It's been awhile since I last touched Calculus so I'm asking for help in a really basic question. I'm trying to prove the following theorem:     Let $a_n$ and $b_n$ be sequences so $\lim_{n\to\infty} a_n\cdot b_n=1$ and $\lim_{n\to\infty} b_n=\infty$. So we get $\lim_{n\to\infty} a_n=0$.   I'm trying to prove this theorem by the limit definition but I can't seem to figure out how to find $\epsilon$. Is it possible to show how?
121525
A. $\lim_{n\to\infty}a_nb_n=?$ given that $\lim_{n\to\infty}a_n=0$ and $ \lim_{n\to\infty}b_n=\infty$  B. $\lim_{n\to\infty}{a_n \over b_n}=?$ given that $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}b_n=0$  C. $\lim_{n\to\infty}{a_n \over b_n}=?$ given that $\lim_{n\to\infty}a_n=\infty$ and $\lim_{n\to\infty}b_n= \infty$    These seem to be difficult situations I come across when I have to evaluate limits.   Some examples,    $\lim_{n\to\infty}n^a\sin(1/n^b)$ with $a,b&gt;1$   $\lim_{n\to\infty}{n^a\over \sin(1/n^b)}$ with $a,b&gt;1$   $\lim_{n\to\infty}{1/n^a\over \sin(1/n^b)}$ with $a,b&gt;1$   $\lim_{n\to\infty} nf(x_n)$ [where we only know that $\lim \ f(x_n) = 0]$    My first question is : Are the two situations  A and B  equivalent ? I feel that they are equivalent as $\lim_{n\to\infty}b_n=\infty \iff \lim_{n\to\infty}{1\over b_n}=0$, The problem is that I'm not sure if we can apply the product rule for limits to take the limit inside.  My second question is : What are some ways to solve such problems ? I only know that we can apply stolz- cesaro in some cases.
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 $\liminf\bigl|\frac{s_{n+1}}{s_n}\bigr|\le \liminf |s_n|^{\frac{1}{n}}\le \limsup |s_n|^{\frac{1}{n}}\le\limsup\bigl|\frac{s_{n+1}}{s_n}\bigr|$
114143
Suppose that $a_n \neq 0$, for every n and that $L = \lim |\frac{a_{n +1}}{a_n}|$ exists. Show that if L &lt; 1, then $\lim a_n = 0$.  What I did so far:  If L &lt; 1 and $L = \lim |\frac{a_{n +1}}{a_n}|$, there exists $n_0$ such that for $n \geq n_0$, $0 &lt; |a_{n+1}| &lt; |a_n|$. That means that the sequence $(|a_n|)_{n \geq n_0}$ is decreasing.   Consider the set $S = \{ |a_0|, ..., |a_{n_0} \}$. S is finite. Let $\beta = \max_{0 \leq i \leq n_0} |a_ i|$. Thus, $(|a_n|)$ is limited (because, for every n, $0 \leq |a_n| \leq \beta).$  Now I have that $(|a_n|)$ is limited and, throwing away a finite number of terms (the $n_0$ firsts) I can assume that it is decreasing. So I know that $(|a_n|) converges.   How can I prove that it converges to $0$?   I also know that if I prove that $ \lim |a_n| = 0$, then I have that $ \lim a_n = 0$.
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Written, in Ex.8 Ch.9.1 of the book Advanced Calculus by P. M. Fitzpatrick :     Suppose that $\sum\limits_{k=1}^\infty a_k$ and $\sum\limits_{k=1}^\infty b_k$ are series of positive numbers such that $$\lim_{n \to \infty} \frac{a_n}{b_n}=l \ \ \ \text{and} \ l&gt;0.$$ Prove that the series $\sum\limits_{k=1}^\infty a_k$ converges iff the series $\sum\limits_{k=1}^\infty b_k$ converges.   Am I correct by the following (sketch of) proof? :  1- For a given $\epsilon_1$ there is $N_1$ such that $\left|\frac{a_n}{b_n} - l\right| &lt; \epsilon_1$ for all $n \ge N_1$.  2- Since the series $\sum\limits_{k=1}^\infty a_k$ converges, for a given $\epsilon_2$ there is $N_2$ such that $\left|b_{n+1}+\dots+b_{n+k}\right|&lt; \epsilon_1$ for all $n \ge N_2$ any for all natural numbers $k$.  3- Define $N = \max {\{N_1,N_2}\}$.   4- From Step (1), $a_{n+k} &lt; (\epsilon_1+l) b_{n+k}$ for all $n \ge N$ any for all natural numbers $k$. [Also, $a_i$'s and $b_i$'s are all positive]. Thus $a_{n+1}+\dots+a_{n+k} &lt; (\epsilon_1+l) (b_{n+1}+\dots+b_{n+k})&lt; \epsilon_3$. Then the convergence of the series $\sum\limits_{k=1}^\infty b_k$ implies the convergence of the series $\sum\limits_{k=1}^\infty a_k$.  5- For the reverse implication, we use the fact that $\lim\limits_{n \to \infty}\frac{a_n}{b_n}=l \iff \lim\limits_{n \to \infty}\frac{b_n}{a_n}= \frac{1}{l}= l' &gt;0$ and repeat the process this time for a and b exchanged.   6- The Quotient Property For Sequences hold for a nonzero limit in the denominator, but since the limit in the numerator also is zero so we may use the The Quotient Property For Sequences.    Thanks.
