I've been set the following question as part of my Real Analysis homework and I'm struggling with it:  Let $(a_n)_n\in\mathbb{N}$ be a sequence of positive numbers. Show that:  $$ \limsup_{n\rightarrow\infty} a_n^\frac{1}{n} \leq \limsup_{n\rightarrow\infty} (\frac{a_{n+1}}{a_n}) $$  I have reasoned that if the RHS takes value $\infty$ then the inequality is automatically satisfied so we can assume RHS $= \lambda &lt; +\infty$  We got a hint in the lecture that went something like this: For $m \geq n$,  $a_m \leq a_n \lambda^{m-n} \Rightarrow a_m^{\frac{1}{n}} \leq (a_n \lambda^{m-n})^\frac{1}{n}$  And then taking supremums a and limsups but I'm confused at this point and I'm not really sure where to take t from here.  Any help would be greatly appreciated.
121598
The purpose of this exercise is to prove that $\lim \frac{n}{(n!)^{1/n}}=e$ when $n$ goes to infinity.   In order to find the limit, the following inequality is used when $n$ goes to infinity with ${a_n}$ a sequence of positive terms:   $$ \liminf \frac{a_{n+1}}{a_n} \le \liminf (a_n)^ {1/n}\le\limsup (a_n)^ {1/n}\le \limsup \frac{a_{n+1}}{a_n}$$  What is the proof of this inequality?
70722
It just seems like I can't wrap my head around this. For a set $A$ how can I show that the supremum of that set $\sup(A)$ is also a limit point of said set? Can I just set $$a=\sup(A)$$ $$b\le a \: \:  \forall \: b \in A$$ and then show that there's a point different from $a$ in any $\epsilon$-neighbourhood for any $\epsilon&gt;0$? Or should I use a completely different approach?
98447
Let $\mathfrak{F} = (A_n)_{n \in \mathbb{N}}$. Prove that $P(\limsup A_n)=1$ if $\forall A \in \mathfrak{F}$ s.t. $P(A) &gt; 0, \sum_{n=1}^{\infty} P(A \cap A_n) = \infty$. (Side question 1 Is second Borel-Cantelli out because we don't know if the $A_n$'s are independent?)  Suppose $\forall A \in \mathfrak{F}$ s.t. $P(A) &gt; 0, \sum_{n=1}^{\infty} P(A \cap A_n) = \infty$, but $P(\limsup A_n)&lt;1$. My prof gave us this convenient hint: Show $\exists M &gt; 0$ s.t. $P(\bigcap_{j=M}^{\infty} A_j^{c}) &gt; 0$.  $1 &gt; P(\limsup A_n)$  $1 - P(\limsup A_n) &gt; 0$  $P(\liminf A_n^{c}) &gt; 0$  By definition of $\liminf A_n^{c}$, $\exists M &gt; 0 $ s.t. $P(\bigcap_{j=M}^{\infty} A_j^{c}) &gt; 0$. (Side questions 2 and 3 Isn't this =1? ...Wait, is that what we're trying to prove? Haha)  $\to 1 - P(\bigcup_{j=M}^{\infty} A_j) &gt; 0$  $\to 1 &gt; P(\bigcup_{j=M}^{\infty} A_j)$  Main question: Now I guess we have to come up with some A s.t.  $\to 1 &gt; P(\bigcup_{j=M}^{\infty} A_j) \geq \sum_{n=1}^{\infty} P(A \cap A_n)$, or is there something else to do?
42323
Let $S\subset\mathbb{R}$ be a non-empty bounded above set. Then there exists a monotone increasing sequence $\{x_n\}\subset S$ such that    $$\lim_{n\to\infty}x_n=\sup S.$$   I'm struggling with creating a formal proof for this problem. I understand that this has to be true, but I'm unsure of how to go about proving it.  I said that $\sup S = C$ and $x_n= C - y/n$ where $y$ is dependent on the size of $S$, but this is just a bunch of jumbled ideas and I can't really think of a way to formalize a proof. Any idea, even totally different from mine would be appreciated.
