10 votes

Testing for convergence - $\int_{0}^{\infty} \frac{x^4}{e^{\sqrt{x}}} dx$

Hint The substitution $x = u^2, dx = 2 u \,du$, transforms the integral to $$2 \int_0^\infty u^9 e^{-u} \,du.$$
Travis Willse's user avatar
4 votes

How to prove that the following series converges?

$f(x) = e^x$ satisfies $f'(x) = e^x < e$ for $0 < x < 1$, so that by the mean-value theorem $$ 0 < e^\frac{1}{n} - e^\frac{1}{n+2} < e \left( \frac 1n - \frac{1}{n+2}\right) = \frac{2e}...
Martin R's user avatar
  • 112k
4 votes

If $f_n\to f$ pointwise on $[0,1]$, then $\sup_{[0,1]}f_n\to\sup_{[0,1]}f$ pointwise?

Presumably the intended question was If $f_n\to f$ pointwise on $[0,1]$, must $\sup_{[0,1]}f_n\to\sup_{[0,1]}f$? If so, the answer is "no". For a counterexample, if we let $$ f_n(x)= \begin{cases} ...
quasi's user avatar
  • 58.6k
4 votes
Accepted

Does $\int_0^{+\infty}\frac{\ln(1+x^n)-\ln(1+x^{n+1})}{(1+x^2)\ln(x)}\, dx$ converge? How?

Define $I$ as the integral that we need to compute and $I_1 = \int_0^{1}\frac{\ln(1+x^n)-\ln(1+x^{n+1})}{(1+x^2)\ln(x)}\, dx$. Then, $$\begin{align*} I &= I_1+\int_{1}^{\infty}\frac{\ln(1+x^n)-\ln(...
Sam's user avatar
  • 1,685
3 votes

How to prove that the following series converges?

We have that $$ \left(1+\frac1n\right)^n\le e \le \left(1+\frac1n\right)^{n+1}$$ and then $$e^\frac{1}{n} - e^\frac{1}{n+2}\le \left[\left(1+\frac1{n-1}\right)^n\right]^\frac1n-\left[\left(1+\frac1{n+...
user's user avatar
  • 154k
3 votes
Accepted

Testing for convergence - $\int_{0}^{\infty} \frac{x^4}{e^{\sqrt{x}}} dx$

You may split your integral in two, one from $0$ to $1$ and other from $1$ to $\infty$. It is obvious the first one is finite due to continuity of the integrand. Furthermore, when $x\to\infty$, $\...
Julio Puerta's user avatar
3 votes
Accepted

Convergence in $L^p_{loc}$ implies convergence of a subsequence in $L^\infty$

Not possible. Take $f_n = \chi_{ [1/n,\ 1+1/n] }$.
daw's user avatar
  • 48.6k
3 votes

How do you show the convergence of $\sum_{n=1}^∞ 2^n\tan(1/n!)$?

When $n\to\infty$, $\displaystyle\frac{1}{n!}\to 0$, so $\displaystyle\lim_{n\to\infty} \displaystyle\frac{\tan\left(\displaystyle\frac{1}{n!}\right)}{1/n!}=1$. Thus, the series has the same character ...
Julio Puerta's user avatar
3 votes
Accepted

On the asymptotic behavior of a sequence[Bulgaria MO 2024 Regional Round, 12,2]

$x_1$ needs to be $>0$ otherwise $x_n = 0$. Using AM-GM inequality you can prove that $x_n^2 \le \sum_{k=1}^{n-1} x_k$ use induction and prove $x_n \le nx_1$ and $y_n$ is bounded. Now let $\beta :...
Kroki's user avatar
  • 12.9k
2 votes

Does the serie converges or diverges?

Let $\displaystyle a_n:=\left(1-\frac{\ln n}{n}\right)^n$ and $b_n:=\dfrac1n$. Then $$\lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty} n\left(1-\frac{\ln n}{n}\right)^n \overset{\star}{=} 1 > 0$...
Alma Arjuna's user avatar
  • 2,730
2 votes
Accepted

Find the radius of convergence of $\sum_{n=1}^\infty {4^n \over n} x^{2n+3}, \sum_{n=1}^\infty 2^n x^{2^n}$.

Here is a slightly different approach to determine the radius of convergence $r$ of the first series. We do not consider it as series with even powers $2n$ of $x$, but as series with powers $n$ of $x^...
Markus Scheuer's user avatar
2 votes
Accepted

Limiting behaviour of state vector under delayed (open-loop) inputs

If your controller is linear, then you can rewrite the closed-loop system into the form $$z(k+1)=M_0z(k)+M_dz(k-d)$$ where $d$ is the constant delay. The state $z$ will contain here the state of the ...
KBS's user avatar
  • 7,000
2 votes

How to prove that the following series converges?