59709
How do I prove that the limit of the sequence exists: $$a_n = \frac1n +\frac 1{n+1} + \frac1{n+2} + \dots + \frac1{2n}$$  I can find the limit as n goes to infinity, $\frac1{2n}$ goes to $0$, so the limit is $0$ right? But how do I prove that the limit exists? Do I have to use the $\varepsilon$ $\delta$ definition of a limit?    For part b) I have to show that the limit is less then $1$ but not less then $\frac12$, I thought the limit was $0$?
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Proof that $\lim{a_n}=L$ when $n$ goes to infinity, then $\{a_n\}$ its a Cauchy Sequence  I start with this hypothesis  $\displaystyle\lim_{n \to\infty}{a_n}=L \Leftrightarrow{\forall{\epsilon}&gt;0}$  $\exists{N}\in{\mathbb{N}}$ such that  $\forall{n}&gt;N \left |{a_n-L}\right |&lt; \epsilon$  The same its true fot any $m&gt;N,\left |{a_m-L}\right |=\left |{L-a_m}\right |&lt; \epsilon$  I will sum the inequalities but I havent any idea how do it... Help me please!!!
19214
Let $\{x_n\}_{n=1}^{\infty}$ be a convergent sequence in a  $(S, ||\cdot ||)$. Find the limit of the sequence \begin{align} \left\{\frac{x_1 + 2x_2 + \cdots + nx_n}{n^2}\right\}_{n=1}^{\infty} . \end{align}  Could you please give me some hints where to start?
48503
How to formally prove that if $\lim \limits_{n \to \infty}a_n=\infty$, then  $\lim \limits_{n \to \infty}\frac{1}{a_n}=0$.  I got confused for how to mix between convergence to infinity and having finite limit. how do I write a proof to link between two things? ( by using $\epsilon $) and not just giving an example.  my proof:   let $M= \frac{1}{\epsilon}$. there exists such a $N\in\mathbb N$ s.t for every $n&gt;N$, it is true that: $$a_n&gt;\frac{1}{\epsilon}$$  we should prove that for every $\epsilon&gt;0$ there exists such a  $N\in\mathbb N$, s.t for every $n&gt;N$ $$\left|\frac1{a_n}\right|&lt;\epsilon$$  so we take the N that satisfies the first conclusion, and that will mean for every $n&gt;N$$$\left|\frac1{a_n}\right|&lt;\epsilon$$
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I intuitively understand proof with limits, but I'm not sure on how to write a formal proof for this example.  For each $n \in \mathbb{N}$, let $a_n$, $b_n$ be real numbers. Also, let $a_{\infty}$, $b_{\infty}$ be real numbers. Suppose that $\lim_{n\to \infty} a_n = a_{\infty}$ and $\lim_{n\to \infty} b_{n} = b_{\infty}$.  Prove that $\lim_{n\to  \infty} (a_n b_n) = a_\infty b_\infty$.
11394
Prove that the $\lim_{n\to \infty} r^n = 0$ for $|r|\lt 1$.   I can't think of a sequence to compare this to that'll work.  L'Hopital's rule doesn't apply.  I know there's some simple way of doing this, but it just isn't coming to me. :(
64925
Let $\{a_n\}$ be a sequence of real number, let $\{n_k\}$ be an increasing sequence of positive integer and let $A_n=a_{n_k}+a_{n_{k+1}}+...+a_{n_{k+1}-1}$ for $k \in N$. Prove that if $\sum_{n=1}^\infty a_n$ converges then $\sum_{k=1}^\infty A_k$ converges. Is the converse true?  I know that $\sum_{n=1}^\infty a_n$ converges implies that $lim _{n_&gt;\infty} a_n =0$, can I conclude that $A_n$ ->$0$ as well, even if I can conclude that, it still doesn't prove that $\sum_{k=1}^\infty A_k$ converges, am I correct?
