Show that the sequence  $\left(\frac{2^n}{n!}\right)$ has a limit.  I initially inferred that the question required me to use the definition of the limit of a sequence because a sequence is convergent if it has a limit $$\left|\frac{2^n}{n!} - L \right|{}{}&lt;{}{}\epsilon$$   I've come across approaches that use the squeeze theorem but I'm not sure whether its applicable to my question. While I have found answers on this site to similar questions containing the sequence, they all assume the limit is $0$.  I think I need to show $a_n \geq a_{n+1},\forall n \geq 1$, so  $$a_{n+1} = \frac{2^{n+1}}{(n+1)!}=\frac{2}{n+1}\frac{2^{n}}{n!}&lt;a_n$$  A monotonic decreasing sequence is convergent and this particular sequence is bounded below by zero since its terms are postive. I'm not sure whether or not I need to do more to answer the question.
1447
Why is     $$\lim_{n \to \infty} \frac{2^n}{n!}=0\text{ ?}$$   Can we generalize it to any exponent $x \in \Bbb R$? This is to say, is     $$\lim_{n \to \infty} \frac{x^n}{n!}=0\text{ ?}$$     This is being repurposed in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: .
7590
I am trying to check whether or not the sequence $$a_{n} =\left\{\frac{n^n}{n!}\right\}_{n=1}^{\infty}$$ is bounded, convergent and ultimately monotonic (there exists an $N$ such that for all $n\geq N$ the sequence is monotonically increasing or decreasing). However, I'm having a lot of trouble finding a solution that sufficiently satisfies me.  My best argument so far is as follows,  $$a_{n} = \frac{n\cdot n\cdot n\cdot \ldots\cdot n}{n(n-1)(n-2)(n-3)\dots(2)(1)} = \frac{n}{n}\cdot \frac{n}{n-1}\cdot  \ldots \cdot \frac{n}2\cdot n$$  so $\lim a_{n}\rightarrow \infty$ since $n&lt;a_{n}$ for all $n&gt;1$. Since the sequence is divergent, it follows that the function must be ultimately monotonic.  This feels a little dubious to me, I feel like I can form a much better argument than that, or at the very least a more elegant one. I've tried to assume $\{a_{n}\}$ approaches some limit $L$ so there exists some $N$ such that  $|a_{n} - L| &lt; \epsilon$ whenever $n&gt;N$ and derive a contradiction, but this approach got me nowhere.  Finally, I've also tried to use the fact that $\frac{a_{n+1}}{a_n}\rightarrow e$ to help me, but I couldn't find an argument where that fact would be useful.
66881
First the textbook proves that all Cauchy sequences are bounded, and so have a convergent subsequence, $\{a_{n_{k}}\}$ that converges to a limit, say $L$. Now we use this to prove that all Cauchy sequences are convergent.  So an $N_1$ exists such that $$\left|a_{n_{k}}-L\right|&lt;\frac{\epsilon}{2}$$ for all $ k &gt; N_1$, and an $N_2$ exists such that $$\left|a_m-a_n\right|&lt;\frac{\epsilon}{2}$$ for all $n,m &gt; N_2 $.  Pick any $k &gt; N_1$ such that $n_k &gt; N_2$. Then for every $n &gt; N_2$, $$\left| a_n - L \right | \leq \left| a_n - a_{n_{k}} \right| + \left| a_{n_{k}} - L \right| &lt; \epsilon/2 + \epsilon/2 = \epsilon$$.  So $$\lim_{n \rightarrow \infty}a_n = L$$  I'm fine with this proof until the last part - I'm confused as to why we can pick an abitrary $k$ like we do? Should the limit not depend only on $n$? Now it appears like it also depends on $k$ and if we pick a $k &lt; N_1$, the inequality isn't true. Can someone clarify this for me please?
5779
I want to calculate the limit of: $$\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $$ or prove that it does not exist. Now I know the result is $4$, but I am having trouble getting to it. Any ideas would be greatly appreciated.
