New answers tagged

4

The following statements For each $\epsilon>0$, there exists $K\in \mathbb{N}$ such that $n\geq K \Rightarrow |a_n-a|\leq \epsilon$ and For each $\epsilon>0$, there exists $K\in \mathbb{N}$ such that $n\geq K \Rightarrow |a_n-a|< \epsilon$ are equivalent. Proof of 1.$\Rightarrow$2. For any $\epsilon>0$, we have $\epsilon/2>0$. ...


0

As André Nicolas wrote, the Cesaro mean, for which the $n$-th term is the average of the first $n$ terms will do what you want. In both your cases, for large $n$, if $(a_n)$ is your sequence, if $b_n = \frac1{n}\sum_{k=1}^n a_k $, then $b_n \to 1$ since the number of $-1$'s gets arbitrarily small compared to the number of $1$'s.


0

The Cesaro mean accomplishes something that is close to your intuition. For the sequence $a_1,a_2,a_3,\cdots$, the Cesaro mean is the limit, if it exists, of the sequence $(b_n)$, where $b_n=\frac{a_1+\cdots+a_n}{n}$. If the limit of $(a_n)$ is $a$, then the Cesaro mean of the sequence $(a_n)$ is $a$. But the Cesaro mean of the sequence $(a_n)$ may exist ...


2

Most methods of summing divergent series can be adapted to making divergent sequences converge. For instance, Cesaro summation replaces the $n^\text{th}$ term of a series with the mean of the first $n$ terms. You could do the same: replace the $n^\text{th}$ term of your sequence with the mean of the first $n$ terms. Then you would be able to show ...


0

Note that for $0 < x < 1$, $0 < x^{n+1} < x^n$. Thus it is a monotone sequence which is bounded below, so it converges. Take the limit to get that $\lim_n x^n = 0$ for $0 < x < 1$ (I will leave it to you to prove this if you do not know this result already). Try out the other pieces by plugging in $x = 0$ and $x=1$ directly to see what you ...


0

You shouldn't have $x\to\infty$, as $x\in [0,1]$. For fixed $x$ you should ask yourself what $x^n$ does as $n$ gets large. There are two cases: $0\le x<1$ and $x=1$.


0

If I understand your question The infinite sum of cn phi n (x) = cn summation phi n (x) because cn constant take the limit


3

Does that mean it converges to $\frac{3}{2}$? Here is a closed form of the series: $$ \sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}=\color{blue}{\frac5{12}}. $$ Proof. By the absolute convergence, one is allowed to write $$ \begin{align} \sum_{k=1}^{\infty} \frac{1}{(4+(-1)^k)^k}&=\sum_{k=1}^{\infty} \frac{1}{(4+(-1)^{2k})^{2k}}+\sum_{k=1}^{\infty} ...


1

For a sequence of real (or complex) numbers, bounded is the correct term. For a sequence of functions, the notions of bounded and uniformly bounded are distinct. Each individual function may be bounded without the sequence being uniformly bounded.


4

Notice that the terms of this series are positive and we have $$\frac{1}{(4+(-1)^k)^k}\le 3^{-k}$$ and the geometric series $\sum 3^{-k}$ is convergent. Use comparison to conclude.


2

Use the $\;k\,-$ th root test: $$\lim\sup_{k\to\infty}\sqrt[k]{\frac1{4+(-1)^k)^k}}=\lim\sup_{k\to\infty}\frac1{4+(-1)^k}=\frac13<1$$ and thus the series converges (observe it is a positive series)


2

Maybe it's interesting to see a proof of the convergence. From PNT we have that $$p_{n}\sim n\log\left(n\right) $$ as $n\rightarrow\infty $ so $$p_{p_{n}}\sim p_{n}\log\left(p_{n}\right)\sim n\log^{2}\left(n\right) $$ and we can observe that, using the integral test, that ...


2

Let $(x_n)_{n=1}^\infty$ be a bounded sequence in $E$. We claim that $X := \{x_n\}_{n=1}^\infty$ is totally bounded. Indeed, suppose this were not the case. Then there is some $\epsilon >0$ with the following property: For every finite set $F \subseteq X,$ one can find $x \in X$ such that $d(x,y) \geqq \epsilon$ whenever $y \in F$. We construct a ...


1

Here is an approach to solving: Suppose $E$ has that property. First prove: For any bounded sequence $\{x_k\}_{k=1}^{\infty}$ in $E$ and for any $\epsilon>0$, there exists a point $x_n$ that is within a distance $\epsilon$ of infinitely many other points in the sequence. Next: Use this property to construct a Cauchy sequence.