99539
Let $(a_n)^{\infty}_{n=m}$ and $(b_n)^{\infty}_{n=m}$ be convergent sequences of real numbers. Let $x$ and $y$ be the real numbers $x:=\lim\limits_{n\to\infty}a_n$ and $y:=\lim\limits_{n\to\infty}b_n$. Show that the sequence $(\max(a_n,b_n))$ converges to $\max(x,y)$; in other words: $$\lim_{n\to\infty}\max(a_n,b_n)=\max\bigl(\lim_{n\to\infty}a_n,\lim_{n\to\infty}b_n\bigr)$$  I was not able to prove it and would appreciate your help.
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How can I show   $\lim_{n \to \infty} a_n = 0$  $a_n = {1.3.5 ... (2n-1)\over 2.4.6...(2n)}$  I have shown that $a_n$ is monotonically decreasing. I thought to shown sequence is bounded from below then it automatically would converge and hence my question will be solve. But I'm unable to show its boundedness... Or there maybe another method to prove this. Thanks
81067
I'm taking Calculus II next semester so I don't have all the tools from it to back me up but something came up in a discussion this evening that asked me to prove that the following sequence converges. As I'm new to formal math I'm hoping someone can verify my proof, especially step $6$. I must prove that $$\lim_{n\to\infty}P_n=b$$ where $$P_{n+1} = P_n + \frac{b-P_n}{2}$$ $$P_0 = a$$ and $a,b \in \mathbb{R}$ such that $a&gt;b$ My proof goes as follows:   We must prove that $$\lim_{n\to\infty}P_n=b$$  That is, $\forall\epsilon&gt;0$ there exists a $\delta$ such that there exists an $n$ satisfying $$n&gt;\delta \iff b-P_n&lt;\epsilon$$  First, let us prove that $P_n$ is bounded by $b$ by contradiction. Let us choose an $i$ such that $P_i &lt; b$ and $P_{i+1}&gt;b$. Then we have $P_{i+1} = P_i + \Delta P$ where $\Delta P &gt; b-P_i$. but this is impossible because by the definition of the recurrence $\Delta P = \frac{b-P_i}{2} &lt; b-P_i$. Therefore, $P_i$ is bounded from above by $b$.  Additionally, it is trivial to show that $P_{i+1} &gt; P_i$.  With this in mind, choose an arbitrary $\epsilon$. We need to show that there exists an $n$ such that $P_n &gt; b-\epsilon$. Assume the contrary, that $P_n$ is bounded by $\epsilon$.  Using a similar argument to what which was used before, assume $P_n$ is bounded by $\epsilon$ and then there exists a $P_i$ such that $P_{i+1}\not &gt; P_i$. But this results in a contradiction with (4)  Therefore, $P_n$ is not bounded by $\epsilon$ and we have proved what we want to prove.
108257
Find the limit of the sequence $a_n=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}.$  The limit is supposed to be $e^{-1}$ and there are a couple of posts in MSE proving it. But here is a proof I encountered showing the limit is $1.$ Please tell me what's going wrong here.      Proof:  We can write $a_n=\sqrt[n]{n!}(b_n-1),$ where $b_n=\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}.$ Hence $$a_n=n\cdot\frac{\sqrt[n]{n!}}{n}\cdot\frac{b_n-1}{\ln{b_n}}=\frac{\sqrt[n]{n!}}{n}\cdot\frac{b_n-1}{\ln{b_n}}\cdot\ln{b_n^n.}$$   But $\lim \limits_{n\to \infty}\frac{\sqrt[n]{n!}}{n}=e^{-1},$ so $b_n\to1$ as $n\to \infty.$ On the other hand, $$\lim \limits_{n\to \infty}b_n^n=\lim \limits_{n\to \infty}\frac{(n+1)!}{n!}.\frac{1}{\sqrt[n+1]{(n+1)!}}=e.$$   Also $$\lim \limits_{n\to \infty}\frac{b_n-1}{\ln{b_n}}=\lim \limits_{n\to \infty}\frac{1}{\ln{[1+(b_n-1)]^{1/(b_n-1)}}}=\frac{1}{\ln e}=1$$         And the author goes on to say that the limit is $e^{-1}.$ But doesn't it say that the limit is $1?$ I must have overlooked something so obvious.
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.
40641
Using only the delta definition of a limit, how can we prove that the sequence $\{a_n\}$, where $a_n = \sin n$, as $n$ tends to infinity does not have a limit?  Thanks!