# Tag Info

1

I believe there are tons of such functions. One of them is: $$e^x \cos (2\pi x)$$

1

We begin with the equation $$e^{-x}=x-1 \tag 1$$ Now let $y=x-1$. Then, we can write $(1)$ as $$ye^y=e^{-1} \tag 2$$ Therefore, solution to $(2)$ in terms of Lambert W Function is given by $$y=W\left(\frac1e\right) \tag 3$$ whereupon solving for $x$ reveals $$x=1+W\left(\frac1e\right)$$ Using the Taylor Series for the Lambert W yields ...

0

The 1st one is $-\ln 2$ (Taylor series). The 2nd one is $$f(x)=\sum_{n=1}^{\infty}n\cdot x^n$$ $$\int \frac{f(x)}{x}dx=\sum_{n=1}^{\infty}x^n=\frac{x}{1-x}$$ $$\therefore f(x)=x\cdot\left(\frac{x}{1-x}\right)'=\frac{x}{(1-x)^2}$$ $$\sum_{n=1}^{\infty}n\cdot\left(\frac12\right)^n=\frac{\frac12}{(1-\frac12)^2}=2$$ So the total sum is $2-\ln2$.

2

Hint: Think about what the definition of $\sum$ is.

2

$$\sum_{x=1}^{20}8x^{3}=8\sum_{x=1}^{20}x^{3}=8(1^3+2^3+3^3+...+20^3)=8(44100)=352800$$

3

For this question you can use the formula $$\sum_{x=1}^{n}x^{3}=(\frac{n(n+1)}{2})^{2}$$ Using this formula where n = 20 you get 44100. Then you multiply by 8 to get 352800.

3

Hint: A proof without using the integration concepts. Show that $$\frac{1}{n + 1} < \log\left(1 + \frac{1}{n}\right) < \frac{1}{n}. \tag{1}$$ You might need the facts that the sequence $\left(1 + \frac{1}{n}\right)^n$ increasingly converges to $e$ and the sequence $\left(1 + \frac{1}{n}\right)^{n + 1}$ decreasingly converges to $e$. Use $(1)$, show ...

