For each natural number $n$, let $$ x_n \colon= \frac{1}{\sqrt{n}} \left( \frac{1}{\sqrt{1}}  + \frac{1}{\sqrt{2}}+ \cdots + \frac{1}{\sqrt{n}} \right). $$  Then is the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ convergent? And if so, then how to find the limit of this sequence?   I'm just not sure how to proceed.
106235
Evaluate the limit : $$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)$$  I can use the sandwich principle, certain convergence criteria, Cesaro mean theorem, limit arithmetic.. things around this area.  Any help would be greatly appreciated, thanks!  Sorry for not elaborating more at the beginning, rookie first-post mistake I suppose. :)
104844
Let $y_n$ be a sequence of real numbers. Prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges.  Let us suppose that $y_{n}$ goes to a limit $L$. Then for all $\varepsilon &gt;0$, for sufficiently large $n$, $|y_n -L| &lt; \varepsilon$ .  But $|y_n - L| &lt; 2|x_n-\frac{L}{3}|+|x_n - \frac{L}{3}|$  But then how to proceed?
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.
3190
I am given this problem: let $a\ge0$,$b\ge0$, and the sequences $a_n$ and $b_n$ are defined in this way: $a_0:=a$, $b_0:=b$ and $a_{n+1}:= \sqrt{a_nb_n}$ and $b_{n+1}:=\frac{1}{2}(a_n+b_n)$ for all $n\in\Bbb{N}$    To prove is that both sequences converge and that they have the same limit. I don't know how to show this. I have spent 2 hours on this, no sign of success
72225
I have to compute the limit of this sequence ${\textstyle\sum_{k=1}^n}\frac1{\sqrt{n^2+k}}$ as $n\rightarrow\infty$. First I was thinking about some Riemann sum and and forced the $n^{2}$ outside the square root but the function was not so pleasant.
42804
\begin{align}f(x) &amp;= \dfrac{e^{1/x}}{x^2}\\ x_{n+1} &amp;= f\left( \dfrac{1}{x_n}\right)\end{align} Show that $x_n$ is convergent and that its limit is $0$.    It is very easy to find the limit from the recurrence relation. Just plugging $l$ instead of the terms of the sequence. But how do I show it is convergent?   I tried to evaluate the difference $x_{n+1}-x_n$ but its a messy calculation... Also I tried using the function monotony. But I don't see how it tells me something about the function
73834
Find the limit of: $$\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$$   I'm pretty sure it goes to zero since $(n+1)^n &gt; n^n$ but when I input large numbers it goes to $0.36$.  Also, when factoring: $$n^{1/n}\left(\frac{1}{1+\frac1n}\right)^n$$ it looks like it goes to $1$.   So how can I find this limit?
31497
Given a sequence $\{x_n\}, \ n\in\Bbb N$: $$ x_n = \frac{(2n)!!}{(2n+1)!!} $$ Show that $x_n$ converges.  I'm wondering why I'm getting a seemingly wrong result (assuming the problem statement asks to prove convergence): $$ \begin{align} x_n &amp;= \frac{(2n)!!}{(2n+1)!!} \\  &amp;= \frac{2\cdot 4\cdot 6\cdots (2n-2)\cdot(2n)}{3\cdot 5\cdot 7\cdots (2n-1)\cdot(2n+1) } \\ &amp;= \frac{2\cdot 4\cdot 6\cdots (2n-2)\cdot(2n)}{3\cdot 5\cdot 7\cdots (2n-1)\cdot(2n+1)} \cdot \frac{2^n\cdot n!}{2^n\cdot n!} \\ &amp;= \frac{4^n (n!)^2}{(2n+1)!} \\  &amp;= \frac{4^n (n!)^2}{(2n+1)\cdot (2n)!} \\ &amp;=\frac{4^n}{2n+1}\cdot \frac{(n!)^2}{(2n)!} \end{align} $$ By Binomial coefficients: $$ {2n\choose n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{(n!)^2}  $$ Thus: $$ x_n = \frac{4^n}{2n+1}\cdot \frac{1}{{2n \choose n}} $$ Doesn't $\frac{4^n}{2n+1}$ grow faster than $\frac{1}{2n\choose n}$ is declining? Shouldn't $x_n$ diverge in that case?
