Let me start with a simpler version of my actual problem. Say I have a partition $\mathbb{N} = \cup_k I_k$ where the $I_k$ are disjoint, and an absolutely convergent series $\sum_n x_n$ in $\mathbb{R}^d$. I want to show the identity $\sum_n x_n = \sum_k \sum_{i\in I_k} x_i$. In this simple situation I can show this using Fubini's theorem, but actually I'm interested in the situation of Banach space valued series that converge unconditionally, so I'd appreciate a more elementary approach that is better suited for generalization.
121942
I'm reading that summability and unconditional convergence are the same. (At least if the index set is countable, so unconditional convergence makes sense.)  Then unconditional convergence is nothing else than convergence of the net of all rearranged partial sums. Thus uniqueness of the limit becomes fairly easy as the limit of every net in a Hausdorff space is unique.  But the usual proof deals with separating functionals on Banach spaces, which is a much more serious technique. So I think there must be a hook in my reasoning. Can somebody resolve this issue? Thanks a lot!  Best Regards, Freeze_S
111764
Let $x_1= a(&gt;0) , x_{n+1} = x_n+\frac{1}{x_n}$ for $n \in \mathbb{N}$. Determine whether $(x_n)$ converges or diverges.   We have $x_1= a &gt; 0, x_2= a+\dfrac{1}{a},  (\dfrac{1}{a} &gt;0)$  which implies that $x_2 &gt; x_1$  Then assume for induction: $x_k &lt; x_{k+1} &lt; a+1 $ (since $\dfrac{1}{x_k} &lt; 1$)  $$x_k + \frac{1}{x_k} &lt; x_{k+1}+ \frac{1}{x_{k+1}} &lt; a + 1$$  $$\implies x_{k+1} &lt; x_{k+2} &lt; a + 1$$  This shows that that $x_k &lt; x_{k+1} \Rightarrow  x_{k+1} &lt; x_{k+2}$. Therefore $x_n &lt; x_{n+1}$ for all $n \in  \mathbb N$.  It follows by the monotone convergence theorem that $x_n$ is increasing and bounded.
21003
Prove or disprove: if $\sum\limits_{n=0}^\infty a_n$  converges and $\lim\limits_{n→\infty}⁡b_n =0$, then $\sum\limits_{n=0}^\infty a_n  b_n$ converges.  I tried this example $$a_n= b_n = \frac {(-1)^n} {\sqrt {n+1}},$$ this will show that $$\sum_{n=0}^\infty a_n  b_n = \sum_{n=0}^\infty \frac {1} { {n+1}},$$ which is the Harmonic series, diverges.   However, I tried every convergent test but I couldn't show that $$\sum_{n=0}^\infty \frac {(-1)^n} {\sqrt {n+1}}$$ converges.
62646
$$\mathop {\lim }\limits_{n \to \infty } {{{a_n}} \over {{b_n}}} = 1$$ Prove the statement implies $\sum {{a_n},\sum {{b_n}} } $ converge or diverge together. My guess the statement is true.    if $\sum{{a_n}}$ diverges, then $\mathop {\lim }\limits_{n \to \infty } {a_n} \ne 0$    So,  $$\eqalign{   &amp; \mathop {\lim }\limits_{n \to \infty } {a_n} = L \ne 0  \cr    &amp; {{\mathop {\lim }\limits_{n \to \infty } {a_n}} \over {\mathop {\lim }\limits_{n \to \infty } {b_n}}} = 1 \Rightarrow {L \over {\mathop {\lim }\limits_{n \to \infty } {b_n}}} = 1 \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {b_n} \ne 0 \cr} $$  therefore,  $\sum {b_n}$ also diverges.    What I was not managed to do is proving that the two series converges together. Or maybe the statement is not always true?
