I need to find where this sequence tends to: $$\frac{1}{n} \sqrt[n]{n(n+1)(n+2)...(2n)}$$  My answer is $2$, by using a trick for the $n^{th}$ root: I take the inside and find where $\dfrac{A_{n+1}}{A_n}$ goes, and this is where the original sequence goes.
31566
I'd like a hint to show that:  $$\lim \frac{1}{n} \sqrt[n]{(n+1)(n+2) \cdots 2n} = \frac{4}{e} .$$  Thanks.
78161
I'm trying to find the limit of $a_n = \left(1-\frac{1}{n^2}\right)^n$ for $n \rightarrow \infty$.  It seems that the limit is $1$, since $a_n = 0.999...$ for large $n$. The presentation $a_n = \frac{(n^2-1)^n}{n^{2n}}$ and expanding was my first idea, but I couldn't get the result from there. Any ideas?
93786
Let $ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}$ . I would like to know whether the given sequence converges.   I see  that,  $ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}= \sum_{k=1}^{n} \frac{1}{n+\frac{k}{n}}.$ When $n$ gets sufficiently large the contribution by the $ \frac{k}{n} $ term is diminishing and $ a_{n} &lt; \sum_{k=1}^{n} \frac{1}{n} = 1 $.   Thank you.
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.
95156
Determine the following limit $$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)\ldots (n+n)} $$ Idea #1 the nth root could be rewritten as $$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{\frac{(n+n)!}{n!}} $$ Noticing that $n^{-1}$ is an infinitesimal, if the n-th root were bounded, the limit would be $0$.  But hoping that $\sqrt[n]{\frac{(n+n)!}{n!}}$ is bounded is a pipe dream, this clearly tends to $\infty$.  Idea #2 $$\lim_{n\to\infty}\ln\left (\frac{1}{n}[(n+1)\ldots (n+n)]^\frac{1}{n}\right ) = \lim_{n\to\infty}\left [-\ln n+\frac{1}{n}\left (\ln (n+1)+\ln (n+2)+\ldots +\ln (n+n) \right )\right ]$$ but trying to bring it to a common denominator doesn't help either, what I would have is $$\lim_{n\to\infty}\frac{1}{n} \left [-n\ln n+\ln (n+1)+\ldots +\ln 2+\ln n\right ]$$ Which doesn't bring me any closer to simplification. The solution of this must something entirely different. I would like a hint/point me in the direction I should go.
88374
It is known that when $n\rightarrow\infty$ the sequence $$\frac{n}{\sqrt[n]{(n!)}}$$ has limit $e$ but I don't know how to prove it's monotonicity. After a short calculus using WolframAlpha I found that this sequence is actually increasing. I tried to compare $2$ consecutive members but I couldn't manage to show something. It is obvious that $$\frac{n^n}{n!}$$ is increasing but that don't help us much (I think)
19214
Let $\{x_n\}_{n=1}^{\infty}$ be a convergent sequence in a  $(S, ||\cdot ||)$. Find the limit of the sequence \begin{align} \left\{\frac{x_1 + 2x_2 + \cdots + nx_n}{n^2}\right\}_{n=1}^{\infty} . \end{align}  Could you please give me some hints where to start?
50312
I'm given two sequences:  $$a_{n+1}=\frac{1+a_n+a_nb_n}{b_n},b_{n+1}=\frac{1+b_n+a_nb_n}{a_n}$$  as well as an initial condition $a_1=1$, $b_1=2$, and am told to find: $\displaystyle \lim_{n\to\infty}{a_n}$.  Given that I'm not even sure how to approach this problem, I tried anyway. I substituted $b_{n-1}$ for $b_n$ to begin the search for a pattern. This eventually reduced to:  $$a_{n+1}=\frac{a_{n-1}(a_n+1)+a_n(1+b_{n-1}+a_{n-1}b_{n-1})}{1+b_{n-1}+a_{n-1}b_{n-1}}$$  Seeing no pattern, I did the same once more:  $$a_{n+1}=\frac{a_{n-2}a_{n-1}(a_n+1)+a_n\left(a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})\right)}{a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})}$$  While this equation is atrocious, it actually reveals somewhat of a pattern. I can sort of see one emerging - though I'm unsure how I would actually express that. My goal here is generally to find a closed form for the $a_n$ equation, then take the limit of it.  How should I approach this problem? I'm totally lost as is. Any pointers would be very much appreciated!  Edit:  While there is a way to prove that $\displaystyle\lim_{n\to\infty}{a_n}=5$ using $\displaystyle f(x)=\frac{1}{x-1}$, I'm still looking for a way to find the absolute form of the limit, $\displaystyle\frac{1+2a+ab}{b-a}$.
