Let $f:[0,1]\rightarrow [0,1]$ be a continuous function and $x_1\in [0,1]$. If $x_{n+1}= \frac{\sum_{i=1}^n f(x_i)}{n}$ then prove or disprove that $\{x_n\}$ is convergent. I have proved  If $f$ is constant then the limit exists.  $\{(n-1)x_n\}$ is monotonic.  By compactness $\{x_n\}$ has limit points   But I am unable to show whether the sequence is monotonic or not and it has one limit point or more than one limit points. I come to know that a similar question was discussed in the following link  But some steps in that solution are not clear to me
121934
I have tried a little bit which as follows- Since $f(x_n)\in[0,1]$, $\{f(x_n)\}$ has a convergent subsequence say $y_n=f(x_{r_n})\ \forall n\in\Bbb{N}$ Let, $\lim y_n=l\implies \lim \frac{y_1+y_2+\cdots+y_n}{n}=l\implies \lim \frac{f(x_{r_1})+f(x_{r_2})+\cdots+f(x_{r_n})}{n}=l$ But I am getting no idea to proceed and prove the convergence of $\{x_n\}$. I have also tried to prove the sequence to be cauchy which goes- $x_{m+1}-x_{n+1}={\sum_{i=1}^m f(x_i)\over m}-{\sum_{i=1}^n f(x_i)\over n}\le {\sum_{i=1}^m f(x_i)\over n}-{\sum_{i=1}^n f(x_i)\over n}$ (since $m\ge n)$ $\implies |x_{m+1}-x_{n+1}|\le \frac{|f(x_m)|+|f(x_{m-1})+\cdots+|f(x_{n+1})|}{n}\le\frac{m-n}{n}$ (since $f([0,1])\subseteq [0,1])$ Now, what will I get if I tend $m,n\to\infty$? But I need to do something more in the 2nd case since I have nowhere used the continuity of $f$.  Can anybody give an idea to prove it? Thanks for assistance in advance.
123238
Let $x_n$ be a sequence of real numbers.  Show that if $\sum_{n=1}^{\infty}x_n $ converges, then $\lim_{n\rightarrow\infty}\sum_{k=n}^{\infty}x_k=0.$  This is what I have done. We have by the Cauchy Criterion, for every $\epsilon&gt;0$ there is an $N\in\mathbb{N}$ such that $n\ge N \ \implies |\sum_{k=n}^{\infty}x_k|&lt;\epsilon.$ Taking the limit as $n \rightarrow \infty$ we get the desired result.  Are there any problems with my proof?
87463
Assume that $(a_n)$ is a sequence of positive real numbers satisfying $a_{n+1}=\frac{2}{a_n+a_{n-1}}$ for $n=2,3,\dots$.  Prove that $(a_n)$ is convergent and find the limit.    I have no idea how to prove convergence. I tried to prove that $(a_n)$ is bounded and monotone, but both trials failed.  If convergence is proven, the limit is easy: if $g=\lim a_n$, then $g\ge 0$ because all $a_n$'s are positive, the equivalent relation $a_{n+1}(a_n+a_{n-1})=2$ gives $g(g+g)=2$, so finally $g=1$.
115884
I ask for some help with this question:  Prove or provide counter example:  If $\sum\limits_{n=1}^\infty na_n$ converges then $\sum\limits_{n=1}^\infty na_{n+1}$ also converges.  I tries this way:  If $\sum\limits_{n=1}^\infty na_n$ converges then $na_n \to 0$, therefore $a_n \to 0$.  There are 3 possible cases:  1) If $a_n &gt;0 $ and $a_n$ is monotonic decreasing sequence then $na_{n+1}&lt;na_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.  2) If $a_n &gt;0 $ and $a_n$ is not monotonic decreasing sequence : it is not possible that $a_{n+1}&gt;a_n$ because in this case $a_n \to \infty$, therefore it must be $a_{n+1} \le a_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.  3) If $a_n$ is sign-alternating series. There I have a problem to find a solution.  Thanks.
