Recent content by shrub_broom

No, just for fun.

Maybe introduce a parametric factor can be help.

First you need to prove that ##\frac{sin(x)}{x}## is trivial at ##x = 0##, to calculate out integral##I = \int_0^{\infty} \frac {sin(x)} {x}##, you can first solve the integral ##\int_0^{\infty} \frac{e^{-ax} sin(x)} {x} dx##, let ##I(a) = \int_0^{\infty} \frac{e^{-ax} sin(x)} {x} dx##, then ##...

For every function limit ## lim_{x -> x_0} f(x) = A ##, if there is a series s.t. ##lim_{n -> \infty} a_n = x_0##, then ## lim_{n -> \infty} f(a_n) = A##. By the way, ## lim_{x -> x_0} f(x) = A ## iff ##forall~a_n~s.t.~ lim_{n -> \infty} a_n = x_0, lim_{n -> \infty} f(a_n) = A##. This is what I...

I believe you are talking about the Heine Theorem.

yes, the xn/n is not monotonous

Well, I still forget to correct some typo.

Exactly i think there is no need to prove this hypothesi.Write the right of the inequality in the form of defination we get ##a+\epsilon+\frac{Nx_1}{n}##. Although N is determined by##\epsilon##, if we choose a big enough ##p##, then##\frac{Nx_1}{n}##can be any positive real number, which means...

Exactly i think there is no need to prove this hypothesi.Write the right of the inequality in the form of defination we get ##a+\eplison+frac{Nx_1}{n}##. Although N is determined by##\eplison##, if we choose a big enough ##p##, then##\frac{Nx_1}{n}##can be any positive real number, which means...

i think it should be ##\frac{x_n}{n}\leq\frac{px_N+qx_1}{pN+q}\leq \frac{x_N}{N} +\frac{Nx_1}{n}##

for any m,n. xm+n≤xm+xn since a≤2b. if divide the right side,it becomes greater and the inequality is still true.

i get a proof. suppose that the limit inferior of the sequence is a, and then prove that the limit superior is no greater than a. to prove so, for any positive ε, there exist N xN/N ≤ a+ε. and for any n∈ℕ∧n≥N, we have n =p N+q p,q∈ℕ and q is less than N. hence, xn=xpN+q≤pxN+N x1; xn/n≤a+ε+Nx1/n...

the sequence is not monotonous. Let x2n=a and x2n+1=b; a≤b≤2a.

Well, since for each m∈ℕ, we get - well i just made a mistake. and i want to mean xn/n is bounded by x1. sorry for that fault

Homework Statement suppose that 0≤xm+n≤xm+xn for all m,n∈ℕ, prove that the limit of xn/n exists when n tends to infinity. Homework EquationsThe Attempt at a Solution I get that xn is bounded by zero and x1. And I guess that xn is monotonous but i find it hard to prove. Or maybe there is...