Let $(U_n)$ be a sequence such as $|u_{n+1}-u_n|&lt;\frac{1}{n^2}$ show that (U_n) is a convergent sequence.  I m stuck with this question, I showed that $\lim\limits_{n \to \inf} u_{n+1}-u_n =0$ Can you help me ?
122547
Let ${x_n}$ be a sequence in $R$ such that $|x_{n+1} - x_n|&lt; \frac{1}{n^2}$ for all $n \in N$.  Show that the sequence is convergent.  If it were $|x_{n+1} - x_n|= \frac{1}{n^2}$, could take help of the fact $\sum \frac{1}{n^2}$ is convergent and of triangular inequality.  But what to do here? Please help. Thanks in advance.
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!
52022
This problem is form the Romanian GM.  Let $(x_{n})_{n\geq1}$ be a sequence of real numbers defined by $x_{1}=1$ and $$x_{n+1}=x_{n}+ \frac{\sqrt{n}}{x_{n}}$$ for all $n \geq 1$. Study the convergence of the sequence $$(\frac{x_{n}}{n^{\frac{3}{4}}})_{n\geq1}$$  I only managed to show from the given relation that $(x_{n})_{n\geq1}$ is strictly increasing and tends to $+\infty$.
39500
Let $$ a_{n+1} =  \frac{1}{k} \left(a_{n} + \frac{k}{a_n}\right) ; k&gt;1, a_1&gt;0 $$ The problem is to show that it converges.  Attempt: The sequence is not monotone but it has a lower bound. It seems that odd terms subsequence and even term subsequence are monotonic sequences (I wrote some basic code to make this observation) though I am not able to prove it analytically. I also know that if odd subsequence and even subsequence converge to same limit then the sequence also converges. So, that tells me that I am on the right track.  Please provide any hints.
48063
I have a question in Analysis. Knowing that $x_{2n}$, $x_{2n-1}$, $x_{3n}$ converge, how can I show that $x_{n}$ converges?
16407
How do I prove that if $|a_{n+1}|/|a_n| \leq |b_{n+1}|/|{b_n}|$ and $\sum |b_n|$ is convergent, then $\sum |a_n|$ is convergent? I know that if I can get $|b_{n+1}/{b_n}| &lt; 1$ then I can use the ratio test to say $\sum a_n$ is convergent, but I don't know if this the proper way to approach the problem.
90392
If $x_1$, $x_2$ are arbitary real numbers, $\alpha\in (0,1]$ and  $x_n=\alpha x_{n-1}+(1-\alpha)x_{n-2}$ for every positive integer $n$ (>2), show that the sequence $\{x_n\}$ is convergent. (Given $x_1&lt;x_2$)  I cannot rearrange to test convergency. Please help me by solving the problem.   Edit  I want to prove by using the property below:     If $\{x_{2n}\}$ and $\{x_{2n-1}\}$ converges to same limit $l$ then $\{x_{n}\}$ converges to $l$.
59518
If I have a sequence {$a_n$} that has the property of $\lim(a_{n+1}-a_n)=0$, does that make it a Cauchy Sequence. I think it doesn't because $a_n = \sum_{k=1}^n \frac{1}{k}$ is a counter example.   However, by definition, there exists a $M$ such that if $n \geq M$ then $|a_{n+1}-a_n| &lt; \frac{\epsilon}{m-n}$  Hence, we have $|a_m - a_{m-1}|+.....+|a_{n+1}-a_n|&lt;|a_m -a_n| &lt;\epsilon$  This proof doesn't work because I cannot be sure I can find a fixed $M$ which might change according to n.  But I wonder if I have an additional condition that says $a_n$ is bounded, I think then the proof works and that I should be able to find a fixed $M$. However, I don't know how to justify this. Maybe I am wrong. Can someone kindly help me figure out 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 ?
22786
Let $a_n$ be a sequence such that: $ a_{n+1}-a_n  \ge \frac{1}{n}$. prove that $\lim_\limits{n \to \infty} a_n = \infty$.  SOLUTION: I want to prove this by disproving Cauchy's test. which means, I want to prove that:  There is an $\epsilon&gt;0$ , such that for each $N$, there is $n&gt;N$ and $p \in$  $\mathbb N$ such that:$|a_{n+p}-a_n| \ge \epsilon$.  but for some reason I couldn't find that $\epsilon$. the "$\frac{1}{n}$" made my job harder.   To conclude: I want to disprove that $a_n$ converges by disproving that its Cauchy's and from $ a_{n+1}-a_n  \ge \frac{1}{n}$ we get that the sequence is monotonic increasing, thus, this will prove that: $\lim_\limits{n \to \infty} a_n = \infty$.  Am I on the right track? are all the conclusions that I've reached are right? How do I really complete this? any kind of help would be appreciated.
