I came across the claim* that given a sequence of positive numbers that converges to zero, it is always possible to find a sequence of integers, so that the product sequence converges to any desired real number. I.e. given $\{ x_n \}$, $x_n&gt;0$, $x_n \to 0$, and $t\in \mathbb{R}$, $\exists$ $\{ k_n \}$, $k_n\in \mathbb{Z}$ such that $k_n x_n\to t$.  I am having a bit of difficulty showing this, any tips? Thanks!  *Appears in lemma 3.43 of Brian C. Hall - Lie Group, Lie Algebras, and Representations.
120780
Suppose $\delta_n$ is a sequence of positive numbers decreasing to $0$. Let $t&gt;0$ be a fixed real number. Then can we find some integers $N_n$ such that $N_n \delta_n \rightarrow t$? We don't know anything about the $\delta_n$ other than the fact that they go to $0$.
104844
Let $y_n$ be a sequence of real numbers. Prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges.  Let us suppose that $y_{n}$ goes to a limit $L$. Then for all $\varepsilon &gt;0$, for sufficiently large $n$, $|y_n -L| &lt; \varepsilon$ .  But $|y_n - L| &lt; 2|x_n-\frac{L}{3}|+|x_n - \frac{L}{3}|$  But then how to proceed?
79337
Suppose $a_0$ is an arbitrary positive real number. Define the sequence $\{a_n\}$ by $$a_{n+1}=\frac{a_n+2}{a_n+1}$$ for all $n\geq0$. I have to prove that $\{a_n\}$ converges.  My attempt: If  $a=\lim_{n\to\infty}{a_n}$ exists, then it should be a solution to $$a=\frac{a+2}{a+1}$$ which is $\sqrt2$. Thus I need to show that $|\sqrt2 - a_n|$ gets arbitrarily small for large $n$. I tried to prove that $|\sqrt2-a_n|&lt;|\sqrt2-a_{n-1}|$ but couldn't.
3701
I was wondering if the following is true. Suppose a series $a_n$ is greater than 0 for all positive integer n, and that $\sum \frac {a_n}n$ converges, then is $\displaystyle \lim_{m\to \infty}\sum_{n= 1}^m {a_n \over m+n} = 0$? It seems to be true because if $\sum {a_n\over n}$ converges, then that means that ${a_n\over n }\to 0$ for $n\to \infty$. This means, neglecting $n$, ${a_n \over m+n}$ will also tend to 0, and thus the summation would be equal to 0, but I don't know if this is true.
63598
If $a_n$ is a sequence such that   $$a_1 \leq a_2 \leq a_3 \leq \dotsb$$   and has the property that $a_{n+1}-a_n \to 0$, then can we conclude that $a_n$ is convergent?   I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in  to a similar question that does not require the sequence to be increasing.  $$0, 1, \frac12, 0, \frac13, \frac23, 1, \frac34, \frac12, \frac14, 0, \frac15, \frac25, \frac35, \frac45, 1, \dotsc$$  This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $\pm\frac1m$ and $m$ increases the further we go in this sequence.    So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge?  Or is this still not sufficient?
10995
How do I find:  $$\prod_{n=1}^\infty\; \left(1+ \frac{1}{\pi ^2n^2}\right) \quad$$  I am pretty sure that the infinite product converges, but if it doesn't please let me know if I have made an error. Also, could I have a nice explanation as to how would someone arrive to the answer.  Thanks alot.
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.)
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.
88350
$(z_n)_{n\in \mathbb{N}}$ is a complex valued sequence such that subsequences $(z_{2n})_{n\in \mathbb{N}}$ and $(z_{2n+1})_{n\in \mathbb{N}}$ converges to same limit.  How could I show that the sequence $(z_n)_{n\in \mathbb{N}}$ converges to the same limit using $\epsilon$, $N$ definition?    For every $\epsilon &gt;0$,  $|z_{2n}-l|&lt;\epsilon$ whenever $n&gt;N_1$, and   $|z_{2n+1}-l|&lt;\epsilon$ whenever $n&gt;N_2$.  We want to show that for every $\epsilon &gt;0$, there exists $N$ such that $|z_{n}-l|&lt;\epsilon$ whenever $n&gt;N$. I tried something like  $|z_{n}-l|\le|z_n-z_{2n}|+|z_{2n}-l|$ but wasn't sure what to do next
80461
I thought about sequences, and I had an idea.     Let's define a sequence $(a_i)$ such that    $$a_0 = c \qquad\text{and}\qquad a_{n+1}=\sin a_n$$   where $c$ is arbitrary constant. Let's also define $$S_N := \sum_{i=0}^Na_i=c\;+\;\sin c\;+\;\sin\sin c\;+\;\cdots\;+\;\underbrace{\sin\sin\cdots\sin}_{N\text{ times}} c$$      Is $\lim_{N\to\infty} S_N$ convergent or divergent?   The problem is that I have no idea how I can, or if it is even possible to, determine $a_n$. Or if it is possible to show whether the sum converges.