2

Your link to the OEIS says that the series converges: "$\displaystyle\sum_{n\ge1} {1\over a(n)}$ converges. In fact, $\displaystyle\sum_{n>N} {1\over a(n)} < {1\over \ln(N)}$, by the integral test. --- Jonathan Sondow, Jul 11 2012."


1

$$\lim_{n\to\infty}\frac{a_n}{b_n}=0\implies a_n\le b_n\;,\;\;\text{for all}\;\;n>N\;,\;\;\text{for some}\;\;N\in\Bbb N$$ and now apply the usual comparison test.


1

Maybe it would help to consider that in cases like this if the integral of one function diverges and a similar function converges, then there is some cutoff in between where on one side it converges (gets smaller fast enough) and on one side it diverges (doesn't get smaller quite fast enough). For the functions you asked about consider 1/x^p as p decreases ...


2

One may recall that, as $M \to +\infty$, we have $$ \begin{align} \int_1^M \color{blue}{\frac1{x^2}}\:dx&=\left[ -\frac1x\right]_1^M=1-\color{blue}{\frac1M} \to \color{blue}{1}, \\\\ \int_1^M \color{red}{\frac1{x}}\:\:dx&=\left[ \:\ln x\:\right]_1^M=\color{red}{\ln M} \to \color{red}{+\infty}. \end{align} $$ Thus your question might be equivalent to ...


1

$$0 \le (x - y)^2 = x^2 - 2xy + y^2$$ $$xy \le \frac 12 x^2 + \frac 12 y^2$$ That is, $$|a_jb_j| \le \frac 12 |a_j|^2 + \frac 12 |b_j|^2 = \frac 12 a_j^2 + \frac 12 b_j^2$$ $$\sum_{k = 1}^{\infty} |a_jb_j| \le \frac 12 \sum_{k = 1}^{\infty} a_j^2 + \frac 12 \sum_{k = 1}^{\infty} b_j^2 \lt \infty$$ And we're done. This can also be proved from Cauchy-Schwarz ...


2

Hint: Look up Cauchy-Schwarz inequality. Edit: By the Cauchy-Schwarz inequality, we have $$ \sum_{i=1}^n |a_ib_i|\le \sqrt{\sum_{i=1}^n a_i^2}\sqrt{\sum_{i=1}^n b_i^2}\ . $$ By your assumption, $\sqrt{\sum_{i=1}^n a_i^2}$ and $\sqrt{\sum_{i=1}^n b_i^2}$ converge so the let hand side also converges and the following relation holds: $$ \sum_{i=1}^{\infty} ...


2

$\begin{array}\\ \sum_{i=1}^{n}\frac{1}{n+i} &=\sum_{i=n+1}^{2n}\frac{1}{i}\\ &=\sum_{i=1}^{2n}\frac{1}{i}-\sum_{i=1}^{n}\frac{1}{i}\\ &\to (\ln(2n)+\gamma)-(\ln(n)+\gamma)\\ &=\ln(2)\\ \end{array} $


2

Martin Argerami gave the simplest solution using integrals. Using another method and admitting that you already heard about harmonic numbers $$S_n=\sum_{i=1}^n\frac1 {n+i}=H_{2 n}-H_n$$ Now, using their asymptotic expansions (given in the Wikipedia page) $$S_n=\log (2)-\frac{1}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and ...


3

$$ \sum_{i=1}^n\frac1 {n+i} = \sum_{i=1}^n\frac1n\,\frac1 {1+\frac in} \to\int_0^1\frac1 {1+x}\,dx=\log2. $$


0

For $$a_n={(\frac{3}{k})}^n$$, we know if $k \le 3 $, it can not pass the divergent test, i.e. $\lim_{n\to\infty} a_n =0$. So the answer is between 4 and 5. For $$a_n=\frac{(3−k)^n}{n+3}$$, when $k=5$ it is $$a_n=(-1)^n\frac{2^n}{n+3}$$. it cannot pass alternating series test, i.e. $\lim_{n\to\infty} a_n =0$. But it can pass the test with k=4. So the ...


1

For the punctual convergence consider the two cases $x \le 0$ and $x > 0$. You should find $f=|x|$. But you are applying correctly Dini's theorem. This part is correct. But to justify a bit more you should say that $f$ and $f_n$ are continuous.


1

Since an infinite geometric series converges iff $\;|r|<1\;$ , with $\;r=$ the series fixed ratio, the first given series already tells you that the first three options are false. For the second series you can use the $\;n-$th root test for the absolute value: ...


