region of summation into intervals of the form $$\pi k < \frac{x}{n^2} < \pi(k+1),$$ for $k\in\mathbb{Z}_{\geq0}$, and write the sum as $$\sum_{n\geq 1}\sin \frac{x}{n^2} = \sum_{k\geq0} \sum_{\frac{x}{\pi(k+1)} < n^2 < \frac{x}{\pi k}} \sin ... 3 Since you speak of the ratio test, maybe you're only concerned with whether the series converges rather than with what the sum is. That is the most that the ratio test can give you. If that is what you're concerned with, then you can say$$ \sum_{k=1}^\infty \frac 1 {k(k+1)} \le \sum_{k=1}^\infty \frac 1 {k^2} $$and that converges, by a simple integral ... 5$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots+(\frac{1}{k}-\frac{1}{k+1})$$Do you see how to do it now? 4 It surely isn't significantly different, but it can be done as follows: The first step is the same:$$ \sum_{n=1}^{x} n=\frac{x(x+1)}{2}=\binom{x+1}{2}\implies \frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\sum_{n=2}^{x}\frac{1}{\binom{n+1}{2}}=\sum_{n=1}^{x-1}\frac{1}{\binom{n+2}{2}} $$Now, take a look at the following integral:$$ ...

2

As has been mentioned, $(1+n^2)^{-1/4}\ge\frac1{\sqrt{2n}}$ which diverges by the integral test. Since $\cos(x+\pi)=-\cos(x)$, we have that \begin{align} \sum_{k=n}^{n+11}\cos\left(\frac{k\pi}6\right) &=\sum_{k=n}^{n+5}\left[\cos\left(\frac{k\pi}6\right)+\cos\left(\frac{k\pi}6+\pi\right)\right]\\ ... 2 The Dirichlet beta function may satisfy you :\beta(x):=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^x}$$with the table of values : \begin{array} {c|c} n&\beta(n)\\ \hline 1&\frac {\pi}4\\ 2&K\\ 3&\frac {\pi^3}{32}\\ 4&\beta(4)\\ 5&\frac {5\,\pi^5}{1536}\\ \end{array} with K the Catalan constant. Here the n even cases are the ... 1 I suspect the formula for n even you are talking about comes from the riemann zeta function seen here. In the same articles you can read that a general formula for \zeta(2k+1) is still an open problem. 2 I'm not sure there is much which can be done radically differently. The proof can be recast as an induction - which is what you might do after computing the first few sums by hand as \frac 13, \frac 12=\frac 24, \frac 35, \frac 23=\frac 46, \frac 57 The n^{th} term is \cfrac 2{(n+1)(n+2)} We can prove that the sum of the first n terms is \cfrac ... 2 Let's prove by induction that$$\sum_{m=2}^n \frac1{\sum_{k=1}^m k} = \frac{n-1}{n+1}$$The base case n=2 is true. Now the induction step:$$\sum_{m=2}^{n+1} \frac1{\sum_{k=1}^m k}=\sum_{m=2}^n \frac1{\sum_{k=1}^m k}+\frac1{\sum_{k=1}^{n+1}k}=\frac{n-1}{n+1}+\frac2{(n+1)(n+2)}=\frac{n^2+n}{(n+1)(n+2)}=\frac{n}{n+2}$$Therefore the answer is n=2012. (I ... 0 Expanding on mjqxxxx's answer: s_k=y_{3k-2}+y_{3k-1}+y_{3k}=\frac{1}{\sqrt{4k-3}}+\frac{1}{\sqrt{4k-1}}-\frac{a}{\sqrt{2k}}. By the binomial theorem, \begin{array}\\ (4k-c)^{-1/2} &=(4k)^{-1/2}(1-\frac{c}{4k})^{-1/2}\\ &=(4k)^{-1/2}(1-\frac{c}{4k}\binom{-1/2}{1}+O(k^{-2}))\\ &=(4k)^{-1/2}(1+\frac{c}{8k}+O(k^{-2}))\\ \end{array}  Therefore ... 0 Hints: you can verify via direction multiplication of the RHS's below that$$ (x-1)^2=(x^2-2x+1),\\ (x^{n+1}-1)=(x-1)(x^n+x^{n-1}+\cdots+1). $$Once you have done that, you can see that both sides of the your original (supposed) equality can be written as$$ (x-1)^2(x^n+x^{n-1}+\cdots+1). $$2 Okay, I think I can guess at the pattern with the most recent addition$$ \begin{align} &\sum_{n=0}^\infty\left(\frac1{\sqrt{4n+1}}+\frac1{\sqrt{4n+3}}-\frac{a}{\sqrt{2n+2}}\right)\\ &=\sum_{n=0}^\infty\frac{\sqrt{(4n+3)(2n+2)}+\sqrt{(4n+1)(2n+2)}-a\sqrt{(4n+1)(4n+3)}}{\sqrt{(4n+1)(4n+3)(2n+2)}}\\ ...