108813
I have the following sequences $(a_n)$, $(b_n)=(a_{2n})$, and $(c_n) = (a_{2n-1})$.  I have to show that:   if $(b_n)$ and $(c_n)$ both convergence to $a$, then $(a_n)$ also converges to $a$; if $(a_n)$ converges then $(b_n)$ and $(c_n)$ also converge; there exist $(a_n)$ not converging but with $(b_n)$ and $(c_n)$ converging.   I am not sure how to solve this since I only know that $(b_n)$ > $(c_n)$ > $(a_n)$ on the terms how the sequences are defined.
100616
$f_n(x) = x - x^n$ for $x\in [0,1]$. Does the sequence converge pointwise or uniformly on $[0,1]$?   We also have to prove why it is or isn't pointwise or uniformly convergent.   I have worked out that on $x\in[0,1)$, the limit as $n$ approaches infinity is $x$. And when $x=1$ the limit is $0$.   So this means its pointwise convergent right? But how do I prove this?  Also how on earth do I prove it is not uniformly convergent? (I'm just assuming it's not).
88350
$(z_n)_{n\in \mathbb{N}}$ is a complex valued sequence such that subsequences $(z_{2n})_{n\in \mathbb{N}}$ and $(z_{2n+1})_{n\in \mathbb{N}}$ converges to same limit.  How could I show that the sequence $(z_n)_{n\in \mathbb{N}}$ converges to the same limit using $\epsilon$, $N$ definition?    For every $\epsilon &gt;0$,  $|z_{2n}-l|&lt;\epsilon$ whenever $n&gt;N_1$, and   $|z_{2n+1}-l|&lt;\epsilon$ whenever $n&gt;N_2$.  We want to show that for every $\epsilon &gt;0$, there exists $N$ such that $|z_{n}-l|&lt;\epsilon$ whenever $n&gt;N$. I tried something like  $|z_{n}-l|\le|z_n-z_{2n}|+|z_{2n}-l|$ but wasn't sure what to do next
64925
Let $\{a_n\}$ be a sequence of real number, let $\{n_k\}$ be an increasing sequence of positive integer and let $A_n=a_{n_k}+a_{n_{k+1}}+...+a_{n_{k+1}-1}$ for $k \in N$. Prove that if $\sum_{n=1}^\infty a_n$ converges then $\sum_{k=1}^\infty A_k$ converges. Is the converse true?  I know that $\sum_{n=1}^\infty a_n$ converges implies that $lim _{n_&gt;\infty} a_n =0$, can I conclude that $A_n$ ->$0$ as well, even if I can conclude that, it still doesn't prove that $\sum_{k=1}^\infty A_k$ converges, am I correct?
99697
I want to prove that the sequence defined by $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ has a limit.  By evaluating the sequence I notice that the sequence is strictly monotonically decreasing starting from $x_2=1.5$.  It seems to suggest itself to prove that the sequenced is bounded by $1\le\big( x_n \big)_{n\ge1} \le 1.5$ and to prove that it is strictly monotonically decreasing starting at $x_2$ which would imply convergence.  How would I proceed and could one prove the existence of the limit without first evaluating the values of the sequence to see how the sequence behaves?
63598
If $a_n$ is a sequence such that   $$a_1 \leq a_2 \leq a_3 \leq \dotsb$$   and has the property that $a_{n+1}-a_n \to 0$, then can we conclude that $a_n$ is convergent?   I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in  to a similar question that does not require the sequence to be increasing.  $$0, 1, \frac12, 0, \frac13, \frac23, 1, \frac34, \frac12, \frac14, 0, \frac15, \frac25, \frac35, \frac45, 1, \dotsc$$  This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $\pm\frac1m$ and $m$ increases the further we go in this sequence.    So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge?  Or is this still not sufficient?
1309
Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.   I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$  I am trying to go ahead of this step.