0

Your question seems to deal with convergence of sequences of functions. In your case, you have the sequence $(f_n)_{n\geq 0}$ where $f_n(x)=|x|^n+(-1)^n$ and you seem to be interested in the function this sequence converges to and in what sense (uniform, pointwise). Hence, the literature is convergence of function sequences.


0

For (b) the series converges pointwise, since $|e^{-k^2x}| \leqslant e^{-kx}$ and $\sum e^{-kx}$ is a convergent geometric series when $x \in (0,1)$ and $0 < e^{-x} < 1$. It is not uniformly convergent since $$\sup_{x \in (0,1)}\sum_{k = n}^{2n}e^{-k^2x} \geqslant \sup_{x \in (0,1)}ne^{-4n^2x} = n,$$ and the RHS does not converge to $0$ as $n \to ...


0

When $n$ goes to infinity (by the way it likes) $1/n$ goes to $0$. When $\alpha$ goes to $0$ (by the way it likes) $p^{\alpha}$ goes to $1$ (because $p^{\alpha}$ is continuous). That is it. And you don't have to use algebra at all.


0

For $p>1$ let $p^{1/n}=1+h_n.$ Then $ h_n>0.$ We show that $h_n\to 0$ as $n\to \infty.$ For $n\geq 2$ we have $$p=(1+h_n)^n=1+h_n\binom {n}{1}+(h_n)^2\binom {n}{2}+...\;>(h_n)^2\binom {n}{2}\implies$$ $$\implies p>(h_n)^2\binom {n}{2}=(h_n)^2 n (n-1)/2\implies$$ $$\implies 2 p/n(n-1)>(h_n)^2\implies$$ $$\implies 2\sqrt p /(n-1)>\sqrt ...


4

Assume $p>1$ then by Bernoulli $$1+nh_n\leq (1+h_n)^n=p$$ and so $$h_n\leq \frac{p-1}{n}$$ It follows that $h_n \to 0$.


2

There are two cases: $0 < p < 1$ and $p > 1$. Here is the second: Let $p^{1/n} = 1+a$. Then $p =(1+a)^n \ge 1+an \gt an $ so $a < p/n$ so $a \to 0$ as $n \to \infty$. If $0 < p < 1$, let $p = \dfrac1{1+a} $ and do a similar thing. As often, nothing original here. As a matter of fact, this question is a duplicate.


5

I'm sure there's a more in-depth version, but the simple answer is: $\lim\limits_{n\to\infty}\sqrt[n]{p}= \lim\limits_{n\to\infty}p^{\frac{1}{n}} = p^0$


1

Because the sequence defining this series is eventually monotonically decreasing (that is, there exists a point after which the sequence is monotonically decreasing), we can apply the Cauchy condensation test: $$\sum_{n=3}^\infty 2^n \frac{\ln(2^n)}{2^n}=\ln(2)\sum_{n=3}^\infty n= \infty$$ Therefore $\sum_{n=1}^\infty \frac{\ln(n)}{n}=\infty$.


2

Denote $\sum a_n x^n$ the given series then $$\left\vert\frac{a_{n+1}}{a_n}\right\vert=\left(1+\frac1n\right)^{-n}\xrightarrow{n\to\infty}\frac1e$$ so by the ratio test, the radius of convergence is $e$.


0

Hint 1: $(n > e) \Rightarrow \ln n > 1$ Hint 2: Neglecting finite number of elements from a series doesn't change it's convergence.


7

Note that $$\frac{1}{n\left(n-1\right)}=\frac{1}{n-1}-\frac{1}{n}$$ What does it tell you?


1

$$\sum_{n\geq 1}\frac{3n-1}{(n+1)^3} = 3\sum_{n\geq 1}\frac{1}{(n+1)^2}-4\sum_{n\geq 1}\frac{1}{(n+1)^3} = 3(\zeta(2)-1)-4(\zeta(3)-1) = \color{red}{3\zeta(2)-4\zeta(3)+1}.$$


8

$$ \sum_{n=1}^\infty\frac{3n-1}{(n+1)^3}\le\sum_{n=1}^\infty\frac{3n}{(n+1)^3}\le\sum_{n=1}^\infty\frac{3n}{n^3}=3\sum_{n=1}^\infty\frac1{n^2}<\infty. $$


0

The idea behind the proof is quite simple once you get read of the useless verbosity. You start by using the fact that $\{n: |u_n-t|<1\}$ is infinite, take $n_1$ in it. Then you can use the fact that $\{n: |u_n-t|<\dfrac12\}$ is infinite, take $n_2$ in it strictly bigger than $n_1$. Proceeding as this by induction, for each $k$ you can choose $n_k$ ...


