I'm trying to solve the second part of the following exercise which is question 32 in Section 1.5 of Real Analysis by Folland.  Suppose $\{\alpha_j\}_{j=1}^{\infty} \subset (0, 1)$. Then:   $\prod_{j=1}^{\infty} (1 - \alpha_j)&gt;0$. if and only if $\sum_{j=1}^{\infty} \alpha_j &lt; \infty$. Given $\beta \in (0, 1)$, exhibit a sequence $\{\alpha_j\}_{j=1}^{\infty}$ such that $\prod_{j=1}^{\infty} (1 - \alpha_j)= \beta$.   How can I prove the second part of the statement? I am pretty sure that it should be solved by induction. But I don't know how to do it.
123356
Suppose $\{\alpha_j\} \subset (0,1)$. Give a sequence ${\alpha_j}$ such that $\prod (1-\alpha_j) = \beta $, where $\beta \in (0,1)$   I was able to prove part a. That is the infinite product of $(1-a_j)$ is positive iff $\sum \alpha_j &lt; \infty $.  However, I have no idea how to exhibit a sequence ${a_j}$ so that the infinite product of $1-\alpha_j$ is some $\beta$ that is in $(0,1)$. Would I need to construct a specific sequence or a more general case?  Any help would be much appreciated as I am really stuck and don't have much of an idea even where to begin.
13671
Let $A_n$ be independent events with $P(A_n) \neq 1$. Show that $\prod_{n=1}^{\infty} (1- P(A_n))=0$ iff $\sum P(A_n) = \infty$  It kind of looks obvious but I really have no idea how to prove it. Can someone give me help?
52093
If $\sum_j^\infty a_j$ and $\sum_j^\infty b_j$ are convergent, show that  $\sum_j^\infty a_j \cdot b_j$ is convergent or provide a counter example where it is not convergent.  If it can be assumed that one of the given series is absolutely convergent, then I can see a proof that the resulting series is convergent. But I don't see a proof or a counter example when both given series are conditionally convergent.
114439
Let $\alpha$ be an irrational real number and let $a_j$ be a sequence of rational numbers converging to $\alpha$. Suppose that each $a_j$ is a fraction expressed in lowest terms: $a_j = \alpha_j / \beta_j$. Prove that the $\beta_j$ tend to $\infty$  Attempt: AFSOC that $\beta_j$ does not tend to infinity, then it is bounded by some $M$. We can also find an interval $\alpha \in (k, k+1)$. Let $\epsilon &gt; 0$, $\exists N_0$ such that for all $N &gt; N_0$ we have $|a_N - \alpha|  &lt; \epsilon$. So $|\frac{p_N}{q_N} - \alpha| &lt; \epsilon$. We try to use the fact that $q$ is bounded but fail to derive a contradiction. I do not know how to use the fact that $\alpha$ is between two integers, although it might not be relevant at all.  Ideas?
67670
How to evaluate the product:  $$\prod _{n=2}^\infty\left(1+\dfrac{1}{n^2}+\dfrac{1}{n^4}+\dfrac{1}{n^6}+\dfrac{1}{n^8}+\ldots\right)$$  $$\left(1+\dfrac{1}{n^2}+\dfrac{1}{n^4}+\dfrac{1}{n^6}+\dfrac{1}{n^8}+\ldots\right)=\dfrac{1}{1-\dfrac{1}{n^2}}$$  How to proceed now?
37120
[Sorry, but I couldn't come up with a better title...]  I'm trying to prove that $$\prod_{i=1}^n (1+a_i) \leq 1 + 2\sum_{i=1}^n a_i$$ if all $a_i$ are non-negative reals with $\sum_{i=1}^n a_i \leq 1$ and it seems I'm stuck.  (And, no, this is not homework!)
64925
Let $\{a_n\}$ be a sequence of real number, let $\{n_k\}$ be an increasing sequence of positive integer and let $A_n=a_{n_k}+a_{n_{k+1}}+...+a_{n_{k+1}-1}$ for $k \in N$. Prove that if $\sum_{n=1}^\infty a_n$ converges then $\sum_{k=1}^\infty A_k$ converges. Is the converse true?  I know that $\sum_{n=1}^\infty a_n$ converges implies that $lim _{n_&gt;\infty} a_n =0$, can I conclude that $A_n$ ->$0$ as well, even if I can conclude that, it still doesn't prove that $\sum_{k=1}^\infty A_k$ converges, am I correct?
