I've got this task to prove that, given $\forall x \geq a$ :  f is monotonically decreasing,  $ f \geq {0}$  $\int_a^\infty f(x)dx $ converges.  Then $\lim\limits_{x\to\infty}xf(x) = 0 $  I proved successfully that $\lim\limits_{x\to\infty}f(x) = 0$ but I don't know if it helps me anyhow.
122181
Let $f(x)$ be a monotone non-increasing function such that $\int_{0}^{\infty}f(x) dx$ converges. Prove: $\lim\limits_{x \to \infty} xf(x)=0$. My question is, why can't I simply contradict any other possibility by using the Integral Limit Comparison Test with $1\over x$? I am, after all, to show that $f(x)=o(x)$ as $x\to \infty$. I really don't understand why monotony is crucial here. I could use some help.
4575
I'm trying to prove  the following claim:     If $f$ is continuous  and $\displaystyle\int_a^\infty |f(x)|\;dx$ is finite then $\lim\limits_{ x \to \infty } f(x)=0$.   Here the counter example of all these sine functions won't do, since  their infinite integral does not converge absolutely.   Any hints for a proof?
117867
How would I go about proving that $\lim_{n\to\infty}\sqrt{n+1}-\sqrt{n}=0$? I have tried to use Squeeze theorem but have not been able to come up with bounds that converge to zero. Additionally, I don't think that converting to polar is possible here.
98473
I want to prove that $xe^{-x}$ is bounded by some constant for all the $x \in [0, \infty)$. So I take derivative of this guy, and found that on $x \in [0, \infty)$, its derivative is positive from $[0, 1)$, $0$ at $x=1$, and negative from (1, in infinity). So  $xe^{-x}$ achieves its maximum in $[0, \infty)$ at $x=1$, which is $1/e$. However, to prove that it's bounded by $1/e$, I need to prove that as $x$ goes to infinity, $xe^{-x}$ converges to $0$. Because otherwise, if it went to some value below $(-1/e)$, $xe^{-x}$ would not be bounded by $1/e$. How can I prove that $xe^{-x}$ converges to $0$? I mean, as $x$ goes to infinity, $e^{-x}$ converges to zero, but at the same time, x diverges to infinity. How can I formally prove that the product term converges to $0$? Do I need to use the epsilon definition of convergence to prove it?  Could you just show me the formal proof of $xe^{-x}$ converges to $0$? Or is there's any other way to approach this question? My professor suggested that I should use the power series for $e^{-x}$...?   Your help is greatly appreciated! Thank you so much!  I hope you enjoy your Saturday!
2992
Hi can you help me solve this exercise? Thanks. Let $f: [0;+\infty) \to \mathbb{R}$ be nonnegative and continuous function. Suppose $\forall a &gt; 0$  $\sum_{n=1}^{\infty} f(na)$ is convergent. Prove that $\int_{0}^{\infty}f(x) dx$ is convergent. I tried to solve it by using the Riemann sum, but for fixed a it doesn't work. I have no other ideas.
114326
Let $f$ be a continuous, nonnegative, real-valued function and $$\int_0^{\infty}f(x)dx&lt;\infty.$$ Prove that $$\lim_{n\to\infty}\frac1{n}\int_0^{n}xf(x)dx=0.$$  A start: If $\lim\limits_{n\to\infty}\int_0^{n}xf(x)dx$ is finite, then it's obvious. Otherwise, perform L'Hopital's rule, we get $\lim nf(n)$, which we want to show is $0$.
57355
I need to prove that :  $$\lim_{n \to \infty} \frac{x^n}{n!}=0$$  I tried to prove it using epsilon-delta, but I still haven't managed to prove it.  I think that maybe with sandwich theorem it possible to prove it.  Any help will be appreciated.
93622
I am trying to prove some theorem, and in the proof of it I have to check the fact that if the sequence $\{a_n\}_{n\in \mathbb{N}}$ converges, then $$\lim_{n\to \infty} n(a_n - a_{n-1}) = 0. $$ I think that this is true because I have tried with many different sequences and it holds, but I do not know how to prove it.  If this is not true, what is a counterexample for it?
50638
Suppose $f$ is positive and increasing on $[0, +\infty)$. Suppose that $F(x)=$$\int_0^x f(t)dt$. Prove that $$\int_0^{+\infty} \frac {1}{f(x)} dx$$ converges if and only if $$\int_0^{+\infty} \frac {x}{F(x)} dx$$ converges.   My thoughts, and attempt at solution -  Since $f$ is increasing, $F(x)=\int_0^xf(t)\,dt\le\int_0^xf(x)\,dt=xf(x).$ We can rearrange that to $\frac1{f(x)}\le\frac{x}{F(x)}.$ Hence if the integral of $\frac{x}{F(x)}$ converges, then so does the integral of $\frac1{f(x)}.$  However, the above working seems incomplete as it bounds the integral from the upper side only. In fact, it shows that if the second integral converges, then the first one may converge, but not vice versa. How do I place a lower bound on the integral, and go about the rest of the solution?   Haven't been able to proceed from here, please help.