1

I will show that $\lim_{n \to \infty} n^m(e-\sum_{k=0}^{n-1} \frac{1}{k!}) = 0$ for any finite $m$. $\begin{array}\\ e-\sum_{k=0}^{n-1} \frac{1}{k!} &=\sum_{k=n}^{\infty} \frac{1}{k!}\\ &=\frac1{n!}\sum_{k=n}^{\infty} \frac{n!}{k!}\\ &=\frac1{n!}\sum_{k=n}^{\infty} \frac{1}{\prod_{j=n+1}^{k} j}\\ &=\frac1{n!}\sum_{k=n}^{\infty} ... 1 Since there are different rules for the even and odd terms, it makes sense to suspect that the even terms will behave differently from the odd terms. So let's investigate the two sets of terms separately. For$n\in\mathbb{N}$, let$b_n = a_{2n-1}$and$c_n = a_{2n}$. From the relations provided, we have $$b_{n+1} = a_{2n+1} = \frac{1}{2}+\frac{a_{2n}}{2} = ... 3 The binomial coefficient$${n+m\choose n}={(n+m)!\over n!m!}$$is a positive integer if m,n\ge0, which implies$${1\over(n+m)!}\le{1\over n!m!}$$so$$e-\sum_{k=0}^{n-1}{1\over k!}=\sum_{k=n}^\infty{1\over k!}=\sum_{m=0}^\infty{1\over(n+m)!}\le{1\over n!}\sum_{m=0}^\infty{1\over m!}={e\over n!}$$0 This is one case where the more general definition of a limit is handy: A sequence s_n converges to L if, for any neighborhood of L, there is some N such that all the s_m with m>N are within this neighborhood. Where the word "neighborhood" refers to some open set containing L. In particular, if you take (L-\varepsilon,L+\varepsilon) ... 1 Divergence is based on the end of the series do the terms settle on a number (Converges) do they go to ±∞ (Diverges) do they bounce between different numbers without settling on one specifically (Diverges) not on how you get there does the distance between specific terms get larger or smaller 2 One says that \displaystyle\sum_{n=0}^\infty \frac 1 {2^n} "converges to 2", NOT that it "converges on 2". One says that \displaystyle\sum_{n=0}^\infty (-1)^n "diverges", but that does not mean it "diverges to" anything. But with \displaystyle\sum_{n=1}^\infty \frac 1 n, one could say that it "converges to \infty" for the same reason one says ... 0 If the series diverges to infinity, it means that as we keep adding more and more terms (we are adding infinitely many terms), the sum just keeps growing and growing without getting closer to any real number. It just gets bigger, eventually bigger than every real number. Here is an example of a series that doesn't diverge.$$\sum \limits_{n = 1}^{\infty} ... 1 I tried to show that the speed of convergence of the sum to e is faster than$1/n,$but with no success. Have you tried Stirling's approximation ? In my opinion, a far more interesting question would have been trying to prove that $$\lim_{n\to\infty}(n+a)\bigg[~e^b-\bigg(1+\dfrac b{n+c}\bigg)^{n+d}~\bigg]~=~b~e^b~\bigg(\dfrac b2+c-d\bigg),$$ ... 0 If$n=1$, then$a_1=1$which is true. Suppose that, for$n=k$and$a_1a_2\cdots a_k=1$, $$a_1+a_2+\cdots a_k\ge k.$$ For$n=k+1$, since$a_1a_2\cdots a_ka_{k+1}=1$, we have that$a_1a_2\cdots a_{k-1}(a_ka_{k+1})=1$implies $$a_1+a_2+\cdots a_{k-1}+a_ka_{k+1}\ge k.$$ Note that we can choose$a_k<1$and$a_{k+1}>1$since if ... 0 If$P(x)=ax^2+bx+c$, then$P(x)=x^2+2x-4$, because$P(1)=-1$,$P(2)=4$,$P(3)=11$gives a system of three linear independent equations in terms of$a,b,c$with the unique solution$(a,b,c)=(1,2,-4)$. The above method proves the uniqueness of such a polynomial. If you only want to find one example, you can use the Lagrange Interpolating Polynomial. ... 0 As a recurrence relation, here would be one way to derive the values:$a_n=a_{n-1}+(2n+3)a_0=-1(The next value is an odd number greater than the previous as the differences go 5,7,9,11,13 for the given sequence.) 3 The reciprocal of the radius of convergence, as given by the Cauchy-Hadamard Theorem is \begin{align} \limsup_{n\to\infty}\left(\frac{n^{\log(n)}}{3^n}\right)^{1/n} &=\frac13\limsup_{n\to\infty}n^{\log(n)/n}\\ &=\frac13\limsup_{n\to\infty}e^{\log(n)^2/n}\\ &=\frac13e^{\limsup\limits_{n\to\infty}\log(n)^2/n}\\[3pt] &=\frac13 \end{align} ... 1 I have benefited from reading @RoryDaulton answer, but I prefer to do the same thing over without using derivatives (although the OP seems to be fine with derivatives, just not sure if this is the way to solve it ... why not). SoA: = \{ b_n:n \in N\}$where$b_n={a_n}^2 - 3{a_n} - 10$and$a_n=4-\dfrac9n$. Note$a_{n+1}-a_n=\dfrac9{n(n+1)}$Let ... 2 Take the$n$th root of the$n$th term to get$n^{(\ln n)/n}(|z+2|/3).$Show$n^{(\ln n)/n} \to 1.$Thus the root test gives convergence if$|z+2|/3 < 1,$divergence if$|z+2|/3 > 1.$You can also check that the series diverges when$|z+2|/3 = 1.$1 To estimate$n^{\ln n}$, you can use $$n^{\ln n} = e^{\ln(n^{\ln n})} = e^{\ln n \ln n} = e^{\ln^2 n}$$ and $$3^n = e^{\ln(3^n)} = e^{n \ln 3}$$ Which should tell you which of these grows quicker (does$n$grow quicker than$\ln^2 n$?) 0 Infine and zero products and powers are the essence of many sequence issues. When factorials appears, Stirling-like formulas and bounds can be useful, for instance: $$\sqrt{2\pi}\ n^{n+1/2}e^{-n} \le n! \le e\ n^{n+1/2}e^{-n}\,.$$ Hence, you get: $$(\sqrt{2\pi})^{1/n^2}\ n^{1/n+1/2n^2}e^{-1/n} \le (n!)^{\frac{1}{n^2}} \le (e)^{1/n^2}\ ... 