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-2

The recurrence is actually $$(4n+11)A_{n+2}=4(4n+7)A_{n+1}-(4n+3)A_n.$$ Notice that the denominator is not constant. That is why it is not a recurrence with constant coefficients.


1

It's clear that $u_k\ge1$ for all $k$ so we get (I think there's a typo in the question) $$u_n=\left(\sum_{k=1}^{\color{red}{n-1}}u_k\right)^{1/2}\ge \sqrt{n-1}$$ hence we get the desired limit of $u_n$, moreover we have $$v_n(u_{n+1}+u_n)=u_n\iff \frac1{v_n}=1+\frac{u_{n+1}}{u_n}\xrightarrow{n\to\infty}2$$ so we get the limit of $v_n$ and ...


2

You're right that such a sequence cannot be convergent. It basically says "There is some $r$ such, for any $\varepsilon>0$ that the sequence $r_n$ is only in $(r-\varepsilon,r+\varepsilon)$ finitely many times". Now, what this is doing is clear if you start choosing $\varepsilon$ to be really big. Like, if we chose $\varepsilon=r$, then $r_n$ could only ...


2

$\dfrac{2!}{1!9!}+\dfrac{2!}{3!7!}+\dfrac{1}{5!5!}$=$\dfrac{2}{10!}$($^{10}C_1$)+$\dfrac{2}{10!}$($^{10}C_3$)+$\dfrac{1}{10!}$($^{10}C_5$) $\implies$$\dfrac{1}{10!}$($^{10}C_1$+$^{10}C_9$+$^{10}C_3$+$^{10}C_7+^{10}C_5$) $\implies$$\dfrac{1}{10!}$(10+10+120+120+252)=$\dfrac{2^9}{10!}$ $\implies$$\dfrac{2^9}{10!}$=$\dfrac{8^a}{2b!}$ $\implies$$a=3, b=5$ If ...


1

Your reindexed summation is correct. Since the sums (outer and inner) for the second series start at $n = 2$ resp. $k = 2$ while in the first they start at $0$, combining the two power series is not totally direct. One way is to split the sums, treat $n = 0$ and $n = 1$ separately, and in the remaining series, for the inner sum $k = 0$ and $k = 1$. A more ...


0

For the sequence $s_n = (1-r^n)\sin\frac{n\pi}2$, we have $|\sin\frac{n\pi}2|\leqslant 1$ for all $n$, so the limiting behavior of the sequence depends on the value of $r$. Clearly if $r=1$ then $s_n=0$ for all $n$ so $\lim_{n\to\infty} s_n=0$. If $0<|r|<1$ then $\lim_{n\to\infty}r^n=0$. Since $\sin\frac{n\pi}2=1$ for $n=1,5,9,\ldots$ and ...


3

$\textbf{Hint:}$ show this by induction on $n\ge0$ $$a_n=2^{2^n}+2^{-2^n},\quad n=0,1,2\dots$$ and notice that $a_k+1=a_{k-1}^2-1=(a_{k-1}-1)(a_{k-1}+1),$ $$1-\frac{1}{a_k}=\frac{a_k-1}{a_k}=\frac{a_{k+1}+1}{a_k+1}\cdot\frac{1}{a_k}$$


0

In given example we're able to factor the denominator. Now we can utilize the fact that $\displaystyle \frac{1}{\alpha+1}=\int_0^1x^{\alpha}\,\mathrm dx$ But note that the integral obtained may or may not be evaluated using elementary methods $$\begin{align} \sum_{n=1}^{\infty}\frac1{(3n-1)(3n+2)} ...


3

$$\sum_{n> N}\frac{1}{n^2}<\sum_{n> N}\frac{1}{n(n-1)}=\cdots?$$ This proves that $$\sum_{n> N}\frac{1}{n^2}<\frac 1 N$$ actually, which might be the intended inequality.