1

So, the terms come in blocks of three, where the sum of the $k$-th block of terms is $$s_k=y_{3k-2}+y_{3k-1}+y_{3k}=\frac{1}{\sqrt{4k-3}}+\frac{1}{\sqrt{4k-1}}-\frac{a}{\sqrt{2k}}.$$ Since the individual $y_i\rightarrow 0$, the sum $\sum_{k=1}^{\infty} y_k$ converges if and only if the sum $\sum_{k=1}^{\infty} s_k$ converges. I suggest you calculate the ...

2

Substitute $x\mapsto e^{-x}$ and then $x\mapsto x/n$: \begin{align}\hspace{-1cm} \int_0^1\frac{\log(1+x)}{1+x}\log(\log(1/x))\,\mathrm{d}x &=\int_0^\infty\frac{\log(1+e^{-x})}{1+e^{-x}}\log(x)\,e^{-x}\,\mathrm{d}x\\ &=\sum_{n=2}^\infty\int_0^\infty(-1)^nH_{n-1}e^{-nx}\log(x)\,\mathrm{d}x\\ ... 0 It looks like it could be a function of the formy=A\sin\left[x-\left(\frac{\pi}2-\alpha\right)y\right]$$where A is the amplitude and \alpha (0<\alpha<\frac\pi2) is the angle where the function is maximum. See my response to another related question here: Skewed Trigonometric Function 0 We may formally express the series in terms of the polylogarithm defined as \displaystyle\text{Li}_p(z):=\sum_{k=1}^\infty \dfrac{z^k}{k^{p}}. Since \sin(kx)=\text{Im}\,(e^{ikx}), this series is the imaginary part of the summation \displaystyle\sum_{k=1}^\infty \dfrac{e^{ikx}}{k^{3/2}}. Therefore we may write ... 5$$a_{1}=1=1^3\\a_{2}=3+5=8=2^3\\a_{3}=7+9+11=27=3^3\\a_{4}=13+15+17+19=64=4^3\\...\\a_{n}=n^3\\so\\a_{n+2}=(n+2)^3$$2 Set A_n=\sum_{k=1}^{n}a_k and \displaystyle A=\lim_{n\rightarrow\infty}A_n. Then$$ \left|\sum_{k=m}^{n}a_k\right|=\left|A_n-A_{m-1}\right|=\left|(A_n-A)-(A_{m-1}-A)\right|\leq \left|A_n-A\right|+\left|A_{m-1}-A\right|\leq 2\epsilon $$and there is no extra A to add to this. 1 Let x=3+2\sqrt{2}\approx 5.8284; then the fractional part of x^n monotonically increases, converging exponentially fast to 1. Other values that work include \phi^2\approx 2.618, or the squares of any of the so-called "silver means" mentioned in the link you provided. 0 We know that$$1-\cos(\frac{\pi}{n})=2\sin^2(\frac{\pi}{n})$$also$$\sin^2(\frac{\pi}{n}) \sim (\frac{\pi}{n})^2, \ \ \ \ \text{as} \ \ \ n\to \infty$$therefore$$\sum\limits_{n=1}^{\infty}n(1-\cos(\frac{\pi}{n}))=\sum\limits_{n=1}^{\infty}2n\sin^2(\frac{\pi}{n})\sim \sum\limits_{n=1}^{\infty}2n (\frac{\pi}{n})^2=\sum\limits_{n=1}^{\infty} ...

0

Hint Use the inequality $1-\cos x\ge \frac{x^2}{3}$ for $-2\le x\le 2.$

0

The Frobenius method is a generalisation of the power series method. It extends the power series method to include negative and fractional powers. It also allows an extension involving logarithm terms. So, it uses an Ansatz with \begin{align*} y(x)&=c_0x^c+c_1x^{\alpha+1}+c_2x^{\alpha+2}+\ldots\\ ...

0

Note that $$\sum_{n=1}^{\infty}1-\cos\left(\pi/n\right)\leq\int_0^\infty 1-\cos\left(\pi/x\right)dx=\pi^{2}/2$$

1

You can prove by various methods that if $x$ is small then $$1-\cos x<\frac{x^2}{2}\ .$$ So if $n$ is large then $$1-\cos\Bigl(\frac\pi n\Bigr)<\Bigl(\frac{\pi^2}2\Bigr)\frac1{n^2}\ ,$$ and you can use a comparison test.

4

You may use the Maclaurin expansion for $\cos x$: $$\cos (x) = 1 -\frac{x^2}{2}+ \mathcal{O}(x^4)$$ giving $$\sum_{n=2}^{\infty}\left( 1-\cos \left(\frac{\pi}{n}\right)\right)=\sum_{n=2}^{\infty}\frac{\pi^2}{2n^2}+\sum_{n=2}^{\infty}\mathcal{O}(\frac{1}{n^4})$$ concluding that your initial series converges.

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