# Tag Info

8

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

8

Hint. Observe that, by telescoping terms, $$\sum_{n=1}^Na_n=\sum_{n=1}^N(b_n-b_{n-1})=b_N-b_0$$ then let $N \to \infty$.

6

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

6

Since $x \mapsto e^{-nx^2}$ is decreasing over $[1,2]$, observe that $$0<\int_1^2e^{-nx^2} dx\leq e^{-n\times1^2}\int_1^2dx=e^{-n}$$ then let $n \to \infty$.

5

Prove that $0\le \ln n \le \sqrt{n}$ first, and $$0\le \frac{\ln n^2}{n^2} =\frac{2\ln n}{n^2} \le \frac{2}{n\sqrt{n}}.$$ $\sum_{n=1}^{\infty}\frac{2}{n\sqrt{n}}$ converges.

4

Observe that $$\frac n{n+1}=\frac1{1+\frac1n}\implies\frac {n^n}{(1+n)^n}=\left(\frac n{1+n}\right)^n=\left(\frac1{1+\frac1n}\right)^n=$$ $$=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e$$ by arithmetic of limits.

4

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

3

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

Your reasoning seems right to me, but conceptually speaking I believe the proof is much more easy. Proof from Rudin's definitions: Put $t^*=\limsup_{n\to\infty} t_n$ and $s^*=\limsup_{n\to\infty}s_n$. For a contradiction, suppose that $s^*>t^*$ and put $\epsilon=\frac{s^*-t^*}{2}, x=s^*-\epsilon$. Notice that $x>t^*$. With the definitions as in ...

2

Your two ideas would have been my first attempts; but I have no idea how to make them work (well, for the first one, I would try with $T$ the flip, but I still wouldn't know how to do it). Let $K\subset B(\ell^2)$ be the compact operators. On $\overline {B(\ell^2)\odot B(\ell^2)}$, consider the ideals $\overline{K\odot B(\ell^2)}$ and $\overline{K\odot ... 2 As$\mathrm{cos}(n\pi) = (-1)^n$we get: $$0\leq \frac{1}{5n^2+2} \leq a_n \leq\frac{1}{5n^2}.$$ 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 ... 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$e$. 2 As the sequence$\;\sup\limits_{k>n} a_k$is non-increasing, the hypothesis means each$\;\sup\limits_{k>n} a_k=\infty$. In other words, for any$A$and any$n$, there exist$k>n$such that$a_k>A$. So let's start with$A=1$, and let$n_1$the smallest$k>0$such$a_k>1$. Then let$n_2$the smallest$k>n_1$such that$a_k>\max(2, ...

2

The "left/right convergence" notion only applies when taking limits in $\mathbb R$, essentially because of the structure of the real line. The set $(a,+\infty)$ intuitively lies on the right side of $a$. Formally, $\displaystyle \lim_{x\to a+} f(x) = l \iff \forall \epsilon>0,\exists \delta>0, \forall x\in \mathbb R, x \in (a,a+\delta)\cap D\implies ... 2 Consider$a_n:=\frac{\ln (n^2)}{n^2}$,$b_n:=\frac{1}{n^{3/2}}$. We have that $$\lim\limits_{n \to \infty}\frac{a_n}{b_n}=\lim\limits_{n \to \infty}\frac{\ln (n^2)}{n^{1/2}}=\lim\limits_{n \to \infty}\big(2\cdot \frac{\ln (n)}{n^{1/2}}\big)=0.$$ Therefore, there exists$N$such that$n>N$implies$\frac{a_n}{b_n}<1$. Hence,$n> N$implies$a_n< ...

2

Convergence only depends on the tail of the series. If $\liminf x_n \ge b > 0$, then there is some $N$ such that for $n > N$, $x_n > b/2$. Let $B = \max(1, 2/b)$. Then for $n > N$, $$\dfrac{1}{1+x_n a_n} \le \dfrac{1}{1 + (b/2) a_n} \le \dfrac{B}{1+a_n}$$ By a Comparison test, $\sum_n 1/(1+x_n a_n)$ converges.