84418
Here's the problem in full. I've stared at it for hours and can't get anywhere, so a hint would be nice.  Suppose that $f:\mathbb N\rightarrow\mathbb R$. If $$\lim_{n\rightarrow\infty}f(n+1)-f(n) = L$$ prove that $\lim_{n\rightarrow\infty}f(n)/n$ exists and equals $L$.  This problem is marked with the word "Cauchy", so I'm guessing Cauchy sequences or something related to them will be necessary, but I can't figure out what.  Edit: It appears as if the Stolz-Cesaro Theorem makes this quite easy. Unfortunately though, we have not discussed this, so I won't be able to use this.
100500
Note that all elements are complex numbers. Suppose we know $$\sum_{n=1}^\infty|a_n|^2\quad\text{and}\quad\sum_{n=1}^\infty a_n$$converge. How can we conclude $\sum_{n=1}^\infty\log(1+a_n)$ converges? This is an effort to prove that $$\prod_{n=1}^\infty(1+a_n)$$converges.  If I knew that $\sum a_n$ was absolutely convergent, then I think I'd be able to use the power series expansion of $\log(1+a_n)$ and interchange summations. However, I don't think this is the intended method of solution. Will someone help me with this homework problem?  Update: Other work I've done so far:  Disregarding finitely many terms if necessary, we can assume $|a_n|&lt;\frac{1}{2}$ for all $n$. Thus, we know $0\leq|a_n-\log(1+a_n)|&lt;|a_n|^2$, so by comparison, we know $\sum_{n=1}^\infty |a_n-\log(1+a_n)|$ converges. Since it is absolutely convergent, it is convergent, so now we have $$\sum_{n=1}^\infty a_n-\sum_{n=1}^\infty\left( a_n-\log(1+a_n)\right)=\sum_{n=1}^\infty\log(1+a_n),$$therefore the series is convergent. Is this correct?
43026
We have a complex number $z$ such that $\Re (z)&lt;0$ and the sequence $ z_n = \prod_{k=1}^n (1+ \frac{z}{k})$. Prove that $\lim_{n \rightarrow \infty} z_n = 0$.    How to do it? I guess it will be easier to prove that $\lim_{n \rightarrow \infty} |z_n|^2 = 0$, it's a trick that usually helps, but even though I thought about if for a longer while, I don't know how to prove it. May somebody help?
56635
I am not able to solve this exercise. I've tried everything.     Exercise Let be $a_n &gt; 0$ a sequence and suppose that $\sum a_n$ converges. Prove that exists a sequence $c_n &gt; 0$ such that $\lim c_n = \infty$ and $\sum c_na_n &lt; \infty$   I don't even know if I have to find an "always-valid" closed form for $c_n$ (maybe in function of $a_n$?) or prove its existence in another way.  Thanks in advance.
45780
question is regarding the same problem. I wish to present my proposed solution and get feedback on my argument, and as such, I claim that it is not a duplicate. (In particular, the other asker actually knew what he was doing, so this question will probably be sufficiently different. It also includes the discussion about the box topology, which the original question lacked.) We are given that $\{x_n\}$ is a sequence of points in the product space $\prod X_\alpha$. We are to show that this sequence converges to $x$ if and only if $\{\pi_\alpha(x_n)\}$ converges to $\pi_\alpha(x)$. Moreover, we are to determine if this holds if we equip $X = \prod X_\alpha$ with the box topology. We first show $\implies$. Suppose, for the sake of contradiction, that $\{\pi_\alpha(x_n)\}$ does not converge to $\pi_\alpha(x)$ for a fixed $\alpha$. Then there exists an open set $U_\alpha$ containing $\pi_\alpha(x)$ such that $X_\alpha - U_\alpha$ contains infinitely many $\pi_\alpha(x_n)$. Since every projection in the product topology is continuous, $\pi^{-1}(U_\alpha)$ is open in $X$. This contradicts the assumption that $\{x_n\} \to x$, since we now have an open set of $X$ where one of the $\{x_{n_i}\}$ clearly do not converge, where $\{x_n\} = (\{x_{n_1}\},\{x_{n_2}\},\cdots)$ We proceed to show $\impliedby$. We are given that $\{\pi_\alpha(x_n)\} \to \pi_\alpha(x)$ for every $\alpha.$ Let $\prod U_\alpha$ be an open set in $X$ containing $x$. By the definition of the product topology, we know that $U_{\alpha} \ne X_\alpha$ for only finitely many $\alpha_1, \alpha_2,\cdots,\alpha_k$. Now, since we know that $\{\pi_\alpha(x_n)\} \to \pi_\alpha(x)$, it follows by the definition of convergence that there exist numbers $N_{\alpha_1},N_{\alpha_2},\cdots,N_{\alpha_k}$ such that $\pi_{\alpha_i}(x_n) \in U_{\alpha_i}$ for $i \in \{1,\cdots,k\}$ and $n \ge N_{\alpha_i}$. Letting $N = \max\{N_{\alpha_1},\cdots,N_{\alpha_k}\}$, we see that $x_n \in \prod U_\alpha$ for $n \ge N$. (Recall that for $\alpha$ not in the finite collection, it equals the whole space $X_\alpha$, where inclusion is trivial.) We conclude that $\{x_n\} \to x$ Lastly, we consider $X$ with the box topology. In $\implies$ we did not use any particular properties of the product topology, aside from the continuity of projections, which also holds for the box topology. As such, I claim that $\implies$ still holds. However, I claim that $\impliedby$ does not hold. I ask for an example that shows this, since I am not able to find any argument I'm happy with. (&quot;Well, our proof used properties of the product topology, so...&quot; is only acceptable for my private notes.)