At infinity, using Taylor expansion, you have: $$e^{\frac1n}-e^{\frac1{n+2}}\sim 1+\frac1n-1-\frac1{n+2}=\frac{2}{n(n+2)}\sim\frac2{n^2} $$ Hence the series is convergent.
Sine of the Time's user avatar
2 votes

Testing for convergence - $\int_{0}^{\infty} \frac{x^4}{e^{\sqrt{x}}} dx$

We can apply infinite series. Namely $$\int\limits_{n^2}^{(n+1)^2 }{x^4\over e^{\sqrt{x}}}\,dx\le {(n+1)^8\over e^n}(2n+1)$$ By the Cauchy $n$-th root test the series $\sum{(n+1)^8\over e^n}$ is ...
Ryszard Szwarc's user avatar
1 vote

Is it always true that if $x_n\to0$, $y_n\to0$ there exist $\epsilon_n\in\{-1,1\}$ such that both $\sum\epsilon_nx_n$and $\sum\epsilon_ny_n$ converge?

Hints for the second question: work in $\mathbb R^2$ with the sequence $z_n = \left(x_n, y_n\right)$ without loss of generality we can assume that, $\left\|z_n \right\| \le 1$. For $k\in \mathbb N$, ...
Kroki's user avatar
  • 12.9k
1 vote
Accepted

Prove that the sequence $\frac{4n^2-3n}{2n^2−1}$ converges to the limit $2$

$|\frac{2-3n}{2n^2−1}|=\frac {3n-2} {2n^{2}-1}\le \frac {3n} {n^{2}}=\frac 3n<\epsilon$ if $n >\frac 3 {\epsilon}$.
geetha290krm's user avatar
1 vote
Accepted

limit of sequence $\{a_n\}$ where $a_n = n!/c^n$, as $n$ tends to infinity

I’d say that the approach you are taking is correct. One way to maybe make it more clear could be the following. We’d like to prove that $a_n\to \infty$. Then, for $n \geq c$, we have $$\frac{n!}{c^n} ...
blomp's user avatar
  • 118
1 vote

Show $nx_n \to 1$ for sequence defined by $x_{n+1}=\frac{x_n}{1+nx_n^2}$

Motivation. Let $y_n = \frac{1}{x_n}$. Then the recurrence relation reads as: $$ y_{n+1} - y_n = \frac{n}{y_n} $$ The continuum analog of the relation is the differential equation $$ y' = \frac{x}{y}. ...
Sangchul Lee's user avatar
1 vote

Show $nx_n \to 1$ for sequence defined by $x_{n+1}=\frac{x_n}{1+nx_n^2}$

Let $a_n = \frac1{nx_n}$ then, \begin{align} a_{n+1} &= \frac1{n+1} \times \frac1{x_{n+1}}\\ &= \frac1{n+1} \left(\frac{1+nx_n^2}{x_n}\right)\\ &= \frac1{n+1} \left(\frac{n}{nx_n} + nx_n\...
Kroki's user avatar
  • 12.9k
1 vote

Show $nx_n \to 1$ for sequence defined by $x_{n+1}=\frac{x_n}{1+nx_n^2}$

Once you have $(nx_n)_n$ converges to a limit $l>0$, suppose $l\neq 1$ : then as $\displaystyle (n+1)x_{n+1}-nx_n=\frac{x_n(1-n^2x_n^2)}{1+nx_n^2} (1)$ you get $\displaystyle (n+1)x_{n+1}-nx_n\sim\...
Fred Bernard's user avatar
1 vote

Show $nx_n \to 1$ for sequence defined by $x_{n+1}=\frac{x_n}{1+nx_n^2}$

Once you have shown that $nx_n \le 1$ for $n \ge 2$ you can use the general form of the Stolz–Cesàro theorem: $$ \frac{1}{\liminf_{n \to \infty} nx_n} = \limsup_{n \to \infty} \frac{1}{nx_n} = \...
Martin R's user avatar
  • 112k
1 vote
Accepted

Convergence radius for Cauchy multiplication series

You were on the right track! Here's a way to successfully apply your idea . . . If $f,g$ are given by \begin{align*} f&= \frac{2-x}{(1-x)(3-x)} = \frac{1}{2(1-x)} + \frac{1}{2(3-x)} \\[4pt] g&...
quasi's user avatar
  • 58.6k
1 vote
Accepted

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be continuous and $A=\{x\in\mathbb{R} | * \}$. If $(x_n)\in A$ and converges to $c$, show $c\in A$.

The definition of continuity is that if $x_n\to c$, then $f(x_n)\to f(c)$. The only requirement is that $x_n$ and $c$ are in the domain of $f$ (otherwise the statement makes no sense because $f$ isn't ...
Bobby Ocean's user avatar
  • 3,163
1 vote

Does $\int_0^{+\infty}\frac{\ln(1+x^n)-\ln(1+x^{n+1})}{(1+x^2)\ln(x)}\, dx$ converge? How?

Mathematica suggests, that the integral does not depend on $n$ ...
Roland F's user avatar
  • 1,597
1 vote

Convergence of a Sequence by definition. Approach issue

You said in a comment that for certain values of $\varepsilon$ I am in no position to continue my argument since I never considered the sign of the right hand side expression in the first place I ...
mathlove's user avatar
  • 138k

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