30657
Following the guidelines suggested in , I am going to post a proposed proof as an answer to the theorem below. I believe the proof works, but would appreciate any needed corrections.  Theorem If a series $\sum_{n=1}^{\infty}a_n$ of real numbers converges then $\lim_{n \to \infty}a_n = 0$
47657
Let $H$ be a Hilbert space with ONB $\{b_1, b_2, \cdots \}$.     I want to prove that the sequence $x_n = \frac{1}{n}\sum_{i=1}^{n^2}  b_i$ converges weakly to $0$.   Notice also that $\|x_n\|=1$.  The claim is easily verified on the basis elements, i.e. for large $n$  $$\langle x_n, b_i \rangle= \frac{1}{n}\to 0$$  but I fail to see why it is true for an arbitrary element $y = \sum_i \langle y, b_i \rangle b_i$. Why does  $$\langle x_n, y\rangle = \frac{1}{n} \sum_{i=1}^{n^2} \langle y, b_i \rangle$$  converge to $0$?
35647
If we have a probability space $(\Omega,\mathcal{F},P)$ and $\Omega$ is partitioned into pairwise disjoint subsets $A_{i}$, with $i\in\mathbb{N}$, then the  says that $P(B)=\sum_{i=1}^{n}P(B|A_{i})P(A_i{})$. This law can be proved using the following two facts: \begin{align*} P(B|A_{i})&amp;=\frac{P(B\cap A_{i})}{P(A_{i})}\\ P\left(\bigcup_{i\in \mathbb{N}} S_{i}\right)&amp;=\sum_{i\in\mathbb{N}}P(S_{i}) \end{align*} Where the $S_{i}$'s are a pairwise disjoint and a $\textit{countable}$ family of events in $\mathcal{F}$. However, if we want to apply the law of total probability on a continuous random variable $X$ with density $f$, we have (): $$P(A)=\int_{-\infty}^{\infty}P(A|X=x)f(x)dx$$ which is the law of total probabillity but with the summation replaced with an integral, and $P(A_{i})$ replaced with $f(x)dx$. The problem is that we are conditioning on an $\textit{uncountable}$ family. Is there any proof of this statement (if true)?
102050
Let $(X,S,\mu)$ be a measure space. Suppose $\mu(X)&lt;\infty$. Put $E_n=\{$ $x\in X:$ $n-1\leq |f(x)|&lt;n \}$ Then, $f\in \mathcal{L}^p$ if and only if $\sum_{n=1}^{\infty}n^p\mu(E_n)&lt;\infty$. My attempt: Notice, $X=\bigcup_{n=1}^{\infty}E_n$. Since $(E_n)_n$ are pairwise disjoint, we deduce by a standard property of measures that $\mu(E_n)&lt;\infty$ for each $n$. As a consequence, $\int |f|^p d\mu=\sum_{n=1}^{\infty}\int_{E_n}|f|^p d\mu$ Now, observe that for each $n$, $n^p\chi_{E_n} \geq |f|^p\chi_{E_n}\geq (n-1)^p\chi_{E_n}$ Upon taking integrals, we obtain: $ \sum_{n=1}^{\infty}(n-1)^p \mu(E_n)\leq\int |f|^p d\mu \leq \sum_{n=1}^{\infty} n^p\mu(E_n)$ So, if $\sum_{n=1}^{\infty}n^p\mu(E_n)&lt;\infty$ then $f\in \mathcal{L}^p$. Now for the reverse, not that if $\mu(E_n)=0$ for each $n$ then we are done. If there is some $n$ such that $\mu(E_n)=0$  if $f\in \mathcal{L}^p$, then $\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)&lt;\infty$. Now the question is how do we determine whether a series coneverges given another one converges? The answer is the $\textbf{limit comparison test}$: Theorem: Let $\sum_{n=1}^{\infty}a_n$ and $\sum_{n=1}^{\infty}b_n$ be two series such that $a_n\geq 0$ and $b_n&gt;0 0$ for all $n$.  Put $c= \lim_{n\rightarrow \infty}\frac{a_n}{b_n}$.  If $0&lt;c&lt;\infty$ then either both series converge or both series diverge.  To invoke this theorem notice that if for all $n$, $\mu(E_n)=0$ then we are immediately done. So, we may assume that for some n , $\mu(E_n)&gt;0$. If for all $n$, this is the case, then we are immediately done. If only finitely many satisfy this property, then we may disregard the ones with 0 measure. Thus, we may assume for each $n$, $\mu(E_n)&gt;0$. In which case, the limit comparison test above allows us to deduce that $c=1$. Since $\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)$ converges, so must $\sum_{n=1}^{\infty}n^p \mu(E_n)$.
