Let $\{y_n\}$ a sequence so that $\sum y_i=+\infty$ and $$y_n&gt;0, \forall n \in \mathbb{N}.$$  Show that if  $$\lim_{n \to +\infty} \frac{x_n}{y_n}=a$$ then $$\lim_{n \to +\infty} \frac{\displaystyle \sum^n x_i}{\displaystyle \sum^n y_i}=a.$$
123186
Let $Y_{n} &gt; 0$ for all $ n\in \mathbb{N} $, with $\sum{Y_{n}}= +\infty$.  If $\displaystyle\lim\limits_{n\rightarrow \infty}\frac{X_{n}}{Y_{n}}= a$ then $\displaystyle\lim\limits_{n\rightarrow\infty}\frac{X_{1}+X_{2}+X_{3}+\dots+X_{n}}{Y_{1}+Y_{2}+Y_{3}+\dots+Y_{n}}= a$  I need a idea for solution.  can i supose that $\lim\limits_{n\rightarrow\infty}Y_{n}$ exist ?
124275
Let $\{a_n\}$ be a sequence of real numbers such that $$\lim_{n \to \infty}\left|a_n+ 3\left(\frac{n-2}{n}\right)^n \right|^{\frac{1}{n}}=\frac{3}{5}.$$  What is the value of $\lim_{n\to \infty} a_n$?
42886
given a sequence $a_n=a^{\frac{1}{n}}$ for $n\in\mathbb{N}^*$, $a\in\mathbb{R},a&gt;1$ then proof that $\lim\limits_{n\to+\infty}a_n=1$ by definition.   proof:  given $a_n=a^{\frac{1}{n}}$ for $a\in\mathbb{R},a&gt;1$.  for $n\in\mathbb{N}^*,n+1&gt;n\Rightarrow \frac{1}{n+1}&lt;\frac{1}{n}$ and because $a&gt;1$ we gets $a^{\frac{1}{n+1}}&lt;a^{\frac{1}{n}}$, since $\frac{1}{n+1}&gt;0$, then $1=a^{0}&lt;a^{\frac{1}{n+1}}&lt;a^{\frac{1}{n}}\Rightarrow 1&lt;a_{n+1}&lt;a_{n}$.  then we proof that $\lim\limits_{n\to+\infty}a_n=1$ we need to show that $\forall\epsilon&gt;0,\exists N,\forall n&gt;N,|a_n-1|&lt;\epsilon$  then for $\epsilon&gt;0$, choose $N=\frac{1}{\log_a(\epsilon+1)}$, since $a&gt;1$ we gets that $$\forall n&gt;N,1&lt;a^{\frac{1}{n}}&lt;a^\frac{1}{N}\Rightarrow0&lt;a^{\frac{1}{n}}-1&lt;a^{\frac{1}{N}}-1\Rightarrow |a^{\frac{1}{n}}-1|&lt;|a^{\frac{1}{N}}-1|=|a^{\log_a(\epsilon+1)}-1|=|\epsilon+1-1|=\epsilon\Rightarrow |a^{\frac{1}{n}}-1|&lt;\epsilon$$  wich implies that $\lim\limits_{n\to+\infty}a_n=1\square$  my proof to the limit is correct?
47786
Hi there I'm trying to solve an exercise which is part of my homework and I would really appreciate a hint how to solve it.  Given the sequence $a_n=1+\frac{1}{2}+...+\frac{1}{n} - \log(n)$ it asks me to prove that it is convergent and to prove that limit when $n$ approaches infinity of $a_n$ is $c$ where $c$ is between $0$ and $1$.  So far I proved its montony and that the sequence is decreasing. I could need a hint on how to bound it though. Thanks in advance.
103599
can anyone help me with this, I got stuck on it.  let $a_n$ and $b_n$ (for $n$ a non-negative integer) be two sequences of real numbers such that $\lim_{n\to \infty}a_n=a$ and $\lim_{n\to \infty}b_n=b$. prove that $\lim_{n\to \infty} \frac{a_0b_n+a_1b_{n-1}+....+a_nb_0}{n}=ab$.
59518
If I have a sequence {$a_n$} that has the property of $\lim(a_{n+1}-a_n)=0$, does that make it a Cauchy Sequence. I think it doesn't because $a_n = \sum_{k=1}^n \frac{1}{k}$ is a counter example.   However, by definition, there exists a $M$ such that if $n \geq M$ then $|a_{n+1}-a_n| &lt; \frac{\epsilon}{m-n}$  Hence, we have $|a_m - a_{m-1}|+.....+|a_{n+1}-a_n|&lt;|a_m -a_n| &lt;\epsilon$  This proof doesn't work because I cannot be sure I can find a fixed $M$ which might change according to n.  But I wonder if I have an additional condition that says $a_n$ is bounded, I think then the proof works and that I should be able to find a fixed $M$. However, I don't know how to justify this. Maybe I am wrong. Can someone kindly help me figure out this problem. Thanks
31778
Problem:      Suppose that for every $n\in\mathbb{N}$, $a_n\in\mathbb{R}$ and $a_n\ge 0$. Given that   $$\sum_0^\infty a_n$$   converges, show that    $$\sum_1^\infty \frac{\sqrt{a_n}}{n} $$   converges.   Source: Rudin, Principles of Mathematical Analysis, Chapter 3, Exercise 7.
