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Convergence or divergence of $a_{n+1}=a_n +\frac{a_{n-1}}{(n+1)^2}$

This is converted from my comment on the limit $a_\infty=\lim\limits_{n\to\infty}a_n$. Once we know it exists, with $g(z):=\sum_{n=1}^\infty a_n z^n$ for $|z|<1$, we have $a_\infty=\lim\limits_{z\...
metamorphy's user avatar
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2 votes

Does this recurrent sequence have a limit?

The question is about a sequence that satisfies the recursion $$ a_k = \frac{k\left(k+2\right)+\left(k+1\right)} {k\left(k+2\right)}\left[a_{k-1} - \frac{k-1}{k\left(k-2\right) + \left(k-1\right)}\,...
Somos's user avatar
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Can someone help me find the hole in my $\epsilon$-$\delta$ argument? Hint needed (Spivak Chapter 5 Problem 20)

Apply the Cauchy criterion to disproof the existence of f(a) for $a \neq 0$ $$\exists \epsilon > 0 \,\, , \forall \delta > 0 : |x_1 - a| < \delta \,\, \text{and} \,\,|x_2 - a| < \delta \...
likely_fail_2202 T_T's user avatar
1 vote

Can someone help me find the hole in my $\epsilon$-$\delta$ argument? Hint needed (Spivak Chapter 5 Problem 20)

“… in which case $l=x$ as to satisfy every $\epsilon>0$.” this is invalid. Once you fix an $\epsilon$, the $\delta$ you get from the definition will depend on $\epsilon$, so the values of $x$ that ...
Alann Rosas's user avatar
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4 votes
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Limit of a function with two variables: $\lim \limits_{(x,y) \to (1,5)} \frac{\arctan (y-5x)}{\lvert x-1 \rvert +\lvert y-5\rvert}$

Along $y=5$, the limit is $\lim_{x\to1}\frac{\arctan(5-5x)}{|x-1|}=-5\lim_{x\to1}\frac{x-1}{|x-1|}$. Obviously, this limit does not exist.
Yves Daoust's user avatar
1 vote

Prove/disprove $\lim\limits_{x\to x_0}g(f(x))=\lim\limits_{y\to y_0}g(y)$

It is not true that, as the sequence $\,(x_n)_{n\in\Bbb N}\subset[a,b]\,$ that converges to $\,x_0\,$ was chosen arbitrarily, it follows that $\displaystyle\lim_{n\to \infty}g(f(x_n))\!=\!\lim\limits_{...
Angelo's user avatar
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0 votes

Prove/disprove $\lim\limits_{x\to x_0}g(f(x))=\lim\limits_{y\to y_0}g(y)$

Here is a simple example. Let $f, g : I \to I$, where $I = [0, 1] \subseteq \mathbb{R}$. Define $f$ by $$f (x) = 0$$ and $g$ by $$g (x) = \begin{cases} 0 & x = 0 \\ 1 & \text{otherwise} \end{...
K. Jiang's user avatar
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0 votes
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$I_n=\int_{0}^{1} \arcsin(1-x^n) \,dx $

First of all you know that the limit exists since the sequenze is monotone. Note that $\arcsin(1-x^n) ≤\pi/2$, so the limit is $≤\pi/2$. Now we need a lower bound: note that in each interval of the ...
user1361032's user avatar
0 votes

Does this recurrent sequence have a limit?

Here is my attempt using the method of generating function: Let $g(x) = \sum_{k \geq 0} a_{k+1}x^{k}$, we have: $\begin{align*} g(x) &= 1+ 0 + \sum_{k \geq 2}\frac{(k+1)*(k+3) + (k+2)}{(k+1)*(k+3)...
Jake ZHANG Shiyu's user avatar
1 vote
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Convergence of series regarding symmetric difference

Yes, rather likely the limit here is in the sense of the set-theoretic limit. And yes, the partial sums in the series are undoubtedly with respect to the symmetric difference $\triangle$, since that ...
Linear Christmas's user avatar
2 votes

Use $\,\varepsilon-\delta\,$ definition of limits to show that $\lim\limits_{(x,y)\to(-1,2)}\frac{x^3+y^3}{x^2+y^2}=\frac{7}{5}$

Alternative approach: This approach presumes that the problem composer's intent is that $~\epsilon~$ and $~\delta~$ be wrestled with, rather than constructing an argument based on continuity of ...
user2661923's user avatar
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0 votes

Does L'Hôpital's work the other way?

