Given the set $A_n =\{(i,j)∈ Z^2:(i,j)=1, 0≤i,j≤n\}$  I am having problems proving that the limit $\lim_{N\to \infty}$  $|A_n|\over N$ exist and to calculate it, any tips?
123182
Let $A_n=\{(i,j)\in\mathbb{Z}^2:\gcd(i,j)=1, \ \  0 \leq i,j\leq n \}$. How to prove the existence of $\lim_{n\to\infty}\frac{|A_n|}{n^2}$, and how calculate this limit?  Thank you!
68710
I am trying to prove that the sequence $$a_n = (n!)^{\frac2n}$$ tends to infinity as $ n \to \infty $.  I've tried different methods but I haven't really got anywhere. Any solutions/hints?
114143
Suppose that $a_n \neq 0$, for every n and that $L = \lim |\frac{a_{n +1}}{a_n}|$ exists. Show that if L &lt; 1, then $\lim a_n = 0$.  What I did so far:  If L &lt; 1 and $L = \lim |\frac{a_{n +1}}{a_n}|$, there exists $n_0$ such that for $n \geq n_0$, $0 &lt; |a_{n+1}| &lt; |a_n|$. That means that the sequence $(|a_n|)_{n \geq n_0}$ is decreasing.   Consider the set $S = \{ |a_0|, ..., |a_{n_0} \}$. S is finite. Let $\beta = \max_{0 \leq i \leq n_0} |a_ i|$. Thus, $(|a_n|)$ is limited (because, for every n, $0 \leq |a_n| \leq \beta).$  Now I have that $(|a_n|)$ is limited and, throwing away a finite number of terms (the $n_0$ firsts) I can assume that it is decreasing. So I know that $(|a_n|) converges.   How can I prove that it converges to $0$?   I also know that if I prove that $ \lim |a_n| = 0$, then I have that $ \lim a_n = 0$.
112903
Is it true that for a sequence $\{a_n\}_{n=1}^\infty\subset\mathbb{C}$ that does not converge to a finite limit $\limsup\sqrt[n]{|a_n|}=\limsup\sqrt[n]{|a_n-a_{n-1}|}$, ($n&gt;1$)?  It can be easily shown that   $\limsup\sqrt[n]{|a_n-a_{n-1}|}\leq \limsup\sqrt[n]{|a_n|}$, but I have problem with proving the opposite inequality (or finding a counterexample).
98766
Let $a_{n+1}=\cos(a_n)$ for $n\ge0$ and $a_0 \in [0,\pi/2]$  Find $\lim_{n \to \infty}a_n$ if it exists.  I drew some sketches and it does seem like the limit exists, it's probably $x$ such that $\cos(x)=x$  I have no idea how to go about solving this, hints would really be appreciated.  Thank you for your time!
110283
Prove that if $x &gt; 0$ and $x_n$ is a sequence with $\lim\limits_{n \to \infty} x_n = x$, then there is a real number $N$ s.t. whenever $n &gt; N$, $x_n &gt; 0$.  This is a homework question and I'm not really sure what methods I should use to prove this, can I get a push in the right direction? I am not expecting a flat out answer as it is a homework problem, but I am stumped!
19573
Let $a &gt; 0$ and let $k \in \mathbb{N}$. Evaluate: $$\lim_{n\to \infty}a^{-nk}\prod ^k_{j=1}\left(a+\frac{j}{n}\right)^n$$  Clueless on this problem. Seek your help.
105434
Good morning, i have a big problem with this proof. I haven't idea about this proof.   Problem:   Suppose $\lim_{n\rightarrow\infty}a_n =L  $ then $\lim_{n\rightarrow\infty}\ \frac{a_1+a_2+\cdots+a_n}n=L  $  I've tried this:  $| a_n-L|&lt;\varepsilon\Rightarrow L-\varepsilon&lt;a_n&lt;L+\varepsilon$ Please help!!
60248
If $ (x_{n})_{n \in \mathbb{N}} $ is a sequence in $ \mathbb{R} $ and $ \displaystyle \lim_{n \to \infty} (2 x_{n + 1} - x_{n}) = x $, then is it necessarily true that $$ \lim_{n \to \infty} x_{n} = x? $$  Could anyone kindly offer suggestions on how to approach this problem? Thanks!
75374
Prove: $\displaystyle\liminf_{n\to\infty}(-a_n)=-\limsup_{n\to\infty}(a_n)$   My general idea was: if $a$ is partial limit (PL) of $a_n$, then $-a$ is a PL of $-a_n$ so it follows that $s$ is the maximal PL iff $-s$ is the minimal PL.   But how do you show it rigorously ?
18978
Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$  Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?  $\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} &lt; \epsilon$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} &gt; \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \sqrt{n} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow n &gt; \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
98232
I found a limit problem which is $\lim_{n\to\infty}{2n \choose n}$ as we know that $\sum_{i=0}^{n}{n \choose i}^2={2n \choose n}$ now we could transform the given problem as $\lim_{n\to\infty} \sum_{i=0}^{n}{n \choose i}^2$ and as $\sum_{i=0}^{n}{n \choose i}^2$ is a polynomial in $n$, therefore, $\lim_{n\to\infty}\sum_{i=0}^{n}{n \choose i}^2=\infty$ Please tell me if my solution is correct or not,and if it is incorrect then please provide me a solution.  Thanks
55529
I'm wondering if I have a sufficient proof of the following:   If $(a_n)$ is a sequence such that $\lim_{n \rightarrow \infty}a_n=A$, then $\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$.   My approach:  For all $\varepsilon &gt; 0$, there exists $N$ such that for all $k&gt;N$, $|a_k -A|&lt;\varepsilon$. So we can break the limit up as follows $$\lim_{n\rightarrow \infty} \frac{a_1+...+a_k}{n} + \lim_{n \rightarrow \infty}\frac{a_{k+1}+...+a_n}{n} \overset{\epsilon \rightarrow 0}{=} 0 + \lim_{n \rightarrow \infty}\frac{nA}{n}=A$$  Is this on the right track, or am I missing something about breaking up the limit in the way I have?
16166
I'm having trouble to prove the following question:  Supose $\{A_n\}_{n\in\mathbb{N}}$ is a family of sets such that $A_1\subset A_2\subset A_3\subset\ldots$ (it's possible to have $A_n=A_{n+1}$). I need to prove that $$\lim_{n\to\infty}A_n = A_1\cup\bigcup_{n=2}^\infty(A_{n}\backslash A_{n-1}).$$  I'm not sure if this is useful, but I proved that $A_{n}\backslash A_{n-1}$ and $A_{n'}\backslash A_{n'-1}$ are disjoint if $n\neq n'$.  Thank you.
39602
Prove that:  $$\lim_{n\to\infty} \sqrt[n]{a^{n} + b^{n}} = \max(a, b)$$  I don't have a clue even how to start the proof here. Any hint is appreciated!
30401
I want to prove the following theorem     Let $\{a_{n}\}$ be a convergent sequence. Prove that $\lim\limits_{n\to \infty}a_{n+1}=\lim\limits_{n\to \infty}a_{n}$   Do I need to show that it is monotone increasing and then use the relation  $$0\leq \lim_{n\to \infty}a_{n+1}- \lim_{n\to \infty}a_{n}\leq 0 \;?$$  Please, I need help on this! Various proofs are welcome! Thank you very much for your time!
95279
I'm self-studying a bit of complex analysis, and I'm attempting to figure out the following.   Suppose $\lim_{n\to\infty}z_n=A$. How can I show that $$ \lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A. $$  Is there a clever way to write the limit to make it more approachable? Thank you.