25693
I have been thinking on and off about a problem for some time now. It is inspired by an exam problem which I solved but I wanted to find an alternative solution. The object was to prove that some sequence of functions converges weakly to zero in $L^2$.   I managed to show (with some help) that the limit $f$ (of a subsequence) satisfies $\int_0^x f \ dm=0$ for all $x&gt;0 $. From this I want to conclude that $f=0$ a.e. I can do this with the fundamental theorem of calculus in its Lebesgue version but there ought to be a more elementary proof.  Can someone here help me out?
98512
I want to show that if $f$ is non-increasing and $f\in L_{1}([a,\infty),m)$ where $m$ is Lebesgue measure then $\lim_{t\to\infty} t f(t)=0$. So far I've been able to show that $f\geq 0$ and that $\lim_{t\to\infty} f(t)=0$. Since monotone functions are differentiable a.e. I thought about using integration by parts but couldn't get anywhere with that. Any hints or suggestions would be greatly appreciated.
30839
Let $a:\mathbb{N}\to\mathbb{R}^+$ be a sequence such that $a_{m+n}\leq a_ma_n$. Prove that $\sqrt[n]a_n$ converges and $$\lim_{n\to\infty}\sqrt[n]a_n=\inf_{n\in\mathbb{N}}\sqrt[n]a_n.$$  So far all I have is $\sqrt[n]a_n\leq a_1$ (from $a_{m+n}\leq a_ma_n$ by induction). If it is true that $\sqrt[n]a_n$ monotone decreasing, we are done but I couldn't show it. Should I do something with $\liminf$ and $\limsup$?
118371
My question is related to:   and   Let $f\colon (0, \infty)\to R$ be  continuous such that  $f(x)\leq f(nx)$ for all positive $x$ and natural $n$.  It was proved that the limit (finite or infinite) in the infinity exists. Do we know if such a function must be (weakly) increasing? I believe that there might be counterexamples.
116728
$\lim_{x \rightarrow \infty} \lim_{n \rightarrow \infty} f_{n}(x) = \lim_{n \rightarrow \infty}\lim_{x \rightarrow \infty}f_{n}(x) $ in case of uniform convergence of $f_{n}(x)$ to $f(x)$  I used this while I was given that $f_{n}(x)$ converges to $f(x)$ uniformly. Is this correct and I was also thinking to prove this!
117760
Consider the sequence of functions  $$ f_n(x) = \frac{x^2}{n} $$ for $x \in \mathbb{R}$  I have shown that $f_n(x)$ converges pointwise towards $0$ for $n \rightarrow \infty$ but I am not sure whether my attempt for showing that it does not converge uniformly towards $0$ for $n \rightarrow \infty$ is correct. Do you mind verifying?   By negation we have that $f_n(x)$ does not converge towards $0$ for $n \rightarrow \infty$ if   $$\exists \epsilon &gt; 0 \forall n \in \mathbb{N} \exists x \in \mathbb{R} \exists n \in \mathbb{N}: n \geq N \Rightarrow |\frac{x^2}{n}| \geq \epsilon$$  Let $\epsilon = 1$. Then for all $n \in \mathbb{N}$ we can find a $x \in \mathbb{R}$ such that for $n \geq N$ that $|x^2/n| \geq 1$. Can I then just pick $x = n^{1/2}$ so we have that $$ |x^2/n| = |(n^{1/2})^2/n| = |n/n| = 1 \geq 1 $$ which means that $f_n(x)$ does not converge uniformly towards $0$ for $n \rightarrow \infty$. Is this ok?
502
I want to show that if $f:(a,b)\to \mathbb R$ is a strictly monotone function and if $x_0\in (a,b)$ such that there exist two sequences $(a_n)$ and $(b_n)$ with $a_n&lt;x_0&lt;b_n$ and $\lim\limits_{n\to \infty}(f(b_n)-f(a_n))=0$, then $f$ is continuous at $x_0$.   By monotonicity we can say that $\lim\limits_{n\to \infty}f(a_n)=f(x_0)=\lim\limits_{n\to \infty}f(b_n)$. Let $(x_n)$ be a sequence in $(a,b)$ such that $x_n\to x$. Here I got stuck. How to show that $f(x_n)\to f(x_0)$? Please help!
119987
Let $f:(a,b)\to \mathbb{R}$ be a strictly increasing function. Does the limit $\lim_{x\to a^+}f(x)$ necessarily exist and is a real number or $-\infty$?  If so, is it true that $\ell=\lim_{x\to a}f(x)\le f(x) \ \ \forall x\in (a,b)$? Please provide proofs.