0

We can indeed write the sum as an integral, after research. Consider: Find: $\psi(1/2)$ By definition: $$\psi(z+1) = -\gamma + \sum_{n=1}^{\infty} \frac{z}{n(n+z)}$$ The required $z$ is $z = -\frac{1}{2}$ so let $z = -\frac{1}{2}$ $$\psi(1/2) = -\gamma + \sum_{n=1}^{\infty} \frac{-1}{2n(n - \frac{1}{2})}$$ Simplify this: $$\psi(1/2) = -\gamma - ...


3

Rewrite the expression as $\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} = \dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}}$. For all $1 \le k \le n$ we have $\dfrac{n}{n+1}\left(1 + \dfrac{k}{n}\right) = \dfrac{1 + \frac{k}{n}}{1+\frac{1}{n}} \le \dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}} \le \dfrac{1 + \frac{k}{n}}{1+0} = 1 + \dfrac{k}{n}$. ...


-1

I think the integral becomes $\int_0^1 1+x dx$. The effect of $k$ in the denominator is to change the sum by a factor between 1 and $1+1/n$, so in the end it does not change the limit.


1

Here is another approach that uses Abel's theorem (see here). Let $f(x) = \sum_{n=1}^\infty (-1)^n {x ^n \over n}$. Since $\sum_{n=1}^\infty (-1)^n {1 \over n}$ converges, Abel's theorem gives $f(1) = \lim_{x \uparrow 1} f(x)$. The radius of convergence of $f$ is 1, we have $f'(x) = \sum_{n=0}^\infty (-1)x^n = {1 \over 1+x}$ and so $f(x) = f(0)+ \int_0^x ...


2

You should have gotten $\displaystyle\sum_{n = 1}^{\infty}\dfrac{(-1)^n}{n} = \sum_{n = 1}^{\infty}(-1)^n(I_{n-1}+I_n)$. (You forgot the summation sign). Now, look at the partial sums: $\displaystyle\sum_{n = 1}^{N}(-1)^n(I_{n-1}+I_n) = -(I_1+I_2)+(I_2+I_3)-(I_3+I_4)+\cdots+(-1)^N(I_{N-1}+I_{N})$. A lot of terms cancel. Now, take the limit as $N \to ...


0

Why not to start with $$A=\sum_{n=0}^{\infty }(-1)^n\Big(\frac{x}{1-x}\Big)^n=\sum_{n=0}^{\infty }\Big(\frac{-x}{1-x}\Big)^n=\sum_{n=0}^{\infty }y^n=\frac{1}{1-y}=1-x$$


2

The proof fails because it treats $k$ as if it were constant with $n$. The proof is already off the rails once you get to $$-k2^{-n}+t<\varepsilon$$ since, from there, you show that this cannot hold for all large enough $n$ - but of course it doesn't hold for all large enough $n$, because that would be equivalent to requiring $t<\varepsilon$, as the ...


0

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} ...


2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} ...


5

Reading the comments, I improve my explanations : when $x$ is close to $0.5$ but less than $0.5$ $0<\frac{x}{1-x}<1$ and hence $\sum_{n=0}^{\infty }\left(\frac{-x}{1-x}\right)^n $ converges. However, observe that when $x$ is close to $1$ or some other number it may happen that series diverges. We should be careful about it. $$\lim_{x\rightarrow ...


4

There are some way how to determine $R=1$, for example: Ratio test We have: $$\lim_{n \to \infty}\frac{|a_{n+1}|}{|a_n|}=\lim_{n \to \infty}\frac{(3n+1)x^3}{(3n+4)}=|x^3|$$ So series converges for $|x|<1$. Root test We have: $$\limsup_{n \to \infty}\sqrt[n]{|a_n|}=\lim_{n \to \infty}|x|^3\sqrt[n]{3n+1}=|x|^3$$


0

The solution is to estimate: $$\sum_{t=1}^\infty e^{-tk} \frac{(tk)^t}{t!}$$ Taylor Series rough estimation $$ f(k) = \sum_{t=1}^\infty e^{-tk} \frac{(tk)^t}{t!}= k + e^{-2k}\frac{2k}{2!} +e^{-3k} \frac{3k^3}{3!}+ e^{-4k} \frac{4k^4}{4!} + e^{-5k} \frac{5k^5}{5!}+ e^{-6k} \frac{6k^6}{6!}$$ if integer $$K > 0$$