2

We have: $\left(1+\dfrac{2}{n!}\right)^{n!} \to e^2$, thus your expression $\to e^{-2}$

2

You want to prove that if $x_n \to x$ in $S$, then $d(x_n,x_0) \to d(x,x_0)$ in $\Bbb R$. One does this as follows: let $\epsilon > 0$. By convergence of $(x_n)_{n\geq 1}$ there is $n_0$ large enough such that $d(x_n,x)< \epsilon$ if $n \geq n_0$. Then: $$|d(x_n,x_0)-d(x,x_0)| \color{red}{\leq} d(x_n,x) < \epsilon$$for all $n \geq n_0$, and so ...

2

We have $$a_n=1+\frac{a_{n-1}}{3+(n-1)}\ge 1+\frac{1}{3+(n-1)} =\frac{3+n}{2+n}.$$

1

Note that $$a_{n}\frac{2+n}{3+n}\geq 1\iff a_{n}\geq\frac{3+n}{2+n}.$$ Since $a_{1}=1$, an easy induction shows that $a_{n}\geq 1$ for all $n$. Thus, your desired inequality follows.

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

Yes, looking at the limit you gave is sufficient: the series diverges. It is a basic (yet fundamental) result that if the series $\sum_n a_n$ converges, then it must be the case that $a_n \xrightarrow[n\to\infty]{}0$.

1

Take a hint: if $\;f\;$ is a differentiable function, then $$\int f'f^m dx=\frac{f^{m+1}}{m+1}+K\;,\;\;m\neq-1$$ If instead zero it had any non zero positive value as lower limit then the integral would converge.

1

This integral is not convergent. One may observe that, for $a>0$, $$\int_a^\infty \frac{1}{x(\ln x)^2} dx=\int_a^\infty \frac{du}{u^2} \qquad \qquad \left(u=\ln(x),\, du=\frac{dx}x\right).$$

1

Let $(a,b)\in S\text{x} S$. By the axiomes of the distance $d$ you have for a point $(x,y)\in S\text{x} S$ $d(x,y)\le d(x,a)+d(a,b)+d(b,y)\Rightarrow d(x,y)-d(a,b)\le d(x,a)+d(b,y)$ Permuting $(y,b)$ and $(x,a)$ you get $$|d(x,y)-d(a,b)|\le d(x,a)+d(b,y)$$ Hence for a given $\epsilon >0$ you can choose $d(x,a)\le \frac {\epsilon}{2}$ and $d(b,y)\le ... 1 The previous inequality shows that $$|f(x)-f_{n_k}(x)| < 2^{-k+1} \qquad \text{for all x \in D \backslash \bigcup_{j=k}^{\infty} E_j,}$$ i.e. $$x \in D \backslash \bigcup_{j=k}^{\infty} E_j \implies x \in \{|f-f_{n_k}| < 2^{-k+1}\}.$$ This is equivalent to $$D \backslash \bigcup_{j=k}^{\infty} E_j \subseteq \{|f-f_{n_k}| < 2^{-k+1}\}.$$ ... 1 Hint. Once you arrive at $$x\sum_{n=0}^\infty (-1)^n(2)^{n-1}(n+2)(n+1)x^n=\sum_{n=0}^\infty (-1)^n(2)^{n-1}(n+2)(n+1)x^{n+1}$$ you can make a change of index, setting$m=n+1$thus$n=m-1$, giving $$\sum_{n=0}^\infty (-1)^n2^{n-1}(n+2)(n+1)x^{n+1}=\sum_{m=1}^\infty (-1)^{m-1}2^{m-2}(m+1)mx^m$$ then collecting the general terms in the two series. Can you ... 1 What you have done so far is fine. Notice as$n\to\infty$,$n!\to\infty$so we can re-parameterize taking some$k=n!$, and take the limit with respect to just$k$instead. Then you have:$\lim_{k\to\infty} \left(\frac{k}{k+2}\right)^k.Can you complete the answer now? 1 Hint: \begin{align*} \left(\frac{n!}{n!+2}\right)^{n!}&=\left(1-\frac{2}{n!+2}\right)^{n!+2}\left(1-\frac{2}{n!+2}\right)^{-2}\\[3pt] \end{align*} Notice that the first factor tends toe^{-2}$while the second tends to$1$as$n\to\infty\$.

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