0

In every metric space $(X,d)$ a sequence $x_n$ converges to $x$ if and only if every subsequence $x_{n_k}$ has a further subsequence converging to $x$ (easy proof by contradiction). All you need to now is thus that convergence in measure is convergence in a metric space (e.g. $d(f,g)=\int \min\lbrace 1,|f(x)-g(x)|\rbrace \, d\mu(x)$ is a suitable metric on ...


0

If $f_k\to f$ in $L^{\infty}$, then for each $n\in\mathbb{N}$ there is an index $k_n$ such that $||f-f_k||_{\infty}<\frac{1}{n}$ for all $k\geq k_n$, hence for each $k\geq k_n$ there is a nullset $N_{k,n}$ such that $$ \sup_{X\setminus N_{k,n}}|f_k(x)-f(x)|<\frac{1}{n} $$ Let $$N=\bigcup_{n=1}^{\infty}\bigcup_{k=k_n}^{\infty}N_{k,n}$$ then $N$ is a ...


1

See Kolmogorov's "Three Series Theorem" if you want the definitive answer on this subject; it gives you results general for all distributions, not just the gamma distribution. https://en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem


1

I will here address the question "First question is, why is this one ($g(x)$ below) so good, and how to find such a fast functional iteration systematically?" Lets take your two example functions: $$f(x) \equiv (cx^re^{-x})^{\frac{1}{r+1}}$$ $$g(x) \equiv \frac{x+1}{\frac{e^x}{c}+1}$$ Expand in a Taylor series about the fixpoint $x = W(c)$. We then ...


1

Assuming you mean $$\sum_{n=1}^\infty n!(2x-1)^n=\sum_{n=1}^\infty n!2^n(x-0.5)^n$$ Now use ratio test: $$\frac{|a_{n+1}|}{|a_n|}=\frac{(n+1)!2^{n+1}}{n!2^n}=2(n+1)>1$$ Inequality holds for every $n$. Hence the radius of convergence is $0$. Formally you have to calculate $$r=\lim_{n\to \infty}\frac{1}{\frac{|a_{n+1}|}{|a_n|}}=\lim_{n\to ...


0

That integral never converges. If $p\geq1$, convergence fails $0$. And if $p<1$, convergence fails at $\infty$ (one just removes small intervals around the zeroes of $\cos x$ to see that at $\infty$ the integral behaves like $\int_1^\infty\frac1{x^p}$).


0

Hints: You have possible problems at $1$ and $2.$ At $1,$ the denominator stays calm but the numerator $\to -\infty.$ But it's only a logarithmic blowup, so that should be OK. At $2,$ you have the battle of zeros: Verify that $|\ln (x-1)|$ is on the order of $|x-2|$ there. Downstairs, just factor $4-x^2$ to see what to do.


2

We have $\vert\sin x\vert \le \vert x\vert$ and to conclude it suffices to prove that the series $\sum \vert x\vert^{\sqrt n}$ is convergent. We have $$n^2 \vert x\vert^{\sqrt n}=\exp(\sqrt n\ln\vert x\vert+2\ln n)\xrightarrow{n\to\infty}0$$ so we have $\vert x\vert^{\sqrt n}\le \frac1{n^2}$ for $n$ large enough and the result follows by comparison with a ...


1

$\sin(1)<0.9$ and $\sum\limits_{k=n^2+1}^{(n+1)^2} a^\sqrt{k}<(2n+1)a^n$. $\sum\limits_{k=1}^\infty (\sin x)^\sqrt{k}\le \sum\limits_{k=1}^\infty (\sin 1)^\sqrt{k}< \sum\limits_{k=1}^\infty 0.9^\sqrt{k}\le \sum\limits_{k=1}^\infty (2k+1)0.9^k= 2\frac{0.9}{(1-0.9)^2}+\frac{0.9}{1-0.9}=189$


1

Since $-1<x<1$, you have $|\sin x|<1$. Then $\log|\sin x|<0$. We have $$ |\sin x|^{\sqrt n}=e^{\sqrt n\,\log|\sin x|}. $$ If we now compare with the integral, $$ \int_1^\infty e^{t\,\log|\sin x|}\,dt=\left.\frac{e^{t\,\log|\sin x|}}{\log|\sin x|}\right|_1^\infty=-\frac{e^{\log|\sin x|}}{\log|\sin x|}<\infty. $$ So the series converges.



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