11822
I'm trying to prove the following:     If $(a_n)$ is a sequence of positive numbers such that $\sum_{n=1}^\infty a_n b_n&lt;\infty$ for all sequences of positive numbers $(b_n)$ such that $\sum_{n=1}^\infty b_n^2&lt;\infty$, then $\sum_{n=1}^\infty a_n^2 &lt;\infty$.   The context here is functional analysis homework, in the subject of Hilbert spaces.  Here's what I've thought:  Let $f=(a_n)&gt;0$. Then the problem reads: if $\int f\overline{g}&lt;\infty$ for all $g&gt;0,g\in \ell^2$, then $f\in \ell^2$. This brings the problem into the realm of $\ell^p$ spaces.  I know the inner product is defined only in $\ell^2$, but it's sort of like saying: if $\langle f,g\rangle &lt;\infty$ for all $g&gt;0,g\in \ell^2$ then $f\in \ell^2$.  I read this as: "to check a positive sequence is in $\ell^2$, just check its inner product with any positive sequence in $\ell^2$ is finite, then you're done", which I find nice, but I can't prove it :P  From there, I don't know what else to do. I thought of Hölder's inequality which in this context states: $$\sum_{n=1}^\infty a_nb_n \leq \left( \sum_{n=1}^\infty a_n^2 \right)^{1/2} \left( \sum_{n=1}^\infty b_n^2 \right)^{1/2}$$  but it's not useful here.
8638
How would I go about finding $\prod_{n=1}^\infty (1+\frac{1}{n^2})$ ? Setting $a_k = \prod_{n=1}^k (1+\frac{1}{n^2})$, the first few partial products are $a_1 = 2$, $a_1 = 2.5$, $a_3 \approx 2.78$, $a_4 \approx 2.591$, $a_{20} \approx 3.50$, and they eventually appear to converge at around $3.639$. (These values were found with a quick python program. So much for my initial conjecture that the product converged to $e$.)   I tried writing $1+\frac{1}{n^2}$ as $\frac{n^2 + 1}{n^2}$, but it wasn't immediately obvious how to prove convergence / find the limit because the product doesn't really factor or telescope. Is there a way to calculate the limit?
45664
Is there any elementary proof that the limit of infinite product $\prod_{k=1}^{\infty}(1-\frac{1}{2^k})$ is not 0?
43710
General condition: $(P_n)_{n\in\mathbb{N}}$ is a sequence of non-zero real numbers. If $\prod_{n=0}^\infty P_n$ exists in reals and is non-zero, then call this infinite product convergent. Otherwise divergent.  I have proven that:  If all $(P_n)_{n\in\mathbb{N}}$ are positive, then its convergence is equivalent to convergence of $\sum_{n = 0}^\infty \log(P_n)$.  If further $(P_n)_{n\in\mathbb{N}}$ are further greater or equal to 1, then the convergence of the product is equivalent to the convergence of $\sum_{n = 0}^\infty (P_n-1)$  Now I am looking for two examples where a)  $(P_n)_{n\in\mathbb{N}}$ are reals. $\sum_{n = 0}^\infty (P_n-1)$ converges but $\prod_{n=0}^\infty P_n$ diverges b) $(P_n)_{n\in\mathbb{N}}$ are reals. $\prod_{n=0}^\infty P_n$ converges but $\sum_{n = 0}^\infty (P_n-1)$ diverges   I have spent a long time on this but failed to find any. I guess it requires complex analysis technique? (Which I don't know) Please help me out. Thank you.
88110
I am given the problem: Let $a_n&gt;0$, prove $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty \sin(a_n)$ converges.  We have this problem in a homework, and I don't believe it can be true. The statement being biconditional implies that when $\sum_{n=1}^{\infty} \sin(a_n)$ converges, so too does $\sum_{n=1}^\infty a_n$, but if we take $a_n=\pi n$ or even $a_n=\pi$, then $\sum_{n=1}^\infty \sin(a_n)$ converges while $\sum_{n=1}^\infty a_n$ diverges. Is my thinking incorrect?