376
Let $a_n$ be a positive real sequence such that the series $\sum a_n$ converges.  I was asked to prove that under such circumstances $\sum a_n^{\frac{n}{n+1}}$ converges.  The previous sum can be rewritten as $\sum \frac{a_n}{a_n^{1/(n+1)}}$.  I can't prove the convergence of this last series... I'm sure comparison test is enough here, but other than $a_n \rightarrow 0$ there is no information that helps.  Thanks for your help.
111182
I have to solve this problem.      Let $\{\lambda_n\}$ be a sequence of real number such that $\lim_{n\rightarrow\infty}\lambda_n=0$ and consider the operator $T:l^p\rightarrow l^p$, $1\leq p\leq \infty$, defined by      $$T(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_n x_n,\ldots\} $$   Prove that $T$ is compact.    My attempt is the following.  I have considered the operator $$ T_N:l^p\rightarrow l^p,\quad T_N(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_N x_N,0,0,\ldots\}. $$ $T_N$ is clearly linear and it is also compact. In fact given any bounded sequence $\{x^{(n)}\}_{n\geq 1}$ in $l^p$, the sequence $\{T_Nx^{(n)}\}_{n\geq 1}$ is bounded in $\mathbb{R}^N\subset l^p$. Then, since every bounded sequence in $\mathbb{R}^N$ has a convergent subsequence, it follows that $T_N$ is compact.   Now, I have to prove that $T$ is compact. For this I proved that $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$. In fact we have: $$ ||T_N-T||=\sup_{||x||=1}||(T_N-T)x||=\sup_{||x||=1}||T_Nx-Tx||=\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}} $$ and, given $\epsilon&gt;0$, there exists $\bar n$ such that $|\lambda_n|&lt;\epsilon$, for every $n&gt;\bar n$. So if we take $N&gt;\bar n$, we get $$ \sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}}\leq\epsilon\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|x_i|^p\right)^{\frac{1}{p}}\leq\epsilon $$ Therefore $||T_N-T||\rightarrow 0$ as $N\rightarrow\infty$ implies that $T$ is compact.   My questions are:  (1) is the proof of the compactness of the operator $T_N$ correct?  (2) is the conclusion correct?  Thanks
59518
If I have a sequence {$a_n$} that has the property of $\lim(a_{n+1}-a_n)=0$, does that make it a Cauchy Sequence. I think it doesn't because $a_n = \sum_{k=1}^n \frac{1}{k}$ is a counter example.   However, by definition, there exists a $M$ such that if $n \geq M$ then $|a_{n+1}-a_n| &lt; \frac{\epsilon}{m-n}$  Hence, we have $|a_m - a_{m-1}|+.....+|a_{n+1}-a_n|&lt;|a_m -a_n| &lt;\epsilon$  This proof doesn't work because I cannot be sure I can find a fixed $M$ which might change according to n.  But I wonder if I have an additional condition that says $a_n$ is bounded, I think then the proof works and that I should be able to find a fixed $M$. However, I don't know how to justify this. Maybe I am wrong. Can someone kindly help me figure out this problem. Thanks
40687
I am starting to learn about the liminf and limsup of a sequence of sets. However, I do not understand the very basics, such as why $\lim\sup A_n=\bigcap_{n\in N}\bigcup_{k&gt;n} A_k,$ and why $\lim\sup A_n=\bigcup_{n\in N}\bigcap_{k&gt;n}A_k.$ Can someone break these definitions down for me?  Also, why is $\lim\inf\subseteq\lim\sup$?
101155
Suppose that a sequence $a_n$ of positive numbers converges to $a$. Show that $$\lim_{n\rightarrow \infty}\left(\prod_{i=1}^{n}a_i\right)^{1/n}=a$$ This seems to be simple using that $x=e^{\log x}$, but I can't go any further after using that.   PS. I'm almost sure that this post is a duplicate, but I wasn't able to find it here. I'm sorry if that's the case.
122062
If the sequence satisfies the property lim$_{n\to \infty}(a_n-a_{n-2})=0$, prove that lim$_{n\to \infty}\frac{a_n-a_{n-1}}{n}=0$.  I am experiencing difficulty showing this rather obvious result. By definition, we know that for any $\epsilon \gt 0$, there is some $N$ such that if $n \gt N$ then $|a_n-a_{n-2}|\lt \epsilon$ My guess is that I need to write $|a_n-a_{n-1}|$ as sums of $|a_n-a_{n-2}|$ from $n$ to $N$. However, I am stuck with forming an inequality here. I would appreciate any solutions or suggestions.
