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2 votes
3 answers
138 views

$|a-b|<\varepsilon\implies a=b$ for uniqueness of limit?

Consider the following statement: Let $a,b\in\mathbb{R}$ be any two real numbers. Then we have that $$\forall\varepsilon>0:|a-b|<\varepsilon\implies a=b.$$ Here is my attempt to prove it: Proof. ...
mahlovic's user avatar
2 votes
1 answer
134 views

Set of finite sums is dense

Let $(a_n)_{n\geq 1}$ be a sequence of non-negative real numbers such that $a_n \to 0$ but $\sum_{n\geq 1} a_n$ diverges. Show that the set of sums $\sum_{n \in S} a_n$, where $S$ ranges over the ...
DesmondMiles's user avatar
  • 2,733
1 vote
1 answer
75 views

$X_n - X_{n-2}\rightarrow 0$, prove that$ \frac{X_n}{n} \rightarrow 0$. [closed]

suppose $X_{n}$ is a sequence of real numbers such that $X_{n}-X_{n-2}\rightarrow 0$. prove that $\frac{X_{n}}{n}\rightarrow 0$.
porsande's user avatar
1 vote
2 answers
52 views

Show that $f(x):=\sum_{n=1}^{\infty}\frac{x}{(1+nx^{2})n^{\alpha}}\rightarrow 0$ as $x\rightarrow 0$, if $\alpha>\frac{1}{2}$.

Consider the function defined by $$f(x):=\sum_{n=1}^{\infty}\frac{x}{(1+nx^{2})n^{\alpha}}.$$ I have showed that, by Weierstrass M-test, when $\alpha>\frac{1}{2}$, the series converges uniformly to ...
JacobsonRadical's user avatar
6 votes
2 answers
1k views

Defining $N$ in the $\epsilon$-$N$ definiton of convergence

In my Real Analysis class we've been spending some time talking about the $\epsilon$-$N$ definition of convergence. The book we are using, Elementary Analysis by Ross, defines convergence as: A ...
jrayers97's user avatar
  • 331
1 vote
2 answers
102 views

Why the following sequence of function does not converge uniformly at $[0, \infty)$

Why the following sequence of function does not converge uniformly at $[0, \infty)$ but converge uniformly for some $a>0, [a,\infty)$ $$f_n(x) := n^2x^2e^{-nx}$$ So I know the limit function $f$ ...
some1fromhell's user avatar