Prove that: $$I=\lim_{n\to\infty}\int_{0}^{1}\int_{0}^{1}...\int_{0}^{1} \frac{x_1^2+x_2^2+ ... +x_n^2 }{x_1+x_2+ ... +x_n} \space dx_1dx_2...dx_n = \frac{2}{3}$$ by considering $x_1, x_2, ... ,x_n$ as continuous variables with uniform distribution over the interval (0, 1) with Chebychev's inequality that $$\lim_{n\to\infty} Prob \space \Big(\frac{2}{3} - \epsilon &lt; \frac{x_1^2+x_2^2+ ... +x_n^2 }{x_1+x_2+ ... +x_n} &lt; \frac{2}{3} + \epsilon \Big) =1$$ for any $ \epsilon &gt;0$
121419
How to prove the following: $$ \lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 \frac{x_1^2+x_2^2+ \cdots +x_n^2}{x_1+x_2+ \cdots +x_n} dx_1 dx_2 \cdots dx_n = \frac23 $$  I would really appreciate if you could help me!
19666
$f$ defined on $[1,\infty )$ is uniformly continuous. Then $\exists M&gt;0$ s.t. $\frac{|f(x)|}{x}\le M$ for $x\ge 1$.  I know f uniformly continuous $\implies \forall\varepsilon &gt; 0s.t.\forall x,y\in [1,\infty)$, $$|f(x)-f(y)|&lt;\varepsilon \space\space\forall |x-y|&lt;\delta$$  By Mean value theorem, $\exists c\in (x,y)\forall x,y\in [1,\infty)$ such that   $$\frac{f(x)-f(y)}{x-y}=f'(c)$$  $$\frac{|f(x)-f(y)|}{|x-y|}=|f'(c)|&lt;M$$ for some $M&gt;0$  $$|f(x)-f(y)|&lt;M|x-y|&lt;M\delta $$  How can I get an expression for $|f(x)|/x$?
48979
Let $X_1, \ldots, X_n$ be i.i.d. continuous uniform random variables on $\left[\vartheta-\frac{1}{2},\vartheta+\frac{1}{2}\right]$ for some $\vartheta \in \mathbb R$.    I am trying to solve $\mathbb E[(T-\vartheta)^2]$ where $T(X) = \frac{1}{2}\left(\min_{1 \le i \le n} X_i + \max_{1 \le i \le n} X_i\right)$ or simplify it such that it becomes clear that it is independent of $\vartheta$.   I have tried to approach this by a variety of conversion but always end up with a term that involes $\mathbb E\left[\left(\min_{1 \le i \le n} X_i\right)\left(\max_{1 \le i \le n} X_i\right)\right]$.  I can't find a way to actually calculate it. Do you have a hint or another way of approaching the task?
29659
Let $\{X_n\}_{n\geq 1}$ be a sequence of random variables s.t $X_n \to a$ in probability, where $a&gt;0$ is a fixed constant. I'm trying to show the following: $$\sqrt{X_n} \to \sqrt{a}$$ and $$\frac{a}{X_n}\to 1$$ both in probability. I'm here to see if my logic was sound. Here's my work     ATTEMPT      For the first part, we have   $$|\sqrt{X_n}-\sqrt{a}|&lt;\epsilon \impliedby |X_n-a|&lt;\epsilon|\sqrt{X_n}+\sqrt{a}|=\epsilon|(\sqrt{X_n}-sqrt{a})+2\sqrt{a}|$$   $$\leq \epsilon|\sqrt{X_n}-\sqrt{a}|+2\epsilon\sqrt{a}&lt;\epsilon^2+2\epsilon\sqrt{a}$$   Notice that    $$\epsilon^2+2\epsilon\sqrt{a}&gt;\epsilon\sqrt{a}$$   It follows then that   $$P(|\sqrt{X_n}-\sqrt{a}|\leq \epsilon)\geq P(|X_n-a|\leq \epsilon\sqrt{a})\to 1 \;\;as\;n\to\infty$$   $$\implies \sqrt{X_n}\to\sqrt{a} \;\;in\;probability$$      For the second part, we have   $$|\frac{a}{X_n}-1|=|\frac{X_n-a}{X_n}|&lt;\epsilon \impliedby |X_n-a|&lt;\epsilon|X_n|$$   Now, since $X_n \to a$ as $n \to \infty$, we have that $X_n$ is a bounded sequence. In other words, there exists a real number $M&lt;\infty$ s.t $|X_n|\leq M$. Thus,   $$|X_n-a|&lt;\epsilon|X_n|\impliedby |X_n-a|&lt;\epsilon M$$   Looking at it in probability, we have   $$P(|\frac{a}{X_n}-1|&gt;\epsilon)=P(|X_n-a|&gt;\epsilon|X_n|)\leq P(|X_n-a|&gt;\epsilon M)\to 0 \;\;as\;n\to\infty$$   I'm pretty confident in the first one, but am pretty iffy on the second. Was my logic sound?
