We were told, in recitation class, about a test for sequences convergence (not series) Which goes as follows:  if $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}=L$ then $\lim_{n \rightarrow \infty} \sqrt [n] {a_n}=L$.  In a previous question I asked:    I was told that this fact is not true. My question is, can anyone think of a counter example for it? Because, If yes, Then I would like to let my tutor know about it, but, I don't want to doubt him befor I am sure of it.  Thank you!
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Today my lecturer put up on the board that:  If $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}$ exists and $a_n&gt;0$ then   $\displaystyle \limsup\limits_{n\to\infty}\left(a_n^{\frac{1}{n}}\right)=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$  however I am not sure why this is true, can somebody give me a hint or something as to how to go about proving this.  thanks for any help
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Note: $x_n$ is a sequence which is not necessarily convergent.    The following was my attempt. Since $\lim_{n\to \infty}a_n=a$ then $\limsup_{n\to \infty}a_n=a$ . Also $\sup(a_nx_n)=\sup(a_n)\sup(x_n)$. Therefore, $\limsup_{n\to \infty}a_nx_n=\limsup_{n\to \infty}a_n\limsup_{n\to \infty}x_n=a\limsup_{n\to \infty}x_n$.    I am not sure if this is correct. So could someone please show me how it is done. Thanks
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I want to prove the following theorem     Let $\{a_{n}\}$ be a convergent sequence. Prove that $\lim\limits_{n\to \infty}a_{n+1}=\lim\limits_{n\to \infty}a_{n}$   Do I need to show that it is monotone increasing and then use the relation  $$0\leq \lim_{n\to \infty}a_{n+1}- \lim_{n\to \infty}a_{n}\leq 0 \;?$$  Please, I need help on this! Various proofs are welcome! Thank you very much for your time!
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Ch.3 #7. Prove that the convergence of $\sum a_n$ implies the convergence of $$\sum \frac{\sqrt{a_n}}{n},$$ if $a_n \geq 0$.      My attempt. If $\sum a_n$ is convergent, then by the root test, $\lim_{n \rightarrow \infty} sup \sqrt[n]{|a_n|} &lt; 1$. This means $|a_n|^{1/n} &lt; 1$. Because $a_n \geq 0$, this means $a_n ^{1/n} &lt; 1$. Now apply the root test to $\sum \frac{\sqrt{a_n}}{n}$ by considering $$\lim_{n \to \infty} sup \sqrt[n]{|\frac{\sqrt{a_n}}{n}|}.$$ $$\sqrt[n]{|\frac{\sqrt{a_n}}{n}|} = \frac{\sqrt{{a_n}^{\frac{1}{n}}}}{\sqrt[n]n}.$$ Because  $a_n ^{1/n} &lt; 1$, certainly the numerator on the left-hand side of the equation above is $&lt;1$. And because $n$ tends to $\infty$, and $n &gt; 1 \rightarrow \sqrt[n]n &gt; 1,$ the denominator is $&gt;1$. Therefore the limit is $&lt;1$ and the root test implies that $\sum \frac{\sqrt{a_n}}{n}$ converges.     If there are any errors in my proof, I think they come from 1) my assumption of the converse of the root test theorem and 2) my misunderstanding of $\lim_{n\to\infty} sup$. Thanks in advance for your help.
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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
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Let $a_1=1$ define $a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n})$ . Show that the sequence converges to $\sqrt{2}$. The idea is to show the sequence is bounded and monotone . But how should I do it?
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I want to show that a sequence $(a_n:n \in \mathbb N)$ converges to a limit $L \in \mathbb R$ if, and only if, an N-th tail of the sequences converges to $L$ for some $N \in \mathbb N$. For each $N \in \mathbb N$, an N-th tail sequence can be defined as $(t_n:n \in \mathbb N)$ where $t_n := a_{N+n}$.  Using the definition of a limit for the sequence $(a_n:n \in \mathbb N)$, we have the following:  For every $0 \lt \epsilon$, there is a natural number $N_{\epsilon} \in \mathbb N$ such that for all $N_{\epsilon} \lt n$, $|a_n - L| \lt \epsilon$. But I'm not sure what to do next in order to show that the $N$-th tail converges. Which natural number should I choose for the $N$-tail? Any help would be much appreciated.
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Here's the problem in full. I've stared at it for hours and can't get anywhere, so a hint would be nice.  Suppose that $f:\mathbb N\rightarrow\mathbb R$. If $$\lim_{n\rightarrow\infty}f(n+1)-f(n) = L$$ prove that $\lim_{n\rightarrow\infty}f(n)/n$ exists and equals $L$.  This problem is marked with the word "Cauchy", so I'm guessing Cauchy sequences or something related to them will be necessary, but I can't figure out what.  Edit: It appears as if the Stolz-Cesaro Theorem makes this quite easy. Unfortunately though, we have not discussed this, so I won't be able to use this.
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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 $\limsup$?
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If $ (x_{n})_{n \in \mathbb{N}} $ is a sequence in $ \mathbb{R} $ and $ \displaystyle \lim_{n \to \infty} (2 x_{n + 1} - x_{n}) = x $, then is it necessarily true that $$ \lim_{n \to \infty} x_{n} = x? $$  Could anyone kindly offer suggestions on how to approach this problem? Thanks!