23637
I am trying to find out if the following sequence $(a_n)_{n\geq 1}$ will either converge or diverge. $$ a_n = (-1)^n + \frac1n $$  If the limit converges, I need to find its limit.   I tried plugging in various numbers for n besides 1, but I am not entirely sure if it really does converge.
23624
Let $ a_{n+1} = \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}), a_1 &gt; 0 $. Show the sequence converge and find it's limit.    I did the following:  We see that $a_n$ is positive for every n obviously (considering $a_1 &gt; 0$). Then:  $$ a_{n+1} - a_n = \frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = \frac{5 - a_n^{3}}{3a_n^{2}}$$  We see that for each $a_n &gt; ^3\sqrt{5}$ the sequence is decreasing. That means the proof collapses to prove  $a_n &gt; ^3\sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n &gt; ^3\sqrt{5}$ then obviously $a_{n+1} = \frac{2a_n^{3} + 5}{3a_n^{2}} &gt; \frac{2(\sqrt[3]{5})^{3} + 5}{3(\sqrt[3]{5})^{2}} = \sqrt[3]{5}$ so we deduced $a_{n+1} &gt; ^\sqrt[3]{5} $.  (It's important to note $a_1$ might be less then $\sqrt[3]{5}$, hence the sequence will decrease from the second element.) So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:   $$\begin{align} L = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}) = \frac{1}{3}(2L + \frac{5}{L^{2}}) =  \sqrt[3]{5}\end{align}$$  I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $\sqrt[3]{5}$, or rather in this case $(\sqrt[3]{5}, \sqrt[3]{5}+\epsilon)$.  Is it correct?  EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $\sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it.
7032
I've spent the better part of this day trying to show from first principles that this sequence tends to 1. Could anyone give me an idea of how I can approach this problem?  $$ \lim_{n \to +\infty} n^{\frac{1}{n}} $$
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 $n\ge N \implies \left|\frac{1}{2n} - 1\right| = |\frac{1-2n}{2n}|$
84155
I need to show that $\sqrt{2}$, $\sqrt{2+\sqrt{2}}$, $\sqrt{2+\sqrt{2+\sqrt{2}}},\ldots$ converges and find the limit.   I started by defining the sequence by $x_1=\sqrt{2}$ and then $x_{n+1}=\sqrt{2+x_n}$. Then I proved by induction that the sequence is increasing and that it is bounded. Then I use the Monotone Convergence Theorem to prove that it converges.  Now, I claim that the limit of the sequence is $2$. So I need to show that for all $\epsilon&gt;0$, there exists $N$ such that for all $n\geq N$ we have $|x_n-2| &lt; \epsilon$. This is kind of where I'm stuck. I don't know how to proceed to prove the convergence to $2$.
82765
My problem is how to prove  $$\frac{1+2}{1}  &gt; \frac{1+3+\frac{3^2}{2!}}{1+2} &gt; \frac{1+4+\frac{4^2}{2!}+\frac{4^3}{3!}}{1+3+\frac{3^2}{2!}}&gt;\cdots $$  i am sure that the sequence is decreasing as limit exp(1).  However, the reason why the sequence is decreasing is not evident.  Thanks if you consider this issue and make me relief.
97685
Consider the following sequence:     $a_{n+2} = \frac{a_{n+1}+a_{n}}{2}$, $a_{1}$ and $a_{2}$ are given.   Write $a_{n}$ as a function of $a_{1}$ and $a_{2}$ and show that its limit is $\frac{a1 + 2a_{2}}{3}$  I think I am loosing myself on algebra here. I can't even do the first part. Any help is welcome, thanks a lot!
115748
The sequence is defined: $$u_1=1, u_{n+1}=1+\dfrac n {u_n}$$  The question asks to find an asymptotic development of 2 terms for $n\to+\infty$. I have got $u_n\sim_\infty\sqrt{n} $, but how to derive the 2nd term?
545
Let $s_n$ be a sequence defined as given below for $n \geq 1$. Then find out $\lim\limits_{n \to \infty} s_n$.   \begin{align}   s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx  \end{align}  I have written a solution of my own, but I would like to know it is completely correct, and also I would like if people post more alternative solutions.