81593
Let $f$ be a real, continuous function defined for all $0\leq x\leq 1$ such that $f(0)=1$, $f(1/2)=2$, and $f(1)=3$. Show that  $$\lim_{n\rightarrow\infty} \int_0^1 f(x^n) dx$$  exists and compute the limit.  Attempt:   Since $f$ is real-valued continuous on $0\leq x\leq 1$ and the boundaries $f(0)=1$ and $f(1)=3$, $f$ is bounded on the interval  and the integral $\displaystyle\int_0^1 f(x^n)dx$ exists for any positive $n$. Thus, we can interchange the limit and the integral and compose the limit, \begin{align*} \lim_{n\rightarrow\infty} \int_0^1 f(x^n)dx&amp;=\int_0^1\lim_{n\rightarrow\infty}f(x^n)dx\\ &amp;=\int_0^1 f\left(\lim_{n\rightarrow\infty} x^n \right)dx\\ &amp;=\int_0^1 f(0)dx=1. \end{align*}  Questions: So, my first question concerns whether interchanging the limit and the integral is correct. It seems it would be justified by the dominated convergence theorem, where $f(x^n)$ is dominated by a function $g(x)=\displaystyle\sup_{0\leq x\leq 1} f(x^n)$ and $f_n(x^n)$, on $0\leq x\leq 1$, converges pointwise to a function $f$ that takes value 1 for $x\neq 1$ and 3 for $x=1$.  My next question is whether the interchange between the limit and composition of the function is allowed. Since the function is continuous and the limit exists at that point, it seems the interchange would be justified.
13292
I must show that $\sum\limits_{n\geq1} \frac{\sin(nx)}{n^2}$ converges $\forall x \in {R}$. Then  if $f(x)=\sum\limits_{n=1}^\infty f_n(x)$, I must prove that  $f(x)$ is continuous for $x\in [0, \pi]$  and  $\int\limits_0^\pi f(x)=2\sum\limits_{n=1}^\infty \frac{1}{(2n-1)^3}$.  I used Weistrass M-test: $\sum\limits_{n\geq1} \frac{\sin(nx)}{n^2}$ converges uniformly because $|\frac{\sin(nx)}{n^2}|\le \frac{1}{n^2}, \forall x\in R$ and $\sum \frac {1}{n^2}$ converges.  I found somewhere that $f(x)=\lim\limits_{n\to\infty}f_n(x)$. Where does it come from? I don't understand. How $f(x)$ looks like? Without $f(x)$ I don't know how to prove the last equality from exercise:(    Later edit: If $ \sum\limits_{n\geq1}f_n\to f$ uniformly , then $f_n\to f$ (punctual converges to $f $- I hope this is correct translation) which means $f_n(x)\to f(x)$ when $n\to \infty ?$   I hope someone could help me again...
92501
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.  I know that by the Bolzano-Weierstrass Theorem, every bounded sequence has a convergent subsequence. But I need some hints as to how to prove the question above. Thanks for your help!
38845
Does the recursive sequence $a_1 = 1, a_n = a_{n-1}+\frac{1}{a_{n-1}}$ converge?  Since the function $x+1/x$ is strictly monotonic increasing for all $x&gt;1$, I don't think that the limit converges, but I'm not sure. Can anybody tell me whether the sequence is converging or not?
76391
If $\{x_n\}$ is a sequence which satisfies $\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c$  where $c$ is a real positive number. Then what can be said about the convergence/ divergence, boundedness/ unboundedness of $\{x_n/n\}$.  Attempt: $\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c$ where $c &gt;0$  => $x_n$ is unbounded and divergent.  However, I am stuck on how to relate this to convergence/divergence of $x_n/n$ . Thanks for the help.
55965
Here is an exercise, on analysis which i am stuck.    How do I prove that if $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$, then the sequence $\{F_{n}(x)\}$ is boundedly convergent on $\mathbb{R}$?
25113
Suppose $f$ is a continuous function at $x = c$ in $[a,b]$.  Prove that for any sequence ${x_n}$ in $[a,b]$ converting to $c$, the sequence $\{f(x_n)\}$ converges to $f(c)$.  That is, $$ \lim_{n\to\infty}f(x_n)= f\left(\lim_{n\to\infty}x_n\right)$$    This proof seems simple but there are a few things that I need to know first. If $\{x_n\}$ converges to $c$, is it sufficient to substitute $c$ in for $\lim_{n\to\infty}x_n$?  Also needing some guidance on the structure of this proof.  Thanks!