23871
Suppose that $f$ is a continuous, non-negative function on the interval $[0,1]$. Let $M$ be the maximum of $f$ on the interval. Prove that $$\lim_{n \to \infty}\bigg[\int_0^1f(t)^n \text{dt}\bigg]^{1/n}=M$$  We wrote out some simple examples to show it worked for functions such as $x^2$. We are having trouble finding how to create a general proof. Thanks for any help!
77964
Using my right to ask for help once again. I can't solve the following problem for quite a long time. Just have no idea how to do that. Here it is. Let $f(x)$ be bounded on any interval $(1,b), b&gt;1$. Then $\lim_{x\to+\infty}{f(x) \over x}=\lim_{x\to+\infty}(f(x+1)-f(x))$. It is from the Russian book "Lekcii po matematicheskomu analizu" by G.I.Arkhipov, V.A.Sadovnichy and V.N.Chubarikov, 2nd ed., Moscow, 2000, page 676. Please, keep in mind, that times, when I did homework had passed long time ago. Now I'm solving problems for pleasure.
40274
I am finding difficulties in finding a counterexample to the following statement. If $f\colon (0,\infty) \to(0,\infty) $ is uniformly continuous, then $$\lim_{x\to \infty} \frac{f(x+\frac{1}{x})}{f(x)} =1.$$
76981
Let $g$ be continuous on the interval $[a, b]$, and a $&lt; c &lt; b$ with $g_{(c)}&lt;0$  Prove, using the epsilon and delta characterisation of a limit, that $\lim\limits_{x \to {c}}\frac{1}{g{(x)}} = \frac{1}{g{(c)}}$ Hence deduce that $\frac{1}{g}$ is continuous at $c$  I know that there exists $g {(x}$) smaller than $\frac{g {(c)}}{2} &lt; 0$ and I can use this information somehow but other than that I am not sure how to approach this question. Is there any special laws I need to follow?
32562
Is $f(x)=x+\frac{x}{x+1}$ uniformly continuous on $(0,\infty)$   Going from the epsilon delta definition we get:   $$\forall x,y&gt;1,\text{WLOG}:x&gt;y \ ,\ \forall\epsilon&gt;0,\exists\delta&gt;0 \ s.t. \ |x-y|&lt;\delta\rightarrow |x+\frac{x}{x+1}-y-\frac{y}{y+1}|&lt;\epsilon$$  But I'm not really sure on how to continue from here.  Alternatively: if $\lim\limits_{x\to \infty}f(x)$ exists on $[0,\infty)$ then the function is uniformly continuous, well, it's easy to see here that $\lim\limits_{x\to \infty}f(x)=\infty$ but is that enough ?
103954
If $f$ Continuous bounded on $\mathbb{R}$，and $$\lim_{h\to0}\sup_{x\in\mathbb{R}}\left|f(x+h)-2f(x)+f(x-h)\right|=0.$$ prove that $f$ uniformly continuous on $\mathbb{R}$. I try to find a $\delta&gt;0$ to make it arbitrary $x_1,x_2\in \mathbb{R}$and$\left|x_1-x_2\right|&lt;\delta$have$$\left|f(x_1)-f(x_2)\right|&lt;\varepsilon.$$but I don't know how to use this complex limit expression.I hope you can give me some tips, thank you!
7416
I'd like a hint to solve the following problem:     Given a $f$ continuos and positive on $[a,b]$ prove that $$\lim_{n\to \infty}  \left[\int_a^b (f(x))^n dx\right]^{{1}/{n}}=\sup_{x\in [a,b]} f(x).$$   Thanks a lot!
13215
$$  \lim_{n\to\infty} n^{1/2} \int_{0}^{1} \frac{1}{(1+x^2)^n}\mathrm{d}x=0 $$  Is my answer correct? But I am not sure of method by which I have done.
37755
Assume that $f: [0,1] \rightarrow \mathbb{R}$ has a continuous $f''$ and $f$ is positive on the interval $(0,1)$ and $0$ at the endpoints.    I want to prove that $$\int_0^1 \Big| \frac{f''(x)}{f(x)} \Big| \,dx &gt;4. $$  I could not make any progress. Since the unit interval is compact, we know that the continuity of $f, f'$ and $f''$ are uniform and they attain their maximum and minimum values. Moreover, I tried to apply the Mean Value Theorem but I failed.
97524
Let $T_n$ be a random variable with $T_n=X_1+...+X_n$ where the $X_i$'s are iid. Further we set $N(t)=max\{ n: T_n\leq n\}$ with the property $\Pr(N(t)&lt;\infty)=1$.  I want to prove that $\mathbb{E}[N(t)]&lt;\infty$.   I have done this so far. Let $F_n$ be the distribution function of $T_n$. $$\mathbb{E}[N(t)]=\sum_{n=1}^{\infty} n P(N(t)=n)=\sum_{n=1}^{\infty} n (F_n(t)-F_{n+1}(t))=\sum_{n=1}^{\infty} F_n(t)$$  Further I know that $0 \leq F_n \leq 1$. But does this implies $\mathbb{E}[N(t)]&lt;\infty$ ? I don't think so... I don't know what I can do further...   Can anybody help me how I can proof it?   Thank you for any help!!!
