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Proving convergence of sums: If $\sum_{n=1}^{\infty}a^2_n$ and $\sum_{n=1}^{\infty}b^2_n$ converge then $\sum_{n=1}^{\infty}a_nb_n$ converges I want to attach a code file to my question, but I am not able to find a link to do it. Is attaching files possible at all? I have a layout similar to:  &lt;div&gt;     &lt;table&gt;     &lt;/table&gt; &lt;/div&gt;   I would like for the div to only expand to as wide as my table becomes. Cannot Upgrade Xcode 11 to Xcode 11.1 I'm unable to upgrade Xcode 11GM to 11.1 for some time on my Mid 2014 MacBook Pro running macOS Catalina. I've installed Xcode from Mac AppStore ages ago (IIRC circa 2014), and upgraded step by step since then. Currently, I'm experiencing the following loop:   AppStore reports that I have an update. I open AppStore, open updates, click "Update" on Xcode. Mac Downloads ~2.5GB first. Installs that patch. Continues to get the full package (~7.6GB). It starts to install. After some churning and install, it reports "Cannot download Xcode now, please try again later."   I have more than enough space on the disk, and nothing is modified in Xcode or in the OS (no homebrew or something similar).  Did anyone experience something similar? Are there solutions to this? Series of sequence with terms over series. Prove or give a counterexample to disprove the following statement about series of real numbers:  If $a_k &gt; 0, s_k = a_1 + \cdots + a_k$ and $b_k = a_k/s_k$ then $\sum a_k$ and $\sum b_k$ either both converge or both diverge. If $a^\phi = 1$ then the order of $a$ divides $\phi$ $H$ is normal subgroup of $G$ iff $gHg^{-1}=H$ for every $g$ of $G$ Given the following definition of normal subgroup     A subgroup $H$ of a group $G$ is said to be normal if, for every $g\in G$: $$gH=Hg$$   I've tried to show that $H\mathrel{\unlhd} G$ if and only if we have $gHg^{-1}=H$ for every $g\in G$:  (if) Let $H\mathrel{\unlhd} G$, $g\in G$; if $l\in gHg^{-1}$, there exists $h\in H$ such that $l=ghg^{-1}$ and since $gh$ = $h'g$ for some $h'\in H$, then $l=ghg^{-1}=h'(gg^{-1}) = h'\in H$. Let $h\in H$; we can write $h$ as $hgg^{-1}$, and since $hg = gh'$ for some $h'\in H$, we obtain $h = hgg^{-1} = gh'g^{-1}\in ghg^{-1}$. So $gHg^{-1}=H$.  (only if) Let $gHg^{-1}=H$, $g\in G$; if we take an $l\in Hg$, $l = hg$ for some $h\in H$, we note immediately that $l=hg=(gh'g^{-1})g=gh'\in gH$. What should I do to show that $gH\subset Hg$?  I'm also wondering if this mess is correct, and if it is possible a more elegant proof of this fact. Reporting Mann-Whitney U-Test without homogenity of variance The Mann-Whitney test requires homogeneity of variance if a median difference is suppossed to be statistically significant.  In case homogenity of variance is not met, but the test is significant:  Which aspects of the test can I report? How to install packages in ubuntu without internet connection? Do immigration checks exist between Northern Ireland and England/Scotland? (not just for flights but also ferry/bus/train) Do immigration checks exist between the Republic of Ireland and England/Scotland? Can ribosomes read ssDNA? My question is whether translation can be done, either naturally or artificially, through a ribosome reading (single-stranded) DNA directly. If not, I would like to know what allows ssRNA to be translated but not ssDNA. Conditional conundrum A friend and I are having a debate about a most perplexing piece of English grammar. Now, I am a native English speaker, and he is Polish - this means that he is rather clued up on the grammatical side of things, whereas I know diddly squat, but know what feels right.   We were discussing Poland's next world cup match tomorrow, when Poland plays against Colombia, and I wrote to him after he me sent a heartwarming video (jokingly, of course), that 'I think threats of stoning if they don't succeed would be more effective.'  He replied that such a sentance should be in the second conditional tense, and thus go like 'I think threats of stoning if they didn't succeed would be more effective.'  So after a lively debate as to whether this is the second, first, or not even a conditional, I have no choice but to turn to the denizens of the interweb, to ask them to apply their grammatical expertise to resolve this vital question.  