Referring to Ben Grossman's interesting application (more of a comment) . Find $\dfrac{w_{n+1}'(x_i)} {w_{n}'(x_i)}$. Product rule: $w_{n+1}'(x_i)=$ $w_{n}'(x_i)(x_i-x_n) +1 \cdot 0$ , and we are done....
Peter Szilas's user avatar
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-1 votes

Use $\,\varepsilon-\delta\,$ definition of limits to show that $\lim\limits_{(x,y)\to(-1,2)}\frac{x^3+y^3}{x^2+y^2}=\frac{7}{5}$

To complement Robert Z's answer, here is a way that makes calculating the limit easier (albeit, it does not use the $\epsilon-\delta$ definition of the limit directly.) Note that the functions $x^{3}+...
roblich mandervach's user avatar
3 votes

Use $\,\varepsilon-\delta\,$ definition of limits to show that $\lim\limits_{(x,y)\to(-1,2)}\frac{x^3+y^3}{x^2+y^2}=\frac{7}{5}$

You should manipulate the RHS in order to find something which goes to zero as $(x,y)\to (-1,2)$. For instance, starting from your work, we have \begin{align*} \left\lvert\frac{x^3+y^3}{x^2+y^2}-\frac{...
Robert Z's user avatar
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2 votes
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Trouble Understanding Difference in Epsilon-Delta Arguments: Why One Works But The Other Fails (Spivak Calculus Problem 5-10c)

It's true that $(\sqrt[3]{x})^3=x$, but $\sqrt{x}$ doesn't even make sense for $x<0$, and hence you can write $(\sqrt x)^2=x$ only for $x>0$.
Martin Argerami's user avatar
2 votes

Trouble Understanding Difference in Epsilon-Delta Arguments: Why One Works But The Other Fails (Spivak Calculus Problem 5-10c)

For each $x\in\Bbb R$, $\sqrt[3]x$ exists (and this was used in the proof). But it is not true that every real number has a square root; only non-negative real numbers have it.
José Carlos Santos's user avatar
5 votes

Limit of a function when 'a' is not in the domain

In a limit $\lim_{x\to a} f(x)$, the number $a$ need not be in the domain of the function, but it does need to be a "limit point" of the domain, i.e. you need to be able to get arbitrarily ...
J. Chapman's user avatar
  • 1,112
2 votes

Is the approach I did to justify this limit equality correct?

By Jacobi’s formula, $\det(A+\Delta A)-\det(A)=\operatorname{tr}(\operatorname{adj}(A)\Delta A)+o(\|\Delta A\|)$. So, when $\Delta A=h\frac{dA}{dt}+o(h)$, $$ \det\left(A+h\frac{dA}{dt}+o(h)\right)-\...
user1551's user avatar
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2 votes

Limit of $\sum_{k=1}^n \frac{1}{k+1/2} - \ln(n+1/2)$

Considering the hint provided by @openspace, I have got the correct result. \begin{align*} \lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k+\frac{1}{2}} - \ln(n+\frac{1}{2}) &= \lim_{n\to\...
ZhouYang's user avatar
  • 105
3 votes
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Limit of $\sum_{k=1}^n \frac{1}{k+1/2} - \ln(n+1/2)$

Let $\psi(z)=\Gamma'(z)/\Gamma(z)$. Then it is known that $$ \psi(z+1)=-\gamma+\sum_{n=1}\left(\frac1n-{1\over n+z}\right), $$ so there is \begin{aligned} \lim_{n\to+\infty}\left(\sum_{k=1}^n{1\over k+...
TravorLZH's user avatar
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1 vote
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Help understanding Spivak's solution, and a verification of my proof. Spivak Chapter 5, Question 3(vi)