120704
Find an example of a function sum $\sum_{n=1}^\infty u_n $ that converges   uniformly on $[a,b]$ and converges absolutely pointwise, but   $\sum_{n=1}^\infty |u_n| $ doesn't converge uniformly on $[a,b]$.   This seems pretty straightfoawrd but it's really hard to find an example. The choice of $u_n$ must be very specific. You can actually prove any power series wont work (due to Abel's theorem)  In addition, in order to prove the uniform convergence, $u_n$ must not satisfy Weierstrass' M-Test (because then $\sum |u_n|$ would converge uniformly aswell)
79633
I'm going over old exam problems and I got stuck on this one. Suppose that $\mathbb{f}\colon \mathbb{D} \to \mathbb{C}$ is analytic and bounded. Let $\{a_n\}_{n=1}^\infty$ be the non-zero zeros of $\mathbb{f}$ in $\mathbb{D}$ counted according to multiplicity. Show that  $$ \sum_{n=1}^\infty \left( 1 - \left|a_n \right|\right)\lt\infty $$  I can understand that $\left|a_n \right|$ goes to $1$ since zeros are isolated, but it doesn't help showing the series is convergent. Any help will be appreciated!
25405
Here is what I know/proved so far:  Let $c_0\subset\ell^\infty$ be the collection of all sequences that converge to zero. Prove that the dual space $c_0^*=\ell^1$.  $Proof$: Let $x\in c_0$ and let $y\in\ell^1$. We claim that $f_y(x)=\sum_{k=1}^\infty x_ky_k$ is a bounded linear functional. Clearly $f_y$ is bounded since $$ |f_y(x)|=\left|\sum_{k=1}^\infty x_ky_k\right|\le\sum_{k=1}^\infty |x_k||y_k|\le||x_k||_\infty\sum_{k=1}^\infty |y_k|=||x_k||_\infty||y||_1. $$ We can also easily see that $f_y$ is linear. Let $x,z\in c_0$ and $y\in\ell^1$ then $$ f_y(x+z)=\sum_{k=1}^\infty (x_k+z_k)y_k=\sum_{k=1}^\infty (x_ky_k+z_ky_k)=\sum_{k=1}^\infty x_ky_k+\sum_{k=1}^\infty z_ky_k=f_y(x)+f_y(z) $$ and for $\alpha\in\mathbb{R}$ $$ f_y(\alpha x)=\sum_{k=1}^\infty \alpha x_k=\alpha\sum_{k=1}^\infty x_k=\alpha f_y(x). $$  Let $\varepsilon&gt;0$ and since $y\in\ell^1$, we know that $\sum_{k=1}^\infty |y_k|$ converges. So there exists $N\in\mathbb{N}$ such that whenever $n&gt;N$ we have  $$ \sum_{k=n}^\infty |y_k|&lt;\varepsilon. $$ Now define the following sequence $x=\{x_k\}_{k=1}^\infty$ as $$ x_k=\begin{cases} \operatorname{sgn}(y_k),&amp;\,k\le N\\ 0, &amp; \,k&gt;N \end{cases}. $$ Thus $x\in c_0$ and   $$ \begin{align} \left|f_y(x)-||y||_1\right|&amp;=\left|\sum_{k=1}^\infty x_ky_k-\sum_{k=1}^\infty |y_k|\right|\\ &amp;=\left|\sum_{k=1}^N\operatorname{sgn}(y_k)y_k-\sum_{k=1}^\infty |y_k|\right|=\left|\sum_{k=1}^N |y_k|-\sum_{k=1}^\infty |y_k|\right|=\left|\sum_{k=N+1}^\infty |y_k|\right|&lt;\varepsilon. \end{align} $$  So we conclude that \begin{equation} \ell^1\subseteq c_0^*. \end{equation} Observe that the above argument also establishes that $||f_y||_*=||y||_1$.  Now let $f$ be any linear functional on $c_0$ and let $\{e_k\}$ be the sequence with a 1 in the $k$-th position and zero elsewhere. Then for any $x\in c_0$ we have $$ |f(x)|=\left|f\left(\sum_{k=1}^\infty e_kx_k\right)\right|=\left|\sum_{k=1}^\infty f(e_k)x_k\right|\le\sum_{k=1}^\infty |f(e_k)|\,|x_k|\le||x||_\infty\sum_{k=1}^\infty |f(e_k)|. $$ Since $f$ is a bounded functional, we must have $\sum_{k=1}^\infty |f(e_k)|$ converging, otherwise $f(x)$ would be unbounded. Thus $\{f(e_k)\}_{k=1}^\infty\in c_0$ and we conclude that  \begin{equation} c_0^*\subseteq\ell^1.  \end{equation} Thus (1) and (2) tell us that  $$c_0^*=\ell^1.$$    QUESTIONS: If we take $c$ to be the collection of sequences that converge to some real number and $c^*$ to be it dual space, I know that $c^*=\ell^1$ as well, but I am not sure how to prove it. Is it enough to observe that if $x\in c$ and that $x\to x'$ then $x-x'\in c_0$, therefore they have the same dual space? I am a little fuzzy here, obviously.  Also, can someone clean up my align environment. I can't figure out how to make it compile properly. The code looks fine on my LaTeX implementation, but it doesn't work here.
58897
I am interested in proving a theorem, which I suppose one may call a sandwich or squeeze theorem for series.  Suppose we have three series: $\sum^{\infty}_{n=1}a_{n}$, $\sum^{\infty}_{n=1}b_{n}$ and $\sum^{\infty}_{n=1}c_{n}$. We know that $\sum^{\infty}_{n=1}a_{n}$ and $\sum^{\infty}_{n=1}c_{n}$ converge; furthermore, let us assume that for all $n\in\mathbb{N}$, the following inequality holds: $a_{n}&lt;b_{n}&lt;c_{n}$. Then, the series $\sum^{\infty}_{n=1}b_{n}$ will also converge.  The only way that I can think of to approach the proof of the above would be via the Cauchy criterion, i.e. showing that  $$\forall_{\varepsilon&gt;0}\,\exists_{n_{\varepsilon}}\,\forall_{m&gt;k&gt;n_{\varepsilon}}\quad|b_{k+1}+...+b_{m}|&lt;\varepsilon.$$  As I understand, in order to show that, we would have to somehow bound $\left|\sum^{m}_{n=k+1}b_{n}\right|$ by $\left|\sum^{m}_{n=k+1}c_{n}\right|$ and/or $\left|\sum^{m}_{n=k+1}a_{n}\right|$. If we assume nonnegativity of the terms of $\sum{}b_{n}$ and $\sum{}c_{n}$, the task becomes trivial. However, without this assumption, I am having problems with finding the right bound.   I would be thankful for some hints on how this could be done, or possibly advice on a different approach to the proof. Thank you in advance.
23507
I just came across the problem of what happens in the boundary cases of .  More concisely, for what $\theta$ will the following sum converge? $$\sum_{n=1}^{\infty} \frac{e^{ni\theta}}{n}$$ It is well known that this sum will diverge for $\theta=0$ and will converge for $\theta=\pi$. In fact, for any odd multiple of an even root of unity, since we have symmetry, for each line through the origin of the complex plane, we can use the alternating series theorem to show that they converge.  Another idea I had was to split  the sum into imaginary and real parts, and if we can prove each of these converges, then we are all set. So now we must prove that both of the following converge $$\sum_{n=1}^{\infty}\frac{\cos(n\theta)}{n}\space, \space \sum_{n=1}^{\infty}\frac{\sin(n\theta)}{n}$$ Now these can be shown to converge easily by Dirichlet's test for any rational multiple of pi, but what about irrational multiples of pi? I know it has been shown to converge with $\theta=1$, but I don't see how to generalize this.
