Let $a \in \mathbb{R}$ and $\lim b_n = b$ (for a real sequence $b_n$). I am trying to show that $\lim a^{b_n}=a^b$ by using the definition of the convergence of a sequence, but I cannot see how to estimate $|a^{b_n} - a^b|$ against $|b_n - b|$ properly.
121580
Let $(x_n)$ and $(y_n)$ be two sequences of real numbers, such that:  $\lim x_n = a &gt; 0$ and $\lim y_n = b \in \mathbb R$  I need to prove that:  $$\lim \left( x_n ^{y_n} \right) = a^b$$  I tried using Cauchy's criterion:  $|x_n-a|&lt;\epsilon$ and $|y_n-b|&lt;\epsilon$, and then we need to prove that $\left| x_n ^{y_n} -a^b\right|&lt;\epsilon$, but I didn't find a way of proving it.
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.
67631
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 $\lim_{n\to  \infty} (a_n b_n) = a_\infty b_\infty$.
31851
How can I prove the convergence of the sequence $b_n=\sqrt[n]{x^n+y^n}$ where $x, y &gt; 0$?  Can I divide it in two cases?  Case 1: $x &gt; y$.  $$ b_n=\sqrt[n]{x^n+y^n} &lt; \sqrt[n]{x^n+x^n} = \sqrt[n]{2 \cdot x^n}=x \cdot \sqrt[n]{2} $$  Case 2: $x &lt; y$.  $$ b_n=\sqrt[n]{x^n+y^n} &lt; \sqrt[n]{y^n+y^n} = \sqrt[n]{2 \cdot y^n}=y \cdot \sqrt[n]{2}$$  Result: Does the sequence converges to $\max(x,y)$?
52746
Does the condition $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?  When I say convergent I mean to $L\in \mathbb{R}$ or to $\infty$ or to $-\infty$  I have a gut feeling this statement is false, just can't find a series which contradicts it.  Thanks a lot
52733
If $x\in \Bbb R$ ; then is the sequence $\{a_n\}$ where $a_1=x$; $a_{n+1}=\cos (a_n)$ convergent?  Obviously $|a_n|\le1$  and hence $(a_n)$ is bounded.  Also $f(x)=\cos x$ is decreasing for $x&gt;0$.  But here $x\in \Bbb R$ .How to proceed here?Please help.
114069
$U(n)$ and $V(n)$ are sequences. $U(n)&gt; 0$  and   $V(n) &gt;0$ such that: ${V(n+1)\over V(n)}≤ {U(n+1)\over U(n)}$.   Prove that:   If $\,\,\lim\limits_{n\to\infty} V(n)= +\infty\,\,$ then $\,\,\lim\limits_{n\to\infty}  U(n) = +\infty$ If $\,\,\lim\limits_{n\to\infty} U(n) = 0 \,\, $  then  $\,\,\lim\limits_{n\to\infty} V(n) = 0$   Can anyone help ?
112943
If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$  Since $(a_n)$ is decreasing and bounded below, by Monotonic Convergence Theorem, $(a_n)$ converges. So, there exists $N$ such that $|a_n-L|&lt; \epsilon$   Since $\sum{a_n}$ is convergent, by the Divergence test, we have $\lim_n{a_n}=0$, which means there exists $N$ such that $|a_n-0|&lt; \epsilon$  Then I get stuck at here.  I try to figure out the statement's meaning by inserting some examples, like $a_n=\dfrac{1}{n^2}$  Can anyone guide me on this question?
80659
Let $(a_k)$ be a sequence of real numbers and let $b_k=\frac{a_1+a_2+\dots a_k}{k}$ for each $k\in \mathbb{N}$.  Prove that if $(a_k)$ converges to $\alpha\in \mathbb{R}$, then the sequence $(b_k)$ also converges to $\alpha$.  My answer: Let $\epsilon &gt; 0$. Since ($a_k$) converges to $\alpha$, then there exists a $N_{\epsilon1}\in\mathbb{N}$ such that $|a_k - \alpha | &lt; \frac{\epsilon}{2}$ for all $k \geq N_{\epsilon1}$. Notice that $\frac{a_1+...+a_n - n\alpha}{k}$ converges to $0$ as $k$ approaches infinity. Thus, there exists $N_{\epsilon2}\in\mathbb{N}$ such that $|\frac{a_1+...+a_n - n\alpha}{k}|&lt;\frac{\epsilon}{2}$ for all $k&gt;N_{\epsilon2}$. Now, let $N= $max$ (N_{\epsilon1},N_{\epsilon2})$. Then for any $k&gt;N$ we get,  $$\begin{array}{rcll} |b_k-\alpha| &amp;=&amp; |b_k - \alpha*1|&amp;\\ &amp;=&amp; |\frac{a_1+...+a_k}{k} - \frac{k\alpha}{k}|&amp;\\ &amp;=&amp; |\frac{a_1+...+a_n}{k} + \frac{[a(n+1)+...+a_k]}{k} - \frac{(k-n)\alpha}{k} - \frac{n\alpha}{k}|&amp; \text{(notice that $N_{\epsilon1}\leq N&lt;k$)}\\ &amp;=&amp; |\frac{a_1+...+a_n - n\alpha}{k} + \frac{[a(n+1)+...+a_k - (k-n)\alpha]}{k}|&amp;\\ &amp;=&amp; |\frac{a_1+...+a_n - n\alpha}{k} + \frac{[(a(n+1) -\alpha) + ... + (a_k - \alpha)]}{k}|&amp;\\ &amp;=&amp; |\frac{a_1+...+a_n - n\alpha}{k} + \frac{(a(n+1) -\alpha)}{k} + ... + \frac{(a_k - \alpha)}{k}|&amp;\\ &amp;\leq&amp; |\frac{a_1+...+a_n - n\alpha}{k}| + |\frac{(a(n+1) - \alpha)}{k}| + ... + |\frac{(a_k - \alpha)}{k}|&amp; \text{(by the triangle inequality)}\\ &amp;=&amp;|\frac{a_1+...+a_n - n\alpha}{k}| + \frac{|(n+1) - \alpha|}{k} + ... + \frac{|a_k - \alpha|}{k}&amp;\\ &amp;&lt;&amp; \frac{\epsilon}{2} + [\frac{\epsilon}{2}/k + ... + \frac{\epsilon}{2}/k]&amp; \\ &amp;=&amp; \frac{\epsilon}{2} + \frac{\epsilon}{2}*\frac{(k-n)}{k}&amp;\\ &amp;&lt;&amp; \frac{\epsilon}{2} + \frac{\epsilon}{2}&amp; \text{(since $k-n&lt;k$)}\\ &amp;=&amp; \epsilon&amp; \end{array}$$ And since we found such an $N$ for arbitrary $\alpha$, then $(b_k)$ converges to $\alpha$ by definition.  I'm not sure i kept all my variables straightened out. How does this look and if you have a better idea let me know!
