Let $\{a_n\}$ be a sequence of real numbers such that $$\lim_{n \to \infty} \mid a_n + 3(\frac{n-2}{n})^n \mid^{\frac1n} = \frac35$$  Then find  $\lim_{n \to \infty} a_n$.  Tried very hard yet not able to crack it. Help Needed.  Anyone able to do the sum??
124275
Let $\{a_n\}$ be a sequence of real numbers such that $$\lim_{n \to \infty}\left|a_n+ 3\left(\frac{n-2}{n}\right)^n \right|^{\frac{1}{n}}=\frac{3}{5}.$$  What is the value of $\lim_{n\to \infty} a_n$?
25796
If $$\left\{x_{n}\right\}\mid x_{1}=5,x_{n+1}=x_{n}^{2}-2,\forall n\geq 1$$ find $$\lim_{n\to\infty}\frac{x_{n+1}}{x_{1}x_{2}\cdots x_{n}}.$$  If someone could help me out with tags, it'd be lovely. I think this is calculus and real-analysis, but I'm not sure--I had the problem scribbled down on a post-it, and I forget where it's from.
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!
64951
Let $(a_n)_{n\in\mathbb{N}}$ be a sequence such that lim inf $|a_n|=0$. Prove that there is a subsequence $(a_{n_{k}})_{k\in\mathbb{N}}$ such that $\sum_{k=1}^\infty a_{n_k}$ converges.  I'm thought about showing that you can make a subsequence that is smaller then $1/n^2$. Is this the right strategy ? Any hints ? Should I use the Cauchy criterion ?
86785
Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence with $\lim\limits_{n\to\infty}x_n = a$, $\{t_n\}_{n\in\mathbb{N}}$ be a sequence with $\lim\limits_{n\to\infty}t_1+t_2+\ldots+t_n = +\infty$.  Prove that  $$\lim_{n\to\infty}\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} = a$$  My idea is use that  $$(t_1+\ldots+t_n)\min\{x_n: n\in\mathbb{N}\}\leq t_1x_1+\ldots+t_nx_n \leq (t_1+\ldots+t_n)\max\{x_n: n\in\mathbb{N}\}$$  But I'm not sure if $\lim\limits_{n\to\infty}\min\{x_n: n\in\mathbb{N}\} = \lim\limits_{n\to\infty}\max\{x_n: n\in\mathbb{N}\} = \lim\limits_{n\to\infty}x_n$.
3816
I have a sequence $(a_n)$ where for each natural number $n$,   $$a_n = (1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})$$ and I want to find its limit as $n\to\infty$.   I obviously couldn't prove it and after several futile attempts decided to post it here.   Here is a list of a few observations which I got from those attempts:   The sequence $(a_n)$ is a strictly increasing sequence. To prove this, I rewrote each element as $$a_n = (1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})= \frac{(n+1)\cdots(n+n)}{n^n}= \frac{(2n)!}{n!n^n}.$$ Then $$\frac{a_{n+1}}{a_n}= \frac{2(n+1)!}{(n+1)!(n+1)^{n+1}}\frac{n!n^n}{(2n)!}=\frac{(2n+1)(2n+2)}{(n+1)^2(1+\frac{1}{n})^n} \to \frac{4}{e}$$ as $n\to\infty$. Since $\frac{4}{e}&gt;1$ we have $a_{n+1}&gt;a_n$ eventually. The limit of this sequence is bounded below by $e$. By replacing $1,2, \ldots, n$ with $1$ in the expression of $a_n$, we get $a_n \geq (1+\frac{1}{n})^n$. And thus $\lim{(a_n)}\geq e$.  $\lim{(a_n)}\geq e^2$ and $\lim{(a_n)}\geq e^3$. The first assertion follows from the fact that $$a_n\geq(1+\frac{1}{n})(1+\frac{2}{n})^{n-1}= \frac{(1+\frac{1}{n})(1+\frac{2}{n})^{n}}{(1+\frac{2}{n})} \to e^2.$$ And the last one follows the same way because $$a_n\geq (1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})^{n-2}.$$   Now I have a gut feeling that for any natural number $k$, one can show that for all sufficiently large natural number $n$, $$a_n\geq (1+\frac{1}{n})\cdots(1+\frac{k-1}{n})(1+\frac{k}{n})^{n-(k-1)}.$$ And therefore for all $k \in \mathbb{N}$, $\lim{(a_n)}\geq e^k$ making the sequence divergent. But I'm really not sure about this approach and I'll appreciate any help towards this end. Thank you.  [Note: As this sequence is quite common, there may be other posts on math.SE asking the same question. I didn't search for them because I just don't know how to search for an expression this big. Though a link related to any previous question concerning this particular sequence will be good enough, I will greatly appreciate if someone takes the trouble to look into my approach/observations and point out where  I'm going wrong.]