OP's analysis is fine, but he makes a small mistake. On the one hand we have $1-\epsilon < \sqrt x < 1 + \epsilon$, which after squaring leads to $(1-\epsilon)^{2} < x < (1+\epsilon)^{2}$. ...
M. Wind's user avatar
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0 votes

Newton approximation in Tao Analysis 1

$$ |f(x) - (f(x_0) + L(x - x_0))| = |f(x) - (f(x_0) + f(x) - f(x_0))| $$ Why would this be the case? Just because we have $$ L = \lim_{\substack{x \to x_0 \\ x \in X \\ x \ne x_0}} \frac{f(x) - f(...
PrincessEev's user avatar
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1 vote
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$A= \{(x,y,z) \in \mathbb{R}^3:x^2+2y^2+z^2 < 4z\}$ limit: $\lim_{n \to \infty}\frac{1}{n} \int_A \frac{y^2z}{ln(x^2+2y^2+n) - ln(n)} \ d\lambda_3$

Sketch: On $A$, $x^2+2y^2<4z-z^2$. The LHS is non-negative, so the RHS must be non-negative which gives $z \in [0,4]$. Thus, \begin{align*} A &\subset B := \{(x,y,z): z \in [0,4], x^2+2y^2<\...
Sounak's user avatar
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1 vote

Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$

Let $\;x_n=\dfrac n{n+1}\;$ for any $\;n\in\Bbb N\,.\quad(\implies\lim\limits_{n\to\infty}x_n=1\;)$ It results that, $\quad n(x_n-1)=-x_n\;\;$ for any $\;n\in\Bbb N\,.$ Moreover, $\begin{align}b_n&...
Angelo's user avatar
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0 votes
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How to show that the limit of a sequence is not equal to some value?

I will clarify my comment. By definition, $l$ is the limit of the sequence $a_n$ if for every $\varepsilon>0$ you can find $N$ such that $\left|a_n-l\right|<\varepsilon$ for all $n>N$. Thus, ...
Davide Masi's user avatar
0 votes

Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$

$$\frac{a_{n+1}}{a_n}=\frac{2n+1}{2n+2}$$ $$L:= \lim\limits_{ n \to \infty} n \left(\frac{2n+1}{2n+2} \left(\frac{n}{n+1}\right)^{\frac 1 3}\right)= \lim\limits_{ n \to \infty} n \left(\frac{2n+1}{2n+...
pie's user avatar
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2 votes

Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$

The expression can be represented as $${n\over 2(n+1)}-\left ({n\over n+1}\right )^{4/3}\,[-(n+1)]\left [\left (1-{1\over n+1}\right )^{2/3}-1\right ]$$ The limit of the last two factors is equal by ...
Ryszard Szwarc's user avatar
3 votes

Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$

Let me call your while expression $A_n$. Notice, that there is the identity $\Gamma(1/2 + n) = \sqrt{\pi} \frac{(2n)!}{n! 4^n}$. Then you have $$a_n = \frac{\Gamma(n+1/2)}{\sqrt{\pi} n!}$$ and yields ...
dForga's user avatar
  • 121
3 votes

Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$

Using $A^3-B^3 = (A-B)(A^2 +AB + B^2)$, we get $$ \begin{split} \frac{n+1/2}{n+1} - \sqrt[3]{\frac{n}{n+1}} &= \frac{\frac{(n+1/2)^3}{(n+1)^3}- \frac{n}{n+1}} {\left(\frac{n+1/2}{n+1}\right)^2 +...
Sewer Keeper's user avatar
  • 1,470
3 votes
Accepted

Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$

Note that: $$n\left[\frac{n+1/2}{n+1}-\left(\frac{n}{n+1}\right)^{\frac{1}{3}}\right] =\frac{n}{n+1}\left[(n+1/2)-\sqrt[3]{n(n+1)^2}\right].$$ $$\lim_{n\to\infty}n\left[\frac{n+1/2}{n+1}-\left(\frac{n}...
Riemann's user avatar
  • 8,545
4 votes

Is there a closed form for the quadratic Euler Mascheroni Constant?