93101
Find the limit of the sequence {$a_{n}$}, given by$$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=\dfrac {1}{3}(1+a_{n}+a^{3}_{n-1}), \ for \ n \ &gt; \ 1$$  My try:  $ a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54$ that is the sequence is incresing and each term is positive. Let the limit of the sequence be $x$. Then $ \lim _{n\rightarrow \infty }a_{n+1}=\lim _{n\rightarrow \infty }a_{n}=x$ $$ \lim _{n\rightarrow \infty }a_{n+1}= \lim _{n\rightarrow \infty }1+a_{n}+a^{3}_{n-1}$$  $\Rightarrow x=\dfrac {1}{3}( 1+x+x^3)$  $\Rightarrow x^3-2x+1=0$  and this equation has three roots $x=\dfrac {-1\pm \sqrt {5}}{2},1$   So the limit of the sequence is $\dfrac {-1 + \sqrt {5}}{2}$.  how can i say that the limit is $\dfrac {-1 + \sqrt {5}}{2}$?
15457
Let $a_1=1$ define $a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n})$ . Show that the sequence converges to $\sqrt{2}$. The idea is to show the sequence is bounded and monotone . But how should I do it?
87031
Let $(x_n)_n$ be a sequence of real numbers such that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0.$$ Show that $(x_n)_n$ must also converge to zero.     $\textbf{My Attempt:}$   My original idea was to use the triangle inequality and the fact that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0$$   to show that the sequence $(x_n)_n$ is Cauchy and hence converges to some real number $a$, which would give me $$\lim\limits_{n\rightarrow \infty}x_n=a=\lim\limits_{n\rightarrow \infty}x_{n+1}\implies a-\dfrac a 2= \lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0\implies a=0.$$ However, I'm having trouble showing that $(x_n)_n$ is Cauchy.   Any suggestions/hints would be greatly appreciated. Thanks!
65779
For the sequence 1/(n-1): x1 is undefined. I think that makes the sequence unbounded. However I can prove from definition that the sequence is convergent. And there is the theorem that all convergent sequences are bounded. So what am I missing?
116515
Is this a valid sequence: $$\left\{\frac1{(n-3)}\right\}?$$ That is, can a sequence have individual terms that are undefined? If so, does this mean that the above sequence is unbounded (since the third term is not smaller than any real number)? [What I do know is that if the above sequence were valid, it would be convergent (by definition), which would in turn mean that it is bounded (by the Boundedness Theorem).]  Subsequent to the main discussion here, I have just come across an old exam question that asks us to show that $\langle\tan(\frac{\sqrt n\pi}4):n∈N\rangle$ is divergent. But this sequence is undefined at say, $n=4$, so like the one above, it isn't even a valid sequence, is it? (I've used a different notation for sequences on Brian M Scott's advice.)
110106
x is a real number between 0 and 1  $ \lim_{n\rightarrow\infty} x ^ {\left(\frac 1 n\right)} = 1 $  SO far i know you need to do this to prove it:  let b>1, prove that $\root n \of b \rightarrow 1$ as $n \rightarrow\infty$  I was given this hint: let $a_n = \root n \of b - 1$ , prove that  $a_n \rightarrow 0$  Thanks!
70203
I need to check if $$\lim_{n \to \infty} \frac{n}{ \sqrt [n]{n!}}$$ converges or not. Additionally, I wanted to show that the sequence is monotonically increasing in n and so limit exists. Any help is appreciated. I had tried taking log and manipulating the sequence but I could not prove monotonicity this way.
20572
Here's Prob. 3, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:      If $s_1 = \sqrt{2}$, and $$s_{n+1} = \sqrt{2 + \sqrt{s_n}} \ \ (n = 1, 2, 3, \ldots),$$ prove that $\left\{ s_n \right\}$ converges, and that $s_n &lt; 2$ for $n = 1, 2, 3, \ldots$.    My effort:   We can show that $\sqrt{2} \leq s_n \leq 2$ for all $n = 1, 2, 3, \ldots$. [Am I right?]  Then we can also show that $s_n &lt; s_{n+1}$ for all $n = 1, 2, 3, \ldots$. [Am I right?]  But how to calculate the exact value of the limit? Where does this sequence occur in applications?
72368
Is the sequence $\Big(\dfrac {\tan n}{n}\Big) $ convergent ? If not convergent , is it properly divergent i.e. tends to   either $+\infty$ or $-\infty$ ? ( Owing to $\tan (n+1)= \dfrac {\tan n + \tan 1}{1- \tan1 \tan n}$ and the non-covergence of   $\Big (\tan n \Big)$ it is easy to see that if $\Big(\dfrac {\tan n}{n}\Big) $ is convergent then the limit must be  $0$ , but I can not proceed further . )