Let $(X,\|\cdot\|)$ be a normed vector space. Show that $X $ is a Banach space if and only if for every succession $\{x_n\}_{n\in \mathbb N} $ such that $\sum_{n=1}^\infty ||x_n|| \lt \infty $, we have that $\sum_{n=1}^\infty x_n $ converges to an element of $X $.  If $X $ is Banach, it's clear that the statement above is true for every succession $\{x_n\}_{n\in \mathbb N} $ such that $\sum_{n=1}^\infty ||x_n|| \lt \infty $, because this implies that $\{S_N\}_N=\sum_{n=1}^N x_n $ is a Cauchy succession in $X $.  However, in the other sense, I don't understand what to do: if I take a Cauchy succession $\{x_n\}_n$ in $X $, I could show that it always converges by showing that $\sum_{n=1}^\infty  (x_n - x_{n-1}) $ converges to some point of $X $. However it is not true in general that $\sum_{n=1}^\infty ||x_n-x_{n-1}|| \lt \infty $, even for a Cauchy succession. Do you have suggestions? Thanks
The following is an Exercise of Conway's Functional Analysis.   Prove that $X$ is a $\,Banach$ space iff whenever $\{x_n\}$ is a sequence in $X$, such that $\sum \| x_n \| &lt; \infty$, then $\sum x_n$ converges.   I easily can show that if $X$ is a Banach space then $\sum x_n$ converges. My problem is showing that $X$ is a Banach space. For this I suppose that $\{s_n\}$ is a Cauchy sequence in $X$, then I want to make a series. For this I do not have any idea. Please help me.
join comma delimited data column Right identity and Right inverse implies a group Let $(G, *)$ be a semi-group. Suppose   $ \exists e \in G$ such that $\forall a \in G,\  ae = a$;  $\forall a \in G, \exists a^{-1} \in G$ such that $aa^{-1} = e$.    How can we prove that $(G,*)$ is a group? On the street, temperatures at night are about -5 &deg;С and about 8 &deg;C during the day. In the apartment it is about 25 &deg;C. So, there is condensation on the windows and I don't want to wipe the windows every time.  How do I prevent condensation on the balcony windows? This is a Q&amp;A that I recently have figured out, so I have posted the answer below.  After installing Windows 10, every time I get a kernel update or I run the update-grub2 it always shows Windows 7 or Windows Recovery Environment instead of Windows 10.  How do I fix this permanently? Fourier transform of the indicator of the unit ball Evaluation of OptionValue[] \selectcolormodel{gray} doesn't work for the duck hair, but it works for the other parts of the ducks, why?  \documentclass{ltugboat}  \usepackage{microtype}  \usepackage{graphicx}  \usepackage{ifthen} \newboolean{grayscale}    \setboolean{grayscale}{true}% for the black &amp; white version    %\setboolean{grayscale}{false}% for the colored version     \usepackage{tikzducks} \usetikzlibrary{positioning}  \newenvironment{mytikz}{%     \begin{tikzpicture}     \ifthenelse{\boolean{grayscale}}{\selectcolormodel{gray}}{}      }     {\end{tikzpicture}}  \usepackage{ifpdf}  \ifpdf  \usepackage[breaklinks,hidelinks]{hyperref}  \else \usepackage{url} \fi   \begin{document}      Why \textbackslash\texttt{selectcolormodel\string{gray\string}} does not work for the duck hair?      Black and white version:          \begin{mytikz}         \begin{scope}[xscale=-1]             \duck[longhair=pink]         \end{scope}         \begin{scope}[xshift=10pt]             \duck[crazyhair=orange]         \end{scope}     \end{mytikz}      Colored version:     \setboolean{grayscale}{false}      \begin{mytikz}         \begin{scope}[xscale=-1]             \duck[longhair=pink]         \end{scope}         \begin{scope}[xshift=10pt]             \duck[crazyhair=orange]         \end{scope}     \end{mytikz} \end{document} 'Some index files failed to download' while updating Determine if an element has a CSS class with jQuery I'm working with jQuery and looking to see if there is an easy way to determine if the element has a specific CSS class associated with it.  I have the id of the element, and the CSS class that I'm looking for.  I just need to be able to, in an if statement, do a comparison based on the existence of that class on the element. What happens to spirits/souls of other creatues than Elves and Men after their death? Instead of the constrained 45 degrees that happens when you press C, is there a specific angle I can get for example 35 degrees instead?   It's currently cut at 45 degrees in the attached image and I need it to be 35. "batch" tag turns into "batch-file" Adjust rounded rectangle radii without creating a new one in Photoshop C6 When I create new rounded rectangles, I see the radii and other information in the properties, but after I work on something else and go back to that layer, the information does not show. How do I adjust the radii of the rounded rectangles without deleting and creating a new one? Which one is correct, or are both of them fine?    "It would take the average person 10 days to read this novel"    Or    "It would take an average person 10 days to read this novel" When you are trying to say that something is automatically filled in, you use the word autofill, or if you were using past tense, autofilled.    I see 3 main ways that people use it:   auto fill / auto filled  auto-fill / auto-filled autofill / autofilled   Note: this also applies to pre fill, pre-fill, and prefill as well.  Which way is the most preferred spelling? Motivated by , can one prove that the order of an element in a finite group divides the order of the group without using Lagrange's theorem? (Or, equivalently, that the order of the group is an exponent for every element in the group?)  The simplest proof I can think of uses the coset proof of Lagrange's theorem in disguise and goes like this: take $a \in G$ and consider the map $f\colon G \to G$ given by $f(x)=ax$. Consider now the orbits of $f$, that is, the sets $\mathcal{O}(x)=\{ x, f(x), f(f(x)), \dots \}$. Now all orbits have the same number of elements and $|\mathcal{O}(e)| = o(a)$. Hence $o(a)$ divides $|G|$.  This proof has perhaps some pedagogical value in introductory courses because it can be generalized in a natural way to non-cyclic subgroups by introducing cosets, leading to the canonical proof of Lagrange's theorem.   Has anyone seen a different approach to this result that avoids using Lagrange's theorem? Or is Lagrange's theorem really the most basic result in finite group theory? Proper submodules of Noetherian modules are intersections of indecomposable modules If $M$ is Noetherian, any submodule $N \neq M$ can be written as a finite intersection of intersection-indecomposable ones.   I just don't know how to begin with the proof of this proposition.   Very kind of you for reading or answering. Inbox notification and reputation score is misaligned in mobile web view The inbox notification indicator and reputation score indicator are misaligned in the mobile web view. Why are disabled apps still running? How can I clip an image via a bezier-path?