15864
A sequence $a_n$ is defined as $a_1=1$ and $a_{n+1}=\sqrt{a_1+a_2+\cdots+a_n}$ .Prove that $ \lim\limits_{n \rightarrow \infty} \frac{a_n}{n}=\frac{1}{2}$  I have no idea how to approach this. But I have a feeling that Cesaro's lemma may come in handy
108257
Find the limit of the sequence $a_n=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}.$  The limit is supposed to be $e^{-1}$ and there are a couple of posts in MSE proving it. But here is a proof I encountered showing the limit is $1.$ Please tell me what's going wrong here.      Proof:  We can write $a_n=\sqrt[n]{n!}(b_n-1),$ where $b_n=\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}.$ Hence $$a_n=n\cdot\frac{\sqrt[n]{n!}}{n}\cdot\frac{b_n-1}{\ln{b_n}}=\frac{\sqrt[n]{n!}}{n}\cdot\frac{b_n-1}{\ln{b_n}}\cdot\ln{b_n^n.}$$   But $\lim \limits_{n\to \infty}\frac{\sqrt[n]{n!}}{n}=e^{-1},$ so $b_n\to1$ as $n\to \infty.$ On the other hand, $$\lim \limits_{n\to \infty}b_n^n=\lim \limits_{n\to \infty}\frac{(n+1)!}{n!}.\frac{1}{\sqrt[n+1]{(n+1)!}}=e.$$   Also $$\lim \limits_{n\to \infty}\frac{b_n-1}{\ln{b_n}}=\lim \limits_{n\to \infty}\frac{1}{\ln{[1+(b_n-1)]^{1/(b_n-1)}}}=\frac{1}{\ln e}=1$$         And the author goes on to say that the limit is $e^{-1}.$ But doesn't it say that the limit is $1?$ I must have overlooked something so obvious.
67395
I do not know how to go about finding the limit of this sequence. I know it diverges to infinity yet I can't find terms to use the squeeze lemma effectively.  $a_n = \frac{n!}{n^{1000}}$    $\lim_{n\rightarrow\infty} \frac{n!}{n^{1000}}$  I know that when $n&gt;1000$, the numerator will "run out" of denominators and the expression will grow, yet I don't know how to formally prove this divergence.  In addition to this, since it is a sequence, no series laws can be applied.   Anyone have an idea on how to approach this type of problem?
25169
I'm studying for qualifying exams and ran into this problem.  Show that if $\{a_n\}$ is a nonincreasing sequence of positive real numbers such that $\sum_n a_n$ converges, then $\lim_{n \rightarrow \infty} n a_n = 0$.  Using the definition of the limit, this is equivalent to showing \begin{equation} \forall \varepsilon &gt; 0 \; \exists n_0 \text{ such that } |n a_n| &lt; \varepsilon \; \forall n &gt; n_0 \end{equation} or \begin{equation} \forall \varepsilon &gt; 0 \; \exists n_0 \text{ such that } a_n &lt; \frac{\varepsilon}{n} \; \forall n &gt; n_0 \end{equation} Basically, the terms must be bounded by the harmonic series. Thanks, I'm really stuck on this seemingly simple problem!
69812
I'm asked to prove the convergence of the sequence $$X_n=\left(1+\frac12\right)\left(1+\frac14\right)\left(1+\frac18\right)\cdots\left(1+\frac{1}{2^n}\right)$$   I proved that it is increasing through the ratio test and then I want to prove that it is bounded.  My question is the following : is it enough to say that $X_n&lt;\left(\frac32\right)^n$ for $n&gt;1$ or not ? Namely, can we say that a sequence is bounded if $\forall$  $n$ ,  $\exists$ a $N$ such that $X_n&lt;N$ or do I have to find a N such that $\forall$  $n$ , $X_n&lt;N$?  In other words, can a bound depend on $n$ ?  Thanks for your help.