Using this answer, it's not hard to show that the limit equals $$ L=\color{blue}{\pi\big(2\gamma-\ln\pi+4\ln\Gamma(3/4)\big)}\approx 2.584981759579253217+ $$ Indeed, we have $L=\lim\limits_{N\to\infty}...
metamorphy's user avatar
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1 vote

Prove of rules of operation with limits that diverge

I will write an answer here so as to not overflow the comments. I apologize for the long answer, I want to convey as much intuition as possible. Your proofs are almost correct. Here are a few tips to ...
Fançois Gatine's user avatar
1 vote
Accepted

Proving $\lim_{y\to 0} y^2 \ln|yx^2|=0$ using sequences

There is no difference between the proofs for the particular sequence $(b_n)$ and a general one: $b_n^{2}\ln (|b_n|x^{2})=b_n^{2}\ln |b_n|+b_n^{2}\ln (x^{2})$ and each of the two terms tends to $0$. [...
geetha290krm's user avatar
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1 vote
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Does there exist a positive sequence with these two properties?

While the answer by @John B is of course correct, I wanted to give a simple explicit construction of such a sequence. Ignoring the $(-1)^n$ for a moment, you essentially want multiples of $\sqrt2$ ...
anankElpis's user avatar
  • 1,549
2 votes

Does there exist a positive sequence with these two properties?

Let's forget first about $(-1)^n$. Given $n\in\mathbb N$, take $c_n>0$ such that $|\{c_n\}-1|<\frac1n$. Of course, $$ |\{m+c_n\}-1|=|\{c_n\}-1|<\frac1n\tag1 $$ for any positive integer $m>...
John B's user avatar
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Showing that the function that returns $1$ on the rationals and $0$ elsewhere has no limit at $0$. Is there anything wrong in this proof?

Added a bit of annotations to the proposed proof attempting to point out some logical flaws. Proof: Suppose that $\lim_{x \to 0; x\in \mathbb R}f(x) = L$. Then for every $\varepsilon > 0$, there ...
Steen82's user avatar
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0 votes

Showing that the function that returns $1$ on the rationals and $0$ elsewhere has no limit at $0$. Is there anything wrong in this proof?

As other comments suggested, I would say that you should consider the following reasoning for the proof of the first part of your post. It is rather classical and concise. Let $u_n=\frac{1}{n+1}$ and $...
Maxime's user avatar
  • 395
0 votes

Evaluating $\lim_{x\to\infty}\left[\cos\frac1x\right]^{h(x)}$, where $h(x)=\frac{x^4+x^2-1}{2x+1}\sin\frac1x$

I found this cool inequality and never managed to used it before, so here it goes: $x \in (0, \frac{\pi}2) \implies \frac{\sin(x)}{x} > \frac 2{\pi}$ of course for $x > \frac 2{\pi}$ we will ...
hellofriends's user avatar
  • 1,990
2 votes

Evaluating $\lim_{x\to\infty}\left[\cos\frac1x\right]^{h(x)}$, where $h(x)=\frac{x^4+x^2-1}{2x+1}\sin\frac1x$

By substituting $t=1/x$ we get $t\to 0^+$ and $$h(1/t)\log \cos t={1+t^2-t^4\over 2t^3+t^4}\sin t\log \cos t \ (*)\\ ={1+t^2-t^4\over 2+t}{\sin t\over t}\,{\cos t -1\over t^2}\,{\log \cos t\over \cos ...
Ryszard Szwarc's user avatar
8 votes

Evaluating $\lim_{x\to\infty}\left[\cos\frac1x\right]^{h(x)}$, where $h(x)=\frac{x^4+x^2-1}{2x+1}\sin\frac1x$

$\textbf{Hint:}$ For large $x$, we have that $$\cos\left(\frac{1}{x}\right) \sim 1 - \frac{1}{2x^2}$$ and $$\frac{x^4+x^2-1}{2x}\sin\left(\frac{1}{x}\right) \sim \frac{x^2}{2}$$ therefore your limit ...
Ninad Munshi's user avatar
0 votes

Showing that $y^2-2 < \delta$ and $2-x^2 < \delta$ $\implies y-x < \delta$

Since $\,\;0<x<\sqrt2<y\;\,$ and $\,\;2-x^2<\delta\;,\;y^2-2<\delta\;,\;$ where $\;\delta>0\;,\;$ it follows that $\begin{cases}2-\delta<x^2<2\\[3pt]2<y^2<2+\delta\end{...
Angelo's user avatar
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4 votes
Accepted

Why can the Binomial Distribution be Approximated by a Normal Distribtuion?