5047
I'm not completely new to convergent series and sequences, but proving the above gave me some troubles. Would appreciate some help :) Show that: $$\lim_{n\to\infty} \frac{\sum\limits^n_{k=1} \frac{1}{k}}{ \ln(n) } = 1$$
70044
I haven't been able to solve this by using the definition of limit. I think maybe trying induction will work, though. Can someone help?  Suppose $\lim \limits_{n \to ∞} a_n=L$. Prove that $\lim\limits_{n \to ∞} \dfrac{a_1+a_2+\cdots+a_n}{n}=L$
106497
I am struggling to prove the following property of limits of sequences:     If $(a_n)_{n\in\mathbb{N}}$ is a sequence such that $\forall n:\ a_n\ne0$ and $\lim\limits_{n\to\infty} a_n = L \ne 0$, then   $$\lim\limits_{n\to\infty} \frac{1}{a_n} = \frac{1}{L}$$   Using the definition of limit, $$\forall \varepsilon&gt;0\ \exists n_0\ \forall n \ge n_0\quad \left|a_n - L\right| &lt; \varepsilon$$ what I have up to now is $$\begin{equation}\label{eq:inv-limit}     \left|{\frac{1}{a_n}-\frac{1}{L}}\right| =      \left|{\frac{L-a_n}{a_nL}}\right| =      \frac{\left|a_n- L\right|}{\left|a_nL\right|} &lt;      \frac{\varepsilon}{\left|a_n\right|\left|L\right|}\,.    \end{equation}$$  But now the coefficient of epsilon is not constant, so that is not sufficient to prove that $\frac{1}{a_n}\to\frac{1}{L}$. How to proceed?
111865
Problem: Let $(x_i)$ be a sequence in (0,1) such that $\lim_{n \to \infty} (1/n)\sum_{i=1}^n x_i^k$ exists for every $k=0,1,2,...$ and $f\in C[0,1]$. Show that $\lim_{n \to \infty} (1/n)\sum_{i=1}^n f(x_i)$ exists.  I've used the fact that $B_n(f)$ converges uniformly to $f$ to get   $\lim_{n \to \infty} (1/n)\sum_{i=1}^n f(x_i) = \lim_{n \to \infty} (1/n)\sum_{i=1}^n \sum_{k=0}^n f(\frac{k}{n}) {n \choose_k }x_i^k (1-x_i)^{n-k}$  And then you can arrange to get   $\lim_{n \to \infty} (1/n) \sum_{k=0}^n \left[ f(\frac{k}{n}) {n \choose_k }\sum_{i=1}^nx_i^k (1-x_i)^{n-k}\right]$  I want to use the fact that $(1-x_i) &lt; 1$ to consolidate some of the equation but I get nowhere.
43921
Suppose $a_n$ and $b_n$ are sequences of positive numbers such that  $\limsup \frac{a_n}{b_n}&lt;\infty$  show that there is a constant $M&gt;0$ s.t $a_n\le Mb_n$ for all $n\ge 1$  My idea:  as i know hat $M=\limsup a_n$ if and only if for all $\epsilon &gt;0$ there is $N\in \mathbb{N}$ so tthat $n\ge N$ implies $a_n&lt;M+\epsilon$
64951
Let $(a_n)_{n\in\mathbb{N}}$ be a sequence such that lim inf $|a_n|=0$. Prove that there is a subsequence $(a_{n_{k}})_{k\in\mathbb{N}}$ such that $\sum_{k=1}^\infty a_{n_k}$ converges.  I'm thought about showing that you can make a subsequence that is smaller then $1/n^2$. Is this the right strategy ? Any hints ? Should I use the Cauchy criterion ?
55799
Consider a convergent sequence $\sum\limits_{n=1}^\infty{a_n}$, with $a_n\ge 0$. Show that the series  $$\sum\limits_{n=1}^{\infty} \frac {\sqrt{{a_n}}}{n}$$ also converges.  Hint: $pq\le \frac 12(p^2+q^2)$
43844
Hope this is a meaningful question, but I'm curious if is possible to show that   $$\lim_{n\to\infty} a^{\frac{1}{n}}=1, \text{where  }a&gt;0$$  using $\delta-\epsilon$ directly or other methods. One method that I am aware of is to use the following:     If $\{s_n\}$ is a nonzero sequence, then $\liminf\bigl|\frac{s_{n+1}}{s_n}\bigr|\le \liminf |s_n|^{\frac{1}{n}}\le \limsup |s_n|^{\frac{1}{n}}\le\limsup\bigl|\frac{s_{n+1}}{s_n}\bigr|$
95279
I'm self-studying a bit of complex analysis, and I'm attempting to figure out the following.   Suppose $\lim_{n\to\infty}z_n=A$. How can I show that $$ \lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A. $$  Is there a clever way to write the limit to make it more approachable? Thank you.
75676
Proposition: Let $(a_n)_{n=1}^\infty$ be a sequence of real numbers. Show that $\limsup_{n\to\infty} (a_n)\leq \sup(a_n)_{n=1}^\infty$.   Any hints on how to prove it will be appreciated.