0

The $n$th term is $(n+1)(n+2)/(3!\cdot 3^n)$. Thus, if we define $$f(x) = \frac1{3!} \sum_{n=1}^{\infty} x^{n+2}{3^n} = \frac{x^3}{6 (3-x)} $$ Then $$f''(x) = \frac1{3!} \sum_{n=1}^{\infty} (n+2)(n+1) \frac{x^n}{3^n} $$ We then seek $1+f''(1)$. The differentiation is straightforward; I get for the stated sum $1+(19/24) = 43/24$.


0

Hint: what are the residues of the function $$ f(z) = \frac \pi{\sin \pi z} $$?


1

For a complex variable $s$, whose real part is greater than zero, the Dirichlet eta function is defined by the series $$\eta(s) := -\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}.$$ In particular, one has that $$\eta(s) = \left(1-2^{1-s}\right)\zeta(s),$$ where $\zeta$ denotes the Riemann zeta function. With that in mind, one need only substitute $s=2$ into the ...


3

We need only to bound the sum $$T(M)=\sum_{k_1\geq k_2\geq 1, k_1\geq M}\frac{1}{(k_1^2+k_2^2)^2}$$ We have: $$T(M)=\sum_{k_1\geq M}\frac{1}{k_1^3}A(k_1)$$ with $$A(k_1)=\frac{1}{k_1}\sum_{k_2=1}^{k_1}\frac{1}{(1+(k_2/k_1)^2)^2}$$ $A(k_1)$ is a Riemann sum, and has as limit $\displaystyle \int_0^1\frac{dx}{(1+x^2)^2}$. Hence $A(k_1)$ is bounded, say by $c$. ...


3

@Jack DAurizio answer is nice,and I have two solution for this \begin{align*}\arctan{\dfrac{1}{2n^2}}&=\arctan{\dfrac{2}{4n^2-1+1}}=\arctan{\dfrac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}}\\ &=\arctan{(2n+1)}-\arctan{(2n-1)} \end{align*} solution 2: ...


11

Since: $$\arctan\frac{3n^2}{2n^4-1}=\arctan\frac{1}{n^2}+\arctan\frac{1}{2n^2}$$ then: $$\sum_{n=1}^{+\infty}\arctan\frac{3n^2}{2n^4-1} = \arg\prod_{n=1}^{+\infty}\left(1+\frac{i}{n^2}\right)+\arg\prod_{n=1}^{+\infty}\left(1+\frac{i}{2n^2}\right).\tag{1} $$ Since, by the Weierstrass product for the $\sinh$ function, $$\frac{\sinh(\pi z)}{\pi ...


2

Let $$\tan x=\frac{1}{n^2} \text{and } \tan y=\frac{1}{2n^2}$$ so we have that $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{\frac{1}{n^2}+\frac{1}{2n^2}}{1-\left(\frac{1}{n^2}\frac{1}{2n^2}\right)}=\frac{3n^2}{2n^4-1}$$ hence we have ...


1

If the space is not complete, absolute convergence actually doesn't imply convergence. Take for example $X = \mathbb{Q}$ (with the norm inherited from $\mathbb{R}$) and choose a sequence of rational numbers $a_n$ such that $$ \sum_{n=1}^\infty a_n $$ is irrational, but $$ \sum_{n=1}^\infty |a_n| = 1. $$ See this question for details. Thus, in $X$, the ...


4

Even though I agree that the solution by @math110 is the best one, I want to add that this can be done using the theory of residues, using the common trick with the cotangent function. I leave it to you to fill in the details, if necessary. We let $$ f(z)=\frac{\pi\cot \pi z}{z^3(z+1)^3}. $$ This function $f$ has poles at all integers. For integers ...