61090
Let $(a_n)_{n=0}^{\infty}$  be a sequence of real numbers satisfying $\sum_{1}^{\infty} |a_n - a_{n-1}| &lt; \infty$   Which of the following conclusions must then be true?  The series $\sum a_nx^n$ converges   Nowhere on $ \mathbb{R}$ Everywhere on $  \mathbb{R}$ On some interval containing $(-1,1)$ Only on $(-1,1)$   What I have tried is Make an inequality with $|a -b| \leq |a| +|b|$ .. but i cant proceed further .. Any sort of help will be deeply appreciable .
48471
For each positive integer $n$, let $x_n=1/(n+1)+1/(n+2)+\cdots+1/(2n)$. Prove that the sequence $(x_n)$ converges.   This is what I have so far..  Let $\epsilon &gt; 0$ be given to us. We must show that there exists an $N \in \mathbb{N}$  such that $n\ge N \implies \left|\frac{1}{2n} - 1\right| &lt; \epsilon$. Choose $N$ to be any positive integer which is larger than $\frac{1}{\epsilon}$. (So $N&gt;\frac{1}{\epsilon}$.) Then $n\ge N \implies \left|\frac{1}{2n} - 1\right| = |\frac{1-2n}{2n}|$
111539
Is there a strictky increasing sequence $\{ a_k\}$  of positive integers such that $\sum 1/{a_k}$ is finite and minimal with respect to this property, in a sense that if $\{ b_k \}$ is another sequence of such, then $a_k \leq b_k$ for all but finitely many values of $k$?  Motivation came from comparing the series $\sum 1/n$ and $\sum 1/{n^2}$, the first of which is divergent but the second of which is not; the sequence for the first is "smaller" than that of the second.  If this question need be motified to make my question clearer, please comment/suggest adit as well.
27721
Let the sequence be defined recursively $X_{n+1}=pX_{n}+q$ for p nonzero and $X_{1}$ arbitrary. In what conditions does the sequence converge?  I bumped into this problem and I don't know where to even begin. I would really appreciate some help.
57722
Not sure how to approach this problem below. I tried assuming the conditional but that doesn't seem to get me anywhere.     If for a sequence of non-negative real numbers $\{a_n\}$, we have $$\sum{a_kb_k}$$ converges whenever $$\sum{b_k^2}$$ converges. then show that $$\sum{a_k^2}$$ converges
13060
Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of real numbers. A well known result is:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ converges to $a\in \mathbb{R}$.   Now, I'm working on an exercise which essentialy asks to prove the following:     The sequence $(x_n)_{n\in \mathbb{N}}$ converges to $a\in \mathbb{R}$ if and only if for every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ there is a (sub)subsequence $(x_{n_{k_j}})_{j\in \mathbb{N}}$ which converges to $a$.   Now, one direction is trivial. If in fact $(x_n)_{n\in \mathbb{N}}$ converges to $a$, then by the result I've mentioned, every subsequence of $(x_{n})_{n\in \mathbb{N}}$ converges to $a$. Let then $(x_{n_k})_{k\in \mathbb{N}}$ be an arbitrary subsequence. In that case, we know that it is a subsequence of itself, and hence we've found a (sub)subsequence which converges to $a$.  The other way around seems quite complicated for me. The reason is as follows: I must assume that every subsequence $(x_{n_k})_{k\in \mathbb{N}}$ has a (sub)subsequence which converges to $a$ and the proof that the original one converges to $a$.  My problem is exactly that "a". If it had all (sub)subsequences converging to $a$, then by the result i've mentionted, the subsequence $(x_{n_k})_{k\in \mathbb{N}}$ would converge to $a$. In that case, we would have every subsequence converging to $a$ and the result would follow.  My idea to prove the claim was then:  First remember that a subsequence is the restriction of the original one $x: \mathbb{N}\to \mathbb{R}$ to an infinite subset $\mathbb{N}'\subset \mathbb{N}$. In that case a (sub)subsequence is the restriction of $\tilde{x} = x|\mathbb{N}'$ to an infinite subset $\mathbb{N}''\subset \mathbb{N}$. But since $\tilde{x}$ is the restriction of $x$ to $\mathbb{N}'$ and since $\mathbb{N}''\subset \mathbb{N}'$, the restriction of $\tilde{x}$ to $\mathbb{N}''$ is itself the restriction of $x$ to $\mathbb{N}''$.  In other words, every (sub)subsequence can be viewed as a subsequence of the original one. Considering this, by hypothesis every subsequence has a (sub)subsequence which converges to $a$. But by this argument, this means that we can replace every subsequence with a convergent one. And then we can use the result I stated.  Is my idea correct? Is this really the way to prove the claim? My only doubt is that I need every subsequence converging to $a$ to use the result. In this case I can't see how I can argue that this implies every subsequence converges to $a$.  How can I finish this proof?
56471
Give either a justification or a counterexample for each of the following statements about real series. $\quad(a)$ If $ka_k\to0$ as $k\to\infty$, then $\sum a_k$ converges. $\quad(b)$ If $\lim\limits_{k\to\infty}\frac{a_{k+1}}{a_k}$ exists and equals $L&gt;1$, then $\sum a_k$ diverges. $\quad(c)$ If $\sum a_k$ converges and $\lim\limits_{k\to\infty}\frac{a_k}{b_k}=1$, then $\sum b_k$ converges.  I think all propositions are false.  But struggling to find counterexamples.  Is there some sort of systematic way of constructing counterexamples for these sorts of series?  Can someone explain their thinking process in giving counter examples for the above?