60475
Monotone Convergence Theorem for general measure: Let $(X,\Sigma,\mu)$ be a measure space. Let $f_1, f_2, ...$ be a pointwise non-decreasing sequence of $[0, \infty]$-valued $\Sigma-$measurable functions, i.e. for every $k\ge 1$ and every $x$ in $X$, $$0 \le f_k(x) \le f_{k+1}(x).$$ Next, set the pointwise limit of the sequence ${f_n}$ to be $f$. That is, for every $x$ in $X$, $$f(x) = \lim_{k\to \infty}f_k(x).$$ Then $f$ is $\Sigma-$measurable and $$\lim_{k\to \infty}\int f_k d\mu = \int f d\mu.$$  I've noticed that when it comes to monotone convergence theorem (either Lebesgue or general measure), usually its definition restricts the monotone function sequences to be nonnegative. I'm not sure why the 'nonnegative' is necessary.
95433
Showing the sequence converges $a_{1}=\frac{1}{2}$,  $a_{n+1}=\frac{1}{2+a_{n}}$.  I already know that if $(a_{n})$ converges then it does to  $\sqrt{2}-1$.But i dont't know how to prove that this sequence cenverges.  EDIT I think that the subsequence $(a_{2n+1})$ is monotonic decreasing and the subsequence $(a_{2n})$ is monotonic increasing.
21219
I need help to prove this in real analysis. I think it uses IMVT, but not sure how to do it.     Let $f(x)$ be a real valued continuous function on $[0,1]$. Show that   $$ \lim \limits_{n\to\infty}(n+1)\int_{0}^{1}x^nf(x)dx=f(1) $$
32518
Let $g:[1, \infty)\rightarrow \mathbb R$ a continuous non-negative function, such that $\int_{1}^{\infty} g(x)\ dx$ converges. Is it true that $\lim\limits_{x\to \infty}g(x)=0$ ?   I tried to find a counter-example but I can't figure a trivial one. I also tried to prove it by the definition of the convergence of $g(x)$ but couldn't show that limit is really $0$.
63070
$\mathbf{Question}:$ Let $f$ be a continuous function on $[0,1]$. Then prove that the limit $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx$ is equal to $f(1)$.   $\mathbf{Attempt}$: First, we try to show that the sequence of functions $\{nx^nf(x)\}_{x\in [0,1]}$ is uniformly convergent to $0$ on the restricted domain $[0,1-\epsilon]$, $0&lt;\epsilon&lt;1$.   Let $\sup_{x\in[0,1]}f(x)= \mathcal{M}$. Then $|{nx^nf(x)}|\leq n(1-\epsilon)^n|\mathcal{M}|$ for any $x$ and $n(1-\epsilon)^n \to 0$ as $n\to \infty$.  Thereby, $\lim_{n \to \infty}\int_0^{1-\epsilon}nx^nf(x)dx=\int_0^{1-\epsilon}{\lim_{n \to \infty} }nx^nf(x)dx=0$  Now, denote $\sup_{x\in [1-\epsilon,1]} f(x)=M(\epsilon)$ and $\inf_{x\in[1-\epsilon,1]}f(x)=m(\epsilon)$.  $\int_{1-\epsilon}^1nx^n\ m(\epsilon)dx\leq\int_{1-\epsilon}^{1}nx^nf(x)dx\leq \int_{1-\epsilon}^1 nx^n\ M(\epsilon)dx$.  Now,  $\lim_{n \to \infty}\int_{1-\epsilon}^1 nx^n\ M(\epsilon)dx =\displaystyle \lim_{n\to\infty}\bigg[ \frac{nx^{n+1}}{n+1}M(\epsilon)\bigg]_{1-\epsilon}^1=M(\epsilon)$ [Similarly the other one is $m(\epsilon)$]  Clearly, as $\epsilon \to 0 , \ \ M(\epsilon), m(\epsilon) \to f(1)$, so $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx=f(1)$.  Is the procedure correct? Kindly verify.
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?