0

This is a standard useful fact. $$\begin{align*} \prod_{k=1}^n\frac{2k-1}2&=\frac{1\cdot 3\cdot 5\cdot\ldots\cdot(2n-1)}{2^n}\\\\ &=\frac{1\cdot 3\cdot\ldots\cdot(2n-1)}{2^n}\cdot\frac{2^n n!}{2^n n!}\\\\ &=\frac{1\cdot 3\cdot\ldots\cdot(2n-1)}{2^n}\cdot\frac{(2\cdot 1)(2\cdot 2)\ldots(2n)}{2^n n!}\;; \end{align*}$$ can you finish it from ...


3

You can faktor out n: $2^m=n\cdot \left( \frac{1+n}{2}+k \right)$ n has to be a faktor of $2^m$. Let it be $2^p$. $2^m=2^p\cdot \left( \frac{1+2^p}{2}+k \right)$ muliplying the equation by 2. $2^{m+1}=2^p\cdot \left(1+2^p+2k \right)$ The equation does not hold, because $1+2^p+2k$ is an odd number. Reffering to the comment of robjohn: If $2^p=1$ and ...


1

$$ \frac{n(1+n)}{2}+nk=n\left(\frac{n+1}{2}+k\right) $$ so if this has to hold $n$ has to be a factor of $2^m$. For any $n\leq2^{\frac{m}{2}}$ it is easy to find values of $k$ that solves the equation.


1

Choose $\epsilon = \dfrac{1}{3} \to \exists N_0 \in \mathbb{N}$ such that if $m, n > N_0 \to |x_n-x_m| < \dfrac{1}{3} $ since its a Cauchy sequence. But $x_n, x_m \in \mathbb{N} \to |x_m-x_n| \in \mathbb{N}\cup \{0\} \to |x_m-x_n| = 0 \to x_m = x_n$. This shows the sequence is a constant after $N_0$, and we can call this constant value $y$. It is easy ...


2

The fact that $0 \not\in \mathbb{N}$ (if you use that convention) is not relevant. What you need to do is choose an $\varepsilon > 0$ such that $|x_n - x| < \varepsilon$ implies that $x_n = x$. In general you can't do this, but in this situation you can. As a hint for which $\varepsilon$ to choose, what is the smallest $|x_n - x|$ can be without ...


1

Note that $$\sum_{k\geqslant 1}\sum_{n\leqslant k}=\sum_{k\geqslant 1}\sum_{n\geqslant 1}[n\leqslant k]=\sum_{n\geqslant 1}\sum_{k\geqslant 1}[n\leqslant k]=\sum_{n\geqslant 1}\sum_{k\geqslant 1}[k\geqslant n]=\sum_{n\geqslant 1}\sum_{k\geqslant n}$$ This is an example of the usefulness of the Iverson bracket.


2

Visualizing things makes the problem much simpler and easier to understand in such cases. Consider the value pairs $(n,m)$ is allowed to take on a 2D grid: $$ \begin{array}{ccccc} (1,1) & & & &\\ (2,1) & (2,2) & & & \\ (3,1) & (3,2) & (3,3) & & \\ (4,1) & (4,2) & (4,3) & (4,4) & \\ \vdots ...


4

JimmyK has ignored the request (in now deleted comments) to add the lower bound proof to his answer, so I will add an answer with the lower bound proof which will resolve the question completely. Just like Jimmy's excellent upper bound proof we use $$ 2014 = \sum_{n=1}^{2014}\left(1 + \frac{1}{y_n}\right)(y_n - y_{n+1}) = \sum_{n=1}^{2014} ...


1

I assume $c$ is not a negative integer, so you don't have trouble with $0$ denominators. If $t_n$ is the $n$'th term we have $$ \dfrac{t_{n+1}}{t_n} = \dfrac{(n+a)(n+b)}{(n+c)(n+1)} = 1 + \dfrac{a+b-c-1}{n} + O(1/n^2)$$ Use Gauss's test.