91145
Let $f$ be a function of bounded variation on $[0,1]$. Let $\{[a_n,b_n]\}_{n=1}^\infty$  such that $(a_n,b_n)$ are pairwise disjoint and $\cup_{n=1}^\infty [a_n,b_n]=[0,1]$. (for example, $[1/2, 1], [0,1/3], [1/3,1/3+1/3^2], \cdots$ ) Can we write  $$ \operatorname{Var}_{[0,1]} f=\sum_{k=1}^\infty \operatorname{Var}_{[a_k,b_k]} f? $$
29945
In this  to a question about a series, a theorem was stated:     If $A= \{a_i \}$ is a set such that $\sum_{i = 1}^{\infty} \frac{1}{a_i}$ converges, then $d(A) = 0$, where $d(A)$ is the  of the set.   My background in number theory is basically zero and all my attempts to prove this have been utterly unseccessful; would anyone please help me, or at least provide me with a hint to prove it?
67083
Regarding this discussion here:   (thank you for the great answers, by the way)  I'm still trying to do Exercise 3.2.2 (b) from  by Dr. Pete Clark.  The question states that for the set $S = \mathbb{Z}^{+}$, the series $\sum_{i\in S}x_{i}$ converges unconditionally if and only if it converges absolutely, i.e, $\sum_{i\in S}|x_{i}| &lt; \infty$.  Thanks to the prior discussion, I get why unconditional convergence isn't simply the same thing as regular convergence in the case of $S = \mathbb{Z}^{+}$.  Unconditional convergence (to $x$) even allows for "cherry picking" of terms in the tail, rather than only being allowed to take them in order, and still reguires the resulting sums to stay within the $\epsilon$-band (of $x$).  (Again, thanks for the detailed answers to my previous question on this.)  My "best" attempt hits a brick wall:  Let $S = \mathbb{Z}^{+}$.  Write $S = S^{+}\cup S^{-}$, where $$S^{+} := \{n\in S : x_{n} \geq 0\}$$ and $$S^{-} := \{n\in S : x_{n} &lt; 0\}$$  $(\Rightarrow)$ Now if I assume that $\sum_{n=1}^{\infty}|x_{n}|$ is not finite ( that is, the series is not absolutely convergent ),  I want to show that the unconditional convergence definition fails.  I want to try to construct arbitrarily large finite subsets $K$ of $S^{+}$ or $S^{-}$ such that $\sum_{i\in K}x_{i}$ is arbitrarily large, which would prove the contrapositive.  I think I can do it if one of $S^{+}$ or $S^{-}$ is finite, but this need not be the case.  Any advice or suggestions?  I feel like I hit a brick wall.  Edit:  I added the solution as an answer below for anyone who might have this same question.  Thanks all for the help.
20372
Claim: Let $X$ be a metric space. If $A,B\in X$ are disjoint, if A is compact, and if B is closed, then $\exists \delta&gt;0: |\alpha-\beta|\geq\delta\;\;\;\forall\alpha\in A,\beta\in B$.  Proof. Assume the contrary. Let $\alpha_n\in A,\beta_n\in B$ be chosen such that $|\alpha_n-\beta_n|\rightarrow0$ as $n\rightarrow \infty$.  Since A is compact, there exists a convergent subsequence of $\alpha_n\;(n\in\mathbb{N})$, $\alpha_{n_m}\;(m\in\mathbb{N})$, which converges to $\alpha\in A$.  We have   $$|\alpha-\beta_{n_m}|\leq|\alpha-\alpha_{n_m}|+|\alpha_{n_m}-\beta_{n_m}|\rightarrow0 \;\;\;as\;\;m\rightarrow\infty.$$ Hence $\alpha$ is a limit point of B and since B is closed $\alpha\in B$, contradiction.  Is my proof correct? I feel as though I am missing something simple which would trivialize the proof.