7641
The problem states:  a sequence of real numbers $a_0, a_1, \dots $ is defined as follows. $a_0$ is an arbitrary real number and for $n \ge 0, a_{n +1} = \lfloor a_n \rfloor \{ a_n \}$ (where $\{x\} = x - \lfloor x \rfloor$).  Prove that for $a_n = a_{n+2}$ for $n$ large enough.  my attempt:  I notice that this is a decreasing sequence because $$\frac{a_{n+2}}{a_{n+1}} = \frac{\lfloor a_{n+1} \rfloor \{ a_{n+1} \}}{ a_{n+1}} &lt; 1$$  since $\{ a_{n+1} \} \le 1$ (we can exclude the case where it's equal to one since then it becomes the null sequence that satisfies the thesis).  Moreover this sequence is bounded between $0$ and $a_0$ so by the monotone convergence theorem it must have a limit.  Taking the limit I obtain $$\ell = \lim \lfloor a_n \rfloor (\ell - \lim \lfloor  a_n \rfloor) \iff \ell = - \lim \lfloor a_n \rfloor^2 / (1 - \lim \lfloor  a_n \rfloor) $$  And so if I prove that there exists $n$ s.t. $ \lfloor a_n \rfloor &lt; 1$ I am done. Because then the limit could only be $0$ and this implies that there exist infinite $a_n = a_{n+2} = 0$.  But I can't seem to prove it. Is my reasoning up to now ok? How could I proceed?  Edit:  Add original text of the problem.
59709
How do I prove that the limit of the sequence exists: $$a_n = \frac1n +\frac 1{n+1} + \frac1{n+2} + \dots + \frac1{2n}$$  I can find the limit as n goes to infinity, $\frac1{2n}$ goes to $0$, so the limit is $0$ right? But how do I prove that the limit exists? Do I have to use the $\varepsilon$ $\delta$ definition of a limit?    For part b) I have to show that the limit is less then $1$ but not less then $\frac12$, I thought the limit was $0$?
99093
There's a bounded sequence $\{{a_n}\} $ s.t. $\lim\limits_{n \rightarrow  \infty}(a_{2n} + 2a_n) = 0$. Try to prove $\lim\limits_{n \rightarrow  \infty}a_n = 0$ (You must prove the existence of its limit before   getting the exact value)   The above problem is located in the chapter about "limsup and liminf", so you may use their property while proving.  EDIT  By the way, I find it hard to do the exercises at the end of each section of my textbook though I DO UNDERSTAND almost everything in that section. So I'm eager to know some other books that can help me to solve problems like the one described above. That would be a great help, thanks!
107313
Let $x_1=1$ and $$x_{n+1} = \frac12\left(x_n + \frac2{x_n}\right).$$ Prove or disprove $(x_n)$ is convergent and show the limit.   When I tried working on it I found the sequence was bounded by square root of 2 and it is was monotone. But apparently the sequence is not bounded by square root of two and is not monotone. But I have no idea why. Any help would be greatly appreciated! Thanks!
68075
Problem: Find with proof limit of the sequence $a_n=1+\frac{1}{a_{n-1}}$ with  $a_1=1$  or show that the limit does not exist.  My attempt:  I have failed to determine the existence. However if the limit exists then it is easy to find it.  The sequence is not  monotonic and I have failed to find any monotonic subsequence subsequence. The sequence is clearly bounded below by $1$. I have observed that the sequence is a  continued fraction so it alternatively increases and decreases.   So, please help me.
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$.
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!
32999
Let $a_n$ be the sequence in $\mathbb{R}$ given by:  $$ a_1 = \sqrt{2}, \; \; a_{n+1} =\sqrt{2+ \sqrt{a_n} }. $$  Show that this sequence converges and find the limit.  I have already proven that this sequence is increasing and bounded by $2$, so the limit exists. Now, when I'm trying to find the limit, say  $L$, I get that $L$ satisfies the equation: $$L^2 -2 - \sqrt{L}=0, $$ but I don't actually know how to solve this equation, so I would like to know if there is an  easier way to find the limit, and in case I must solve the last equation, how can I solve it?  Thanks in advance!