1 OUTLINE: Substitute the expression 4-\frac 9n for a_n into the expression a_n^2-3a_n-10. This will give you a function$$f(n)=\left(4-\frac 9n\right)^2-3\left(4-\frac 9n\right)-10$$Simplify that function expression. Treat it as a function of real numbers f(x) and take its derivative. Use that derivative to analyze the function f. You will find ... 1 No, without the weights it is impossible. If you have the weights, then the answer is:$$\frac{1}{n}\sum_{i=1}^ka_il_i$$But if you don't have the weights, then the averages no longer contain enough information. This is alluded to in the final paragraph. For example, consider the sequences 1, 1, 2, 2 and 1, 1, 1, 2. We could split the first into 1, 1 ... 2 One possible pattern that it could follow is that the denominator always increases by one at each step and the numerator is either one higher than the denominator or one smaller than the denominator depending on the parity of the term. If that were the case, then the sequence with the next several terms will look like: ... 1 A try: Suppose that such a sequence exist. Put \displaystyle na_{n+1}=\sum_{k=1}^n a_k+L+e_n, with e_n\to 0. Replacing n by n+1, we get$$(n+1)(a_{n+1}-a_{n+2})=e_n-e_{n+1}$$As a_n is decreasing, we get that e_n is decreasing too. Now as e_n\to 0, we have e_n\geq 0, and \displaystyle a_{n+1}-a_{n+2}=\frac{e_n-e_{n+1}}{n+1} \leq ... 4 There is no such sequence. Let s_n=\sum_{k=1}^n a_k. By (iii) we can write$$ n(s_{n+1}-s_n) - s_n = \ell + b_n = \ell(n+1-n)+ b_n$$where b_n\to0. Then it follows that for all n\ge1,$$ \frac{s_{n+1}+\ell}{n+1} = \frac{s_n+\ell}n + \frac{b_n}{n(n+1)} = \frac{s_1+\ell}1 + \sum_{k=1}^n \frac{b_k}{k(k+1)} = A+c_{n+1},$$where$$ A = ... 0 Once you have a collection of objects (be it numbers or other abstract entities) it's natural to overlay structure onto those objects (e.g. a topology, an ordering, an algebra of sets, etc.) We can also arrange them into a sequence. So we want to study sequences in general and to understand their properties in a variety of given contexts. Usually, we start ... 1 Zeno's Paradox. Zeno noted that before you can travel to a place, you must first travel halfway to the place, then from there travel half the rest of the way, etc. He reasoned that movement is impossible because you cannot travel an infinite number of steps in a finite time. In the language of sequences, the relevant numbers to Zeno's paradox are:$1/2$, ... 0 Let$f (n)$be the lower threshold for the level$n$. Then the player needs to gain$f (n + 1) - f (n)$points to reach level$n + 1$from the starting point of$n$th level. From what you said I think you want$f (n + 1) - f (n)$to be proportional with$n$. Say$f(n + 1) - f(n) = Cn$for some absolute constant$C$, and essentially,$f (1) = 0$. Telescoping ... 0 I am not entirely sure I understand your question completely, but I hope the following answers it: We have: Total number of levels:$L$Total number of points:$Y$Initial step size:$xL$levels means$L-1$steps from one level to the next, with the following points per step: $$\begin{array}{r|cccccccccccc} Level & 1 & & 2 & & 3 ... 0 Let N be the number of level increases (= 1 less than the total number of levels if the starting level is included in the count). The problem can only be solved in integers if the 'total' number of points T (i.e., the points required to reach the last level) is a multiple of N(N+1)/2 (= the sum of the first N integers) and in that case the ... 1 HINT: (if you're familiar with the gamma function):$$\lim_{n\to\infty}n\left(e-\sum_{k=0}^{n-1}\frac{1}{k!}\right)=\lim_{n\to\infty}n\left(e-\frac{e\Gamma(n,1)}{\Gamma(n)}\right)=\lim_{n\to\infty}n\left(\frac{e\left(\Gamma(n)-\Gamma(n,1)\right)}{\Gamma(n)}\right)=\lim_{n\to\infty}\frac{en\left(\Gamma(n)-\Gamma(n,1)\right)}{\Gamma(n)}=$$... 2 You can try to show that \frac{1}{n!}\leq\frac{2}{2^n}, then \frac{1}{n!}+\frac{1}{(n+1)!}+\dots\leq\frac{4}{2^n}.$$\frac{1}{n!}\leq\frac{1}{1\times2\times3\times\dots\times n}\leq\frac{1}{1\times2\times2\times\dots\times2}=\frac{2}{2^n}$$So ... 2 This follows from the AM-GM inequality. The shortest proof makes use of the Jensen Inequality. The function f(x)= \ln x is concave since f''(x)=-\frac{1}{x^2}<0. Now by Jensen Inequality, we have$$f(\frac{\sum_{i=1}^n a_i}{n}) \ge \frac{1}{n} \sum_{i=1}^n f(a_i)$$Taking the exponential function, we have$$\frac{1}{n} \sum_{i=1}^n a_i \ge ... 1 Assuming$x\in\mathbb{R}$: $$\sum_{n=1}^{\infty}\space\frac{x^2}{(x^2+1)^n}=\lim_{m\to\infty}\sum_{n=1}^{m}\space\frac{x^2}{(x^2+1)^n}=$$ $$\lim_{m\to\infty}\space\left(1-\frac{1}{\left(x^2+1\right)^m}\right)=1-\lim_{m\to\infty}\frac{1}{\left(x^2+1\right)^m}=$$ ... 1 (a) "In particular" is wrong: take$a_{n}=-1+1/n$then$b=-1$and (potentially) every$k\ge 1$works. (b)-(c) Correct. (d) Wrong: (1) and (2) are equivalent. (e)-(f) Correct. 0 Using wikipedia:$f_{n}(z)=exp(z/n) - cos(z) + \frac {i} {n} $is holomorphic for every$k \in \mathbb N$(you can prove that) in sn open conected set,$B(0, \delta)$.$f_{n}(z)=exp(z/n) - cos(z) + \frac {i} {n} $converges to$f(z)=1 - cos(z)$as n aproaches$\infty$. f is holomorfic too. f(0)=0 of order 2 (you can proof that). then$\exists \rho , k \in ...