82532
I have a sequence defined by $$ a_1=1,\quad a_2=1,\quad a_n=\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\text{ for } n\ge2. $$ Now, if $\lim_{n\to \infty}a_n=g$ then $\lim_{n\to \infty}a_n=\lim_{n\to \infty}\Bigl(\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\Bigr)=\frac{2}{g}$ so $g=\sqrt{2}$ or $g=-\sqrt{2}$ but $a_n&gt;0$ so $g=\sqrt{2}$.  Now, how do I prove that it has an actual  limit? Also, it can be proven that $1\le a_n\le2$, and it's not monotonic because $a_4 \gt a_5 \lt a_6$. Also, it's not monotonic after some $N\in\mathbb N$
2069
Question: Consider a sequence $\{a_{n}\}$ such that $a_{1},a_{2}&gt;0$, and for all $n \in \mathbb{N}$ we have:  $$a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$$  Prove that :  $\displaystyle\lim_{n\to\infty}a_{n}$ exists and find this limit.  My work: If this limit exists, let $\displaystyle\lim_{n\to\infty}a_{n}=x&gt;0$; then we have $$x=\sqrt{x}+\sqrt{x}\Longrightarrow x=4$$ But I can't see how to prove the limit of $\{a_{n}\}$ exists. Thank you for you help!
82879
I have to find $$\lim_{n\to\infty}{(\sqrt[3]{n+1}-\sqrt[3]{n})}$$ here what I did $$\lim_{n\to\infty}{(\sqrt[3]{n+1}-\sqrt[3]{n})}$$=$$\lim_{n\to\infty}{\left(\sqrt[3]{n\left(1+\frac{1}{n}\right)}-\sqrt[3]{n}\right)}$$ How can I proceed?
38619
Problem :   Let $a$ be a positive number. Then $$\lim_{n \to \infty}\left[\frac{1}{a+n}+\frac{1}{2a+n}+\cdots +\frac{1}{na+n}\right]$$  Please suggest how to proceed in such limit problems, will be of great help thanks.
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.
111111
If $ a_0 ,a_1 ,\ldots,a_k  $ are real numbers such that  $$ a_0  + a_1  + \cdots + a_k  = 0$$  Find $$  \lim_{n \to \infty } (a_0 \sqrt n  + a_1 \sqrt {n + 1}  +\cdots + a_k \sqrt {n + k} )$$   I just can't see any solution, can anyone help ?
122212
Let $\{a_n\}$ be a sequence of positive real numbers such that   $a_1 =1,\ \  a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.   Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...  (A) $(1,2]$, (B) $(2,3]$, (C) $(3,4]$, (D)$(4,5]$.   Solution attempt:   Firstly, we figure out what $\frac{a_{n+1}}{a_n}$ is going to look like. We get, from the recursive formula, $\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$ (remembering the fact that $a_n&gt;0$, the other root is rejected).   We know that, if $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}&gt;1$, then $\lim a_n \to \infty$.   Further, $(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}&gt;0$. (Again, the other root is rejected due to the same reason).   Hence, $(a_n)$ increases monotonically.   Therefore, the largest value of $\frac{a_{n+1}}{a_n}$ is approximately $1+\sqrt{1+\frac{1}{1}} \approx 2.15$  Now, the sum can be approximated as $\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$ (In actuality, $\mathbb{sum}&lt; 1.3$).  So, option $(A)$ is the correct answer.   Is the procedure correct?   I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.   Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?
83427
I have been trying to solve this problem for awhile, and I can't seem to solve it.Below is the question. I suspect that the series does converge, but I am unable to prove this. Thanks for any help in advance.  Let $(a_n)_{n=1}^\infty$ be a convergent sequence of real numbers. Can it be concluded that $\sum\limits_{n=1}^{\infty}|a_{n+1} - a_{n}|$ converges? Give a proof or a counterexample.
13687
I can show that the following limit exists but I am having difficulties to find it. It is $$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^n}{n^n}$$ Can someone please help me?
81542
I'm doing exercises. In the related book, there is a claim. Is this right? I'm not sure.     For a sequence $\{a_n\}$, there exists a limit $a$ such that $\lim_{n\to\infty} a_n=a$ if and only if for any $p\in \Bbb N$, $\lim_{n\to\infty} |a_{n+p}-a_n|=0$   If not, could you kindly give some counterexamples?  Thanks.
88487
Let we have the following sequence $(a_n)$ such that $$a_n=\sum_{k=1}^n\frac{1}{n+k}$$ How can I prove that $(a_n)$ is increasing bounded sequence, then prove it is convergent and find its limit?