Tag Info

Hot answers tagged

5

If $\{a_n^2\}$ converges, it is bounded. Then $\{a_n\}$ is bounded and monotone. Anything can happen. There is no way to know a priori.


2

Note first that $$\dfrac{2k+1}{(k(k+1))^2}\geqslant \dfrac{2k}{k^2(k+1)^2}= \dfrac{2}{k(k+1)^2}\geqslant \dfrac{2}{(k+1)^3}$$ If we prove that $$\dfrac{1}{n^3}\geqslant \dfrac{2}{3^n} \tag{1}$$ which is equivalent to $$\dfrac{3^n}{n^3} \geqslant 2 \tag{2}$$ then we can conclude that $$\dfrac{2k+1}{(k(k+1))^2}-\dfrac{4}{3^{k+1}}\geqslant 0.$$ The inequality ...


2

Yes, proof of divergence by finding a lower bound is perfectly ok. You have $$\sum_{k=1}^n \frac{1}{k}\to \infty$$ $$\frac13\sum_{k=1}^n \frac{1}{k}\to \infty$$ And finally $$\sum_{k=1}^n\frac{1}{2k+1}>\sum_{k=1}^n\frac{1}{3k}\to\infty$$


2

Here is some additional information which could be helpful. Starting with OPs definition of $H_k^r$: \begin{align*} &H_k^{-1} := 1\qquad& k\geq 1\tag{1}\\ &H_k^r:=\sum_{n=1}^k\frac{1}{H_1^{r-1}+\ldots+H_n^{r-1}}\qquad &k\geq 1, r\geq 0\tag{2}\\ \text{and}\quad&\\ &L_k:=\lim_{r\rightarrow\infty}H_k^r\qquad&k\geq 1\tag{3} ...


2

Recall again the prime number theorem, which gives that $$ p_n \approx n\log n.$$ Then the sum we want can be compared to $$ \sum_p \frac{1}{p \log \log p} \sim \sum_n \frac{1}{n\log n \log \log n}.$$ I claim this latter sum diverges based on the integral test for convergence. Claim: The following series diverges: $$\sum_n \frac{1}{n \log n \log \log ...


2

Long story short, $$ p_n \leq C n \log n $$ by Chebyshev's theorem ($\pi(n)\geq D \frac{n}{\log n}$), hence the series is lower bounded by a multiple of $$ \sum_{n\geq 3}\frac{1}{n\log n \log\log n}$$ that diverges due to Cauchy condensation test (applied twice).


2

You need to fix $x>0$. Then there is $K$ such that $\frac{1}{2^k}<x$, for $k>K$. Then $f_k(x)=0$, for $k>K$. Then $f(x)=0$. For $x\leq0$, $f_k(x)=0$, for all $k$. Then $f(x)=0$. Now, for $L^1$ we can integrate. ...


2

We have: $$\begin{eqnarray*} I_n &=& \int_{0}^{1}x^{n+1}(1-x)(1-x^2)\cdot\ldots\cdot(1-x^n)\,dx \\&\leq& \int_{0}^{1}x^{n+1}(1-x^n)^n\,dx = \frac{\Gamma(n)\,\Gamma\left(1+\frac{2}{n}\right)}{\Gamma\left(2+n+\frac{2}{n}\right)}\leq\frac{1}{n^2},\tag{1}\end{eqnarray*}$$ while on the other hand, if we set: $$ A(n,m)\triangleq\int_{0}^{1} x^n ...


1

If you denote $M = \max\{|a_n|\}$ you have $$\sum_{n=1}^\infty |a_n|^p = \sum_{n=1}^\infty |a_n|^{p-2} |a_n|^2 \le M^{p-2} \sum_{n=1}^\infty |a_n|^2 < \infty$$whenever $p \ge 2$.


1

If $p\geq2$ then $\sum|a_n|^p$ converges since $$\lim\limits_{n\to\infty}\frac{|a_n|^p}{|a_n|^2}=\lim\limits_{n\to\infty}|a_n|^{p-2}=0$$ where p>2. On the other hand if $0<p<2$ then there exists a sequcence such that $\sum|a_n|^2$ is convergent but $\sum|a_n|^p$ is divergent. Formally, for all $p\in(0,2)$ there exists ...


1

Your sequence is going to zero pointwise. On the one hand, $f_k(0) \equiv 0$. On the other hand, if $x>0$ and $k>-\log_2(x)$, then $f_k(x)=0$. Additionally, your sequence is dominated by $f(x) = \frac{1}{\sqrt{x}} \chi_{[0,1]}(x)$, which is in $L^p$ for $p \in [1,2)$. So it is definitely going to zero in $L^p$ for $p \in [1,2)$. It does not go to zero ...


1

As pointed in the other answer, the series is pointwise convergent on $\left(0,\frac{\pi}{2}\right)$ by Dirichlet's test, since: $$\sum_{n=1}^{N}\sin(x)^n \leq \frac{\sin x}{1-\sin x}$$ and $\frac{1}{n}$ decreases towards zero. However, in a left neighbourhood of $x=\frac{\pi}{2}$ the original series behaves like the harmonic series: $$ ...


1

We can use the fact that $$ \sum_{i=1}^n\frac1{\sqrt i}=\sum_{i=1}^n\frac1{\sqrt i}\int_i^{i+1}\mathrm dx\ge\sum_{i=1}^n\int_i^{i+1}\frac1{\sqrt{x}}\mathrm dx=\int_1^{n+1}\frac1{\sqrt{x}}\mathrm dx=2\sqrt{n+1}-2. $$ Similarly, $\sum_{i=1}^ni^{-1/2}\le2\sqrt n-1.$ Hence, $\sum_{i=1}^n i^{-1/2}\sim 2\sqrt n$ as $n\to\infty$ and $$ ...


1

The difference between $\frac 43(2-x)^{3/2}$ and $-\frac43i(x-2)^{3/2}$ is: An overall minus sign. A factor of $i$. The thing raised to the $(3/2)$th power has been negated. Of these (3) is what causes (2), since $(-1)^{3/2}=\pm i$. As for (1), the overall sign of the square root of something is not well defined when we're working with complex numbers. ...


1

Because you are looking for 3's in the last two terms of a sequence cycle of 7, the function defined below will work: $$a_n=\begin{cases} 3 &if \,\,\,\, n \mod 7=6,0\\ 2 &otherwise\\ \end{cases}$$


1

The series doesn't seem to converge anywhere. First, let's look at whether it converges at $x=0$. The power series for $Erf$ is $$ Erf(z)=\frac{2}{\sqrt{\pi}}\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)k!}z^{2k+1} $$ from which $$ \frac{\pi}{2x}Erf\left[\frac{cx}{2}\right]=\frac{c\sqrt{\pi}}{2}\sum_{k=0}^\infty ...


1

The radius of convergence is the distance to the nearest singularity For $(b)$ you can advance as (based on your calculations) using partial fraction $$P(x)= \frac{x}{(1-2x-x^2)}= \frac{A}{x-a} + \frac{B}{x-b} $$ where $a,b$ are the roots of $1-2x-x^2$ and $A$ and $B$ need to be determined. The calculations gives: $$\frac{-1-\sqrt{2}}{2 \sqrt{2} ...



Only top voted, non community-wiki answers of a minimum length are eligible