0

I think that one of the most interesting thing about sequences is that sequences are "easy" to manage. In a first place, when you modeling something, a recurrence relation often arise. When you have this relation, you can compute (by hand or with a computer) all the terms you want. But in a lot of cases, your phenomenon is continuous (e.g. physical ...

3

Relations between causes and effects is a central topic in all branches of science. In quantitative mathematics, it is embodied by the concept of a function. For instance, you might be interested by the relation between the length of an iron rod and the temperature, $L=f(T)$. An interesting special case of functions are those having the set of natural ...

3

Sometimes with a sequence of simple objects you can approximate an object that is complicated. The sequence $\{x_n:n\ge1\}$ defined by $x_1=1$ and $$x_{n+1}=\frac12\Bigl(x_n+\frac2{x_n}\Bigr)$$ converges to $\sqrt 2$ as $n\to\infty$ (see here). So you have a sequence of rational numbers that converges to an irrational number.

9

I don't think it is that clear cut. Motivation is important (you don't want to work on a topic nobody cares about) but sometimes the motivations is just that (a sizable number of) people finds it interesting! Of course you have more chances of finding interested people if the topic "pops up" in various problem, so people think it's worth to study it ...

0

If you want to express $a_k^{(n)}$, the $k$-th term of the sequence for a given value of $n$, as a function of $k$ and $n$, you’ll need to define the function piecewise: it’s clear that $a_k^{(n)}$ is defined differently for $k=1,\ldots,6$, for $k=7,\ldots,n-3$, and for the last three terms. I’ll do the first two pieces and leave the third to you: ...

1

We want to show $\frac{H_{n, - r}}{n^r\left( n + 1 \right) } \le \frac{H_{n - 1, - r}}{n\left( n - 1 \right)^r}$ where $H_{n, a} =\sum_{k=1}^n \frac1{k^a}$, so $H_{n, -r} =\sum_{k=1}^n k^r$. From this, $H_{n - 1, - r} =H_{n , - r}-n^r$, so we want $\frac{H_{n, - r}}{n^r\left( n + 1 \right) } \le \frac{H_{n , - r}-n^r}{n\left( n - 1 \right)^r}$. ...

3

Arguing a little sloppily, where "$\sim$" means "within a factor of", $\begin{array}\\ \frac{1}{n^2}\sum_{k=1}^n \frac{k^2}{(k+1)\log(k+1)} &\sim \frac{1}{n^2}\sum_{k=2}^n \frac{k^2}{k\log(k)} \qquad\text{since we can ignore }k=1\\ &\sim \frac{1}{n^2}\sum_{k=2}^n \frac{k}{\log(k)}\\ &\sim \frac{1}{n^2\log(n)}\sum_{k=2}^n k \qquad\text{since ... 2 This is given by the "multinomial theorem", an extension of the binomial theorem. See https://en.wikipedia.org/wiki/Multinomial_theorem. 1 I think what you want to prove is that $$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \operatorname*{Res}_{\zeta=\zeta_k} \left [\pi \, \cot{(\pi \zeta)} \, f(\zeta)\right ]$$ where$\zeta_k$are the non-integer poles of$f$. You do this by showing that $$\lim_{N \to \infty} \oint_{C_N} d\zeta \, \pi \, \cot{(\pi \zeta)} \, f(\zeta) = 0$$ on the square ... 0 When you multiply a polynomial, you take every permutation of monomials from each product. As a familiar example, $$(x+1)(x+2) = x^2+x+2x+1 \cdot 2$$ In the case given, when considering a monomial in the final product, the first factor decides the one's place, the second factor decides the hundred's place, etc. Since the numbers$0-9$can be used for each ... 2 Hint. You may observe that $$\sum\limits_{n=0}^9z^{10^kn}=\frac{1-z^{10^{k+1}}}{1-z^{10^k}}$$ then recognize a telescoping product: $$\prod\limits_{k=0}^N\sum\limits_{n=0}^9z^{10^kn}=\prod\limits_{k=0}^N\frac{1-z^{10^{k+1}}}{1-z^{10^k}}.$$ Letting$N \to \infty\$ gives the result.

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