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a) Let $a&gt;0$ and the sequence $x_n$ fulfills $x_1&gt;0$ and $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$ for $n \in \mathbb N$. Show that $x_n \rightarrow \sqrt a$ when $n\rightarrow \infty$. I have done it in two ways, but I guess I'm not allowed to use the first one and the second one is incomplete. Can someone please help me?  We already know $x_n \rightarrow \sqrt a$, so we do another step of the iteration and see that $x_{n+1} = \sqrt a$.  Using limit, $x_n \rightarrow x, x_{n+1} \rightarrow x$ (this is the part I think it's incomplete, don't I have to show $x_{n+1} \rightarrow x$, how?), we have that   $$x = \frac x 2 (1 + \frac a {x^2}) \Rightarrow 1 = a/x^2 \Rightarrow x = \sqrt a$$ b) Let the sequence $x_n$ be defined as $x_{n+1}= 1 + \frac 1 {x_n} (n \in \mathbb N), x_1=1$. Show that it converges and calculate its limit. &quot;Tip: Show that sequences $x_{2n}$ and $x_{2n+1}$ monotone convergent to the limit.&quot; I didn't understand the tip, how can this help me? Does it make a difference if the number is odd or even? Thanks in advance!
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A student was recently asked this question by his instructor:  $$\sum_{n=1}^\infty{\left(\sqrt[n]{n}-1\right)}$$  Converge or diverge?  I feel a little dumb for not being able to answer it. The following tests fail to prove convergence or divergence:  nth term test for divergence (limit is 0), ratio test (limit is 1), root test (see ratio test), limit comparison with $\sqrt[n]{n}$ (not sure why I thought that'd work)  Something I did try was using the fact that  $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\;,$$  to rewrite $\sqrt[n]{n}-1$ as  $$\frac{n-1}{n^{1-1/n}+n^{1-2/n}+\cdots+n^{2/n}+n^{1/n}+1}\;.$$  However, I'm not sure what to compare this to. According to wolfram alpha this series "diverges by the comparison test", but comparison to what? There is a similar problem in Baby Rudin, but for $(\sqrt[n]{n}-1)^n$, and a simple nth root test resolves that series [convergence] in a hurry. Any ideas? Have any of you encountered a similar looking series before? Thanks.
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I need to check if $$\lim_{n \to \infty} \frac{n}{ \sqrt [n]{n!}}$$ converges or not. Additionally, I wanted to show that the sequence is monotonically increasing in n and so limit exists. Any help is appreciated. I had tried taking log and manipulating the sequence but I could not prove monotonicity this way.
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I ask for some help with this question:  Prove or provide counter example:  If $\sum\limits_{n=1}^\infty na_n$ converges then $\sum\limits_{n=1}^\infty na_{n+1}$ also converges.  I tries this way:  If $\sum\limits_{n=1}^\infty na_n$ converges then $na_n \to 0$, therefore $a_n \to 0$.  There are 3 possible cases:  1) If $a_n &gt;0 $ and $a_n$ is monotonic decreasing sequence then $na_{n+1}&lt;na_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.  2) If $a_n &gt;0 $ and $a_n$ is not monotonic decreasing sequence : it is not possible that $a_{n+1}&gt;a_n$ because in this case $a_n \to \infty$, therefore it must be $a_{n+1} \le a_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.  3) If $a_n$ is sign-alternating series. There I have a problem to find a solution.  Thanks.
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I'm currently tackling the following infinite series in order to determine if it converges or not:  $$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt{n}}{n}$$  What I've done is rearranged this to make use of the comparison test, shown below:  $$= \frac{\sqrt{n+1}-\sqrt{n}}{n} * (\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}})$$  $$= \frac{(n+1)-(n)}{n(\sqrt{n+1}+\sqrt{n})}$$  $$= \frac{1}{n(\sqrt{n+1}+\sqrt{n})}$$  $$= \frac{1}{n^{3/2}+\sqrt{n+1}} &lt; \frac{1}{\sqrt{n+1}} &lt; \frac{1}{\sqrt{n}} $$  However, $\frac{1}{n}$ is smaller than $\frac{1}{\sqrt{n}}$, therefore:  $$  \frac{1}{n}   &lt; \frac{1}{\sqrt{n}}$$   This implies divergence by the comparison test due to the fact $\frac{1}{n}$ diverges, and if a smaller series diverges any larger series must as well.  Is this right? If I've made any mistakes in any part, please tell me!
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I am trying to conclude about the convergence of $$\sum_{n=1}^{\infty}{\sqrt[n]{n}-1 \over n}$$ If I take ${\sqrt[n]{{\sqrt[n]{n}-1 \over n}}}= {\sqrt[n]{\sqrt[n]{n}-1} \over \sqrt[n]{n}}$ this looks like it converges to $0$ but I am suspicious about the numerator. At the same time, other techniques I know like the ratio test, $n(\sqrt[n]{n}-1)$ and comparisons don't yield meaningful results. Any hints?
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Show that $a_n=\sin(n)$ does not converge     My idea:  Take two subsequences: $a_{n_k}=\sin(\frac {\pi k} 2)$ , $a_{n_l}=\sin(\frac {2\pi l} 3)$  So: $\forall n$ : $\lim_{n\to\infty} a_{n_k}=1$, $\lim_{n\to\infty} a_{n_l}=-1$  So two inifinite subsequence converge to different limits thus the sequence $a_n$ doesn't converge.   Is that correct ?  Edit:   The contrapositive of the defintion of a limit of a sequence:   $\exists\epsilon&gt;0 : \forall n\in N : \exists n&gt;N \Rightarrow |x_n-L|&gt;\epsilon$  Take $\epsilon =1$ and we know that $\sin(n)$ is bounded so lets take it's supermum: 1  $|1-1|&gt;\epsilon=1\Rightarrow 0&gt;1 \Rightarrow$ Contradiction.