1

Since $(k-1)^2+(k-1)+1 = k^2-k+1$, you need to substitute $k' = k-1$ into the 2nd sum: The 1st sum has bounds $1 \le k \le n$, i.e. $0 \le k-1 \le n$, so the 2nd sum has bounds $0 \le k' \le n-1$. Using this substitution, we have: $\displaystyle\sum_{k = 1}^{n}\arctan\dfrac{1}{k^2-k+1} = \sum_{k = 1}^{n}\arctan\dfrac{1}{(k-1)^2+(k-1)+1} = \sum_{k' = ...


0

When you change the dummy variable in a summation, you have to change the limits to make sure you get the same terms. It is the same as substitution in integrals. In your first example of $x=y^2+k$ you did not define which of $y,k$ is the new summation variable. If it is $k$, the range becomes $1-y^2$ to $n-y^2$ If $y$ is the new summation variable, it ...


5

Base of induction Since $a_1=1$ and $a_2=2/3$, the inequality $a_{n+1}\le a_n$ is true for $n=1$. Inductive step If $a_{n+1}\le a_n$, then $a_{n+1}^2\le a_n^2$, since all $a_n$ are positive by definition. Hence $$a_{n+2} = \frac13\left(a_{n+1}^2+\frac1{n+1}\right)\le \frac13\left(a_n^2+\frac1n\right) = a_{n+1}$$ which establishes the inductive step. ...


1

The general statement implied by the title of the question is false. There are real valued sequences with infinitely many limit points. For example, the sequence $$ 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, \dots $$ has all natural numbers as limit points. Even worse, take a bijection $f\colon\mathbb N\to\mathbb Q$. Now $a_n=f(n)$ has every real number as ...


2

You have $s(n)=s(0)0.9^n+90\cdot 0.9^{n-1}+90\cdot 0.9^{n-2}+\dots 90$, where each term after the first comes from the decay of each added $90$. This is a geometric series which we can sum, getting $$s(n)=s(0)0.9^n+90\frac {1-0.9^n}{1-0.9}$$


1

Suppose there is some other limit $D$ so that $D \notin \{A,B,C\}$. Let $\varepsilon = \frac{1}{2}\min\{|D-A|,|D-B|,|D-C|\}$. Note that for this value of $\varepsilon$ there are no values of $a_n$ in the $\varepsilon$ ball of $D$, implying that $D$ is not a limit. This is the basic idea of the proof. The wording could be changed a bit to match the ...


3

We have: $$-\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n}$$ for any $x$ such that $|x|<1$, hence: $$-\frac{\log(1-x)}{1-x}=\sum_{n\geq 1}H_n x^n.$$ and since $\frac{d}{dx}\log^2(1-x) = -2\frac{\log(1-x)}{1-x}$ we have: $$\log^2(1-x) = 2\sum_{n\geq 1}\frac{H_n}{n+1}x^{n+1}\tag{1}.$$ Since, by partial summation: $$\sum_{n=1}^{N}\frac{H_n}{n}= ...


0

Source: Hojoo Lee, "Topics in Inequalities - Theorems and Techniques" (February 25, 2006), p.44.


19

Hint $$\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$$ and $$(a-b)^3=(a^3-b^3)-3ab(a-b)$$ so \begin{align*}\dfrac{1}{n^3(n+1)^3}&=\left(\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}\right)-3\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\left(\dfrac{1}{n}-\dfrac{1}{(n+1)}\right)\\ ...


0

since note $$\left(\dfrac{x_{n+1}}{y_{n+1}}\right)^2=\dfrac{x^2_{n+1}}{x_{n+1}y_{n}}=\dfrac{x_{n+1}}{y_{n}}=\dfrac{1}{2}\left(\dfrac{x_{n}}{y_{n}}+1\right)$$ so let $\dfrac{x_{n}}{y_{n}}=\cos{c_{n}},c_{n}\in(0,\dfrac{\pi}{2})$ use this $$\dfrac{1}{2}\left(\cos{2x}+1\right)=\cos^2{x}$$ so $$\cos{c_{n+1}}=\cos{\dfrac{c_{n}}{2}}\Longrightarrow ...



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