Thanks in advance, Conor “have found” vs. “found” Error to calculate date diff with DateTime class Find the value of $\space\large i^{i^i}$? Is $\large i^{i^i}$ real ? How to find it?   Thank You! Why can I no longer ask questions when all I did was ask some simple questions? I have asked the three questions below, one of which I deleted myself. How can Stack Overflow  block me from asking further questions? It seems Stack Overflow is not for beginners, but just for people with years of experience.   Could anyone tell me what the problem is with my questions and how I can improve them?      (deleted by myself) dynamic does not contain a definition for a property from a project reference Infinite distinct factorizations into irreducibles for an element Consider the factorization into irreducibles of $6$ in $\mathbb{Z}[\sqrt{-5}]$. We have $6=2 \times 3$ and  $6=(1+\sqrt{-5}) \times (1-\sqrt{-5})$, i.e. $2$ distinct factorizations. And,  $$6^2=3 \times 3\times2\times2$$ $$=(1+\sqrt{-5}) \times (1-\sqrt{-5}) \times (1+\sqrt{-5}) \times (1-\sqrt{-5})$$ $$=(1+\sqrt{-5}) \times (1-\sqrt{-5})\times3\times2.$$ More generally, $6^n$ will have $n+1$ distinct factorizations into irreducibles in $\mathbb{Z}[\sqrt{-5}]$ by a simple combinatorial argument. But, can we construct a ring in which there exists an element that has an infinite number of distinct factorizations into irreducibles? To make life harder, can we construct an extension of $\mathbb{Z}$ in which this happens? I have been thinking about this for a while and have managed to find no foothold.. Any help is appreciated. Let random vector $x = (x_1,...,x_n)$ follow multivariate normal distribution with mean $m$ and covariance matrix $S$. If $S$ is symmetric and positive definite (which is the usual case) then one can generate random samples from $x$ by first sampling indepently $r_1,...,r_n$ from standard normal and then using formula $m + Lr$, where $L$ is the Cholesky lower factor so that $S=LL^T$ and $r = (r_1,...,r_n)^T$.  What about if one wants samples from singular Gaussian i.e. $S$ is still symmetric but not more positive definite (only positive semi-definite). We can assume also that the variances (diagonal elements of $S$) are strictly positive. Then some elements of $x$ must have linear relationship and the distribution actually lies in lower dimensional space with dimension $&lt;n$, right?  It is obvious that if e.g. $n=2, m = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, S = \begin{bmatrix} 1 &amp; 1 \\ 1 &amp; 1\end{bmatrix}$ then one can generate $x_1 \sim N(0,1)$ and set $x_2=x_1$ since they are fully correlated. However, is there any good methods for generating samples for general case $n&gt;2$? I guess one needs first to be able identify the lower dimensional subspace, then move to that space where one will have valid covariance matrix, then sample from it and finally deduce the values for the linearly dependent variables from this lower-dimensional sample. But what is the best way to that in practice? Can someone point me to books or articles that deal with the topic; I could not find one. A non-trivial, non-negative, function bounded below by its derivative with $f(0)=0$? I did not know what to search to see if this existed elsewhere. But, I could not find it. Here's the question:  Does there exist a continuously differentiable function, $f: [0,1] \rightarrow [0, \infty)$, such that the following hold? (i) $f(0) = 0$ (ii) $f'(x) \le c f(x)$ for all $x$ ($c$ is a fixed constant) (iii) $f \not\equiv 0$  This was a fun problem someone asked me a long time ago. I have always been convinced there is no such function, but I cannot prove it. I have mainly tried using the mean value theorem (in multiple forms), and re-wording the problem in terms of the integral of a continuous function to no avail. I spent some time trying to use (ii), the definition of derivative, and squeeze theorem, but no luck. After much time trying to prove non-existence, I spent a little time trying to find such a function... also with no luck. This problems ability to avoid being solved has since taken away the original fun and replaced it with a twinge of annoyance.