98882
Let $\{ a_n \}_{n=1}^{\infty}$ be a bounded sequence s.t. $a_n&gt;0,\ \forall n \in \mathbb{N}$. Suppose $\lim_{n \rightarrow \infty} a_na_{n+1}=1$, prove $\limsup_{n \rightarrow \infty} a_n\geq1$.   I tried to prove by contradicition using subsequence, but couldn't work it out.  Any help appreciated.
116157
If $n\geq m$ show that: $(x^m+y^m)^{1/m} \ge (x^n+y^n)^{1/n}$, all numbers being positive real.   Obviously since $n\geq m$, for every $a\in \Bbb Z^+$: $a^{1/m}\geq a^{1/n}$, but does it help?
87463
Assume that $(a_n)$ is a sequence of positive real numbers satisfying $a_{n+1}=\frac{2}{a_n+a_{n-1}}$ for $n=2,3,\dots$.  Prove that $(a_n)$ is convergent and find the limit.    I have no idea how to prove convergence. I tried to prove that $(a_n)$ is bounded and monotone, but both trials failed.  If convergence is proven, the limit is easy: if $g=\lim a_n$, then $g\ge 0$ because all $a_n$'s are positive, the equivalent relation $a_{n+1}(a_n+a_{n-1})=2$ gives $g(g+g)=2$, so finally $g=1$.
43921
Suppose $a_n$ and $b_n$ are sequences of positive numbers such that  $\limsup \frac{a_n}{b_n}&lt;\infty$  show that there is a constant $M&gt;0$ s.t $a_n\le Mb_n$ for all $n\ge 1$  My idea:  as i know hat $M=\limsup a_n$ if and only if for all $\epsilon &gt;0$ there is $N\in \mathbb{N}$ so tthat $n\ge N$ implies $a_n&lt;M+\epsilon$
26570
The last problem I posted had a wrong statement. I recovered the correct one.  Let $(a_n)$ be a sequence of positive real numbers. Prove that $\sum_{n\geq 0}a_n$ converges iff $\displaystyle \sum_{n\geq 0} \frac{a_n}{\sum_{k=0}^n a_k}$ converges  The direct statement is easy to prove using comparison test. I'm (again!) stuck with the converse. I tried summation by part, without success. Note that the convergence of $\displaystyle \sum_{n\geq 0} \frac{a_n}{\sum_{k=0}^n a_k}$ implies $\displaystyle \frac{a_n}{\sum_{k=0}^n a_k} \to 0$ I don't know what to do next ...
2154
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A = -\sup(-A)$.   We know that $-A$ is bounded above. Hence $\sup(-A)$ exists. So to proceed, one would show that $\inf A \leq -\sup(-A)$ and $\inf A \geq -\sup(-A)$? Or $-\inf A &gt; \sup(-A)$ and $-\inf A &lt; \sup(-A)$.  I get the sense that one can use the following fact as well: (i) Let $\alpha = \sup S$. If $\beta &lt; \alpha$ then $\beta$ is not an upper bound for $S$.
96657
What I've done so far: I have proved that this sequence is bounded below by 0, which is a very rough estimate. I know that the infimum is $\sqrt2$.  Anyway, the question first asks me to prove that the sequence decreases monotonically. And I've tried the following:  Suppose $a_{n+1} \le a_n$, then we would get $\frac{1}{a_n} \le \frac{a_n}{2}$. After some algebra, I ended up with $\frac{1}{a_n^2 +2}\le\frac{a_n^2 +2}{2}$. Now, $a_{n+2} - a_{n+1}=\frac{2a_n}{a_n^2 +2}-\frac{a_n^2 +2}{4a_n}$. I was looking for some similarities between these two expressions but it doesn't seem to work out.  A more general question is, if the sequence is defined recursively, what are some common strategies to find the limit? I just started studying analysis and this kind of questions kind of troubles me.  Thank you guys.