109196
Some days ago I answered a question that asked to find  $$\mathop {\lim }\limits_{x \to 0} {x^\alpha }\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^{\alpha  + 1}}}}dt} $$  given that $f$ is continuous in $[0,1]$  I proceeded as follows:  $$\eqalign{   &amp; t = x\cdot u  \cr    &amp; dt = x\cdot du \cr} $$  So this is produces:  $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du} $$  I then thought: "Well, if $f$ is continuous in the closed interval, then it is also uniformly continuous, so I can assume   $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  = f\left( 0 \right)\int\limits_1^\infty  {\frac{{du}}{{{u^{\alpha  + 1}}}}}  = \frac{1}{\alpha }f\left( 0 \right)$$  This turned out to be true. However, I wasn't very comfortable with such "move". So now I'm thinking, one can put  $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du} $$  And then   $$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du} } \right| &lt; \int\limits_1^{\frac{1}{x}} {\frac{{\left| {f\left( {xu} \right) - f\left( 0 \right)} \right|}}{{{u^{\alpha  + 1}}}}du}  &lt; \epsilon \frac{{1 - {x^\alpha }}}{\alpha }$$  However, this is still insufficient since I need to adress the behaviour of the upper limit too. Can someone show me how to adress both behaviours simultaneously?    Would this work?  Let $P$ be the statement that $$\mathop {\lim }\limits_{x \to 0} \int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  = \frac{{f\left( 0 \right)}}{\alpha }$$  Then $P$ is true if and only if  $$ \forall \epsilon  &gt; 0\exists \delta  &gt; 0$$  Such that if $$\left| x \right| &lt; \delta $$ then $$ \left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right)}}{{{u^{\alpha  + 1}}}}du}  - \frac{{f\left( 0 \right)}}{\alpha }} \right| &lt; \epsilon $$  But then   $$\left| {\int\limits_1^{\frac{1}{x}} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \int\limits_1^{\frac{1}{x}} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| &lt; $$  $$\left| {\int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( {xu} \right) - f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du}  + \int\limits_1^{\frac{1}{\delta }} {\frac{{f\left( 0 \right)}}{{{u^{\alpha  + 1}}}}du - \frac{{f\left( 0 \right)}}{\alpha }} } \right| \leqslant $$  $$\varepsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } &lt; $$  And since  $$\frac{{1 - {\delta ^\alpha }}}{\alpha } &lt; \frac{1}{\alpha }$$ $$\epsilon \frac{{1 - {\delta ^\alpha }}}{\alpha } + f\left( 0 \right)\frac{{1 - {\delta ^\alpha }}}{\alpha } - \frac{{f\left( 0 \right)}}{\alpha } &lt; \frac{\epsilon }{\alpha } &lt; \epsilon $$
75574
I have some problems understanding  of this problem:  Given $f,g : \mathbb{R} \rightarrow \mathbb{R}$ - continuous functions with period $1$, prove that  $$\lim _{n \rightarrow \infty} \int_0^1 f(x)g(nx) \, dx = \int_0^1 f(x) \, dx \cdot \int_0^1g(x) \, dx$$  It says there that if we split this integral into $n$ parts to get  $$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} \int_{\frac{r}{n}}^{\frac{r+1}{n}} f(x)g(nx) \, dx$$  we will have   $$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} f\left(\frac{r}{n}\right) \int_{\frac{r}{n}}^{\frac{r+1}{n}}g(nx) \, dx$$  because for large $n$, $f$ is roughly constant over the range.  Could you tell me why $f$ is roughly constant over the range for $n$ large enough?  I would really appreciate all your insight.
64072
Let $X_1(x),X_2(x),\ldots ,X_n(x)$ be $n$ iid random variables. I know that  $$\mathbf{P}(\lim_n X_n(x)=X(x))=1$$ for each $x\in \mathbb{R}$. We also know that the limit random variable $X$ is a.s. constant and considering $X\colon \mathbb{R}\to\mathbb{R}$, it is continuous. I want to show that  $$\mathbf{P}(\lim_nX_n(x)=X(x)\ \text{ for all }x\in \mathbb{R})=1.$$ I thought about using countable additivity of probability on $\mathbb{Q}$ and then applying continuouty to extend the result to the whole $\mathbb{R}$. Since $$\mathbf{P}(\lim_nX_n(x)\neq X(x)\ \text{ for all }x\in \mathbb{Q})=\sum_{x\in \mathbb{Q}}\mathbb{P}(X_n(x)\neq X(x))=0+0+\ldots.$$ Is there anything else I should add to this explanation? Ideas?