For the second factor in step 8, let $v=1/n$ , and take logarithm to get $$ \frac{1}{v} \log\left\lbrace 1 + u(v)\right\rbrace, $$ where $$ u(v)=p\left( \exp\left\lbrace \frac{\sqrt{v}t}{\sqrt{p(1-p)}}...
Zack Fisher's user avatar
0 votes

Showing that $y^2-2 < \delta$ and $2-x^2 < \delta$ $\implies y-x < \delta$

Given $0 < x < \sqrt{2} < y$ (by convention $\sqrt{}$ is positive and so is an approximation) and $0 < x^2 < 2 < y^2$, then $\delta > 0$. Further, $\sqrt{2} < y < \sqrt{2 + \...
bhache's user avatar
  • 1
2 votes

Why can the Binomial Distribution be Approximated by a Normal Distribtuion?

Consider the Binomial PMF: $$ P(X=k)=\binom{n}{k} p^k(1-p)^{n-k} $$ We will use Stirling's Approximation: $$ n!\approx \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n $$ ...for the binomial coefficient: $$ \...
vallev's user avatar
  • 408
0 votes

Prove that sequence $(a_n)_{n\geq 1}$ is not convergent.

Note, this is not a full solution. For any angle $\theta \in (-2\pi, 2\pi)$ the following holds (you can prove it via geometry): $$ 1-e^{i\theta}= \begin{cases} Ae^{i\frac{\theta-\pi}{2}} & \theta\...
zetko's user avatar
  • 319
1 vote

Why $\lim_{x \rightarrow \infty } \frac {P(x)}{Q(e^x)} = 0$ for polynomials $P(x)$ and $Q(x)$?

We will use two properties. if $\deg R<\deg S$ for two polynomials $R(x), S(x),$ then ${R(x)\over S(x)}\underset{x\to\infty}{\longrightarrow} 0.$ $e^t>t>0$ which implies $e^x=(e^{x/m})^m> ...
Ryszard Szwarc's user avatar
1 vote
Accepted

Why $\lim_{x \rightarrow \infty } \frac {P(x)}{Q(e^x)} = 0$ for polynomials $P(x)$ and $Q(x)$?

I'm not sure I understand your objection. In general, if the numerator of a fraction grows sufficiently slower than its denominator, regardless of which infinity that denominator goes to ($+\infty$ or ...
PrincessEev's user avatar
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5 votes

Why $\lim_{x \rightarrow \infty } \frac {P(x)}{Q(e^x)} = 0$ for polynomials $P(x)$ and $Q(x)$?

If $Q(x)=a_0+a_1x+\cdots+a_nx^n$, with $n\in\{1,2,3,\ldots\}$ and $a_n\ne0$, then\begin{align}\lim_{n\to\infty}\frac{Q(e^x)}{e^{nx}}&=\lim_{n\to\infty}a_0e^{-nx}+a_1e^{-(n-1)x}+\cdots+a_{n-1}e^{-x}...
José Carlos Santos's user avatar
1 vote
Accepted

$\lim_{x\to\infty}\exp{(x+iy)}$

$\lim_{x \rightarrow \infty} \exp(x + iy)$ diverges to infinity in a direction on the complex plane that is determined by the value of $y$. We could write that as $\infty \cdot \exp(iy)$ as shorthand ...
ConMan's user avatar
  • 25.7k
1 vote

Proving that $\lim_{h\to 0 } \frac{b^{h}-1}{h} = \ln{b}$

$$\lim_{h\to 0}{b^h-1\over h}$$ First, we define a variable t such that $b^h-1=t$. Therefore, $b^h=t+1$ and $h=\log_b{(t+1)}$. It should be noted that, if we take the limit as $h$ goes to $0$, $t$ ...
Eli Berk's user avatar

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