24384
Define the recursive sequence $(a_n)$ as $a_{n+1} = a_n^2 + a_n$ where $n \in \mathbb{N}$. We want to find $a_1 \in \mathbb{R}$ such that $(a_n)$ converges to a limit that depends on $a_1$.   Attempt: Let $a_1 = k$, then $a_2 = k^2 + k$, $a_3 = (k^2 + k)^2 + (k^2 + k)$, ... clearly if we have |k| > 1 then the sequence $(a_n)$ will diverge. From this we deduce the interval of convergence must be when $|k| \le 1$. Suppose $(a_n) \to L$ then $a_{n+1} = a_n^2 + a_n$ $\implies L = L^2 + L \implies L = 0$. Surely if $k = 0$ this will occur as $a_n = 0$ for all n. But it is not clear to me if there are other $k$ such that the recursive sequence converges.
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$$$\left|\frac1{a_n}\right|&lt;\epsilon$$
71569
So one of my professors proposed a problem to me and it has stumped me for some time now. Here's how it goes:  Suppose you have a sequence $a_n$ of real numbers such that $$\lim_{n\to\infty} a_{n} = 0$$ and suppose the sequence of partial sums $s_n$ is bounded. Prove that $s_n$ converges or give a counterexample.  I'm hoping to figure this out without anyone handing me the complete solution, so if someone could point me in the right direction with a hint, it would be much appreciated.
44746
Let $c&gt;0$ be a real number. I would like to study the convergence of $a_n := c^n n!/n^n$, when $n \to \infty$, in function of $c$.   I know (from  question) that $n!&gt;(n/e)^n$, so that $c^n n!/n^n&gt;1$ for $c ≥e$. But this doesn't imply that the sequence goes to infinity. And I'm not sure what to do for $c&lt;e$. I tried usual tests (D'Alembert...), without any success. I would like to avoid using Stirling approximation.  Thank you for your help!
83427
I have been trying to solve this problem for awhile, and I can't seem to solve it.Below is the question. I suspect that the series does converge, but I am unable to prove this. Thanks for any help in advance.  Let $(a_n)_{n=1}^\infty$ be a convergent sequence of real numbers. Can it be concluded that $\sum\limits_{n=1}^{\infty}|a_{n+1} - a_{n}|$ converges? Give a proof or a counterexample.
66776
Let $(x_n)$ be a bounded sequence of real numbers such that $x_n \ge0$ for each $n \in \mathbb N$. Show that $\liminf(\sqrt{x_n})=\sqrt{\liminf x_n}$.   I assume we need to consider the subsequence of ($\sqrt{x_n}$), but I'm really confused with the  notion on "boundedness". Can I assume the question implies an upper bound on $(x_n)$?  Even then wouldn't the equation be implied? Given how roots generally affect limits.
86785
Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence with $\lim\limits_{n\to\infty}x_n = a$, $\{t_n\}_{n\in\mathbb{N}}$ be a sequence with $\lim\limits_{n\to\infty}t_1+t_2+\ldots+t_n = +\infty$.  Prove that  $$\lim_{n\to\infty}\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} = a$$  My idea is use that  $$(t_1+\ldots+t_n)\min\{x_n: n\in\mathbb{N}\}\leq t_1x_1+\ldots+t_nx_n \leq (t_1+\ldots+t_n)\max\{x_n: n\in\mathbb{N}\}$$  But I'm not sure if $\lim\limits_{n\to\infty}\min\{x_n: n\in\mathbb{N}\} = \lim\limits_{n\to\infty}\max\{x_n: n\in\mathbb{N}\} = \lim\limits_{n\to\infty}x_n$.
7641
The problem states:  a sequence of real numbers $a_0, a_1, \dots $ is defined as follows. $a_0$ is an arbitrary real number and for $n \ge 0, a_{n +1} = \lfloor a_n \rfloor \{ a_n \}$ (where $\{x\} = x - \lfloor x \rfloor$).  Prove that for $a_n = a_{n+2}$ for $n$ large enough.  my attempt:  I notice that this is a decreasing sequence because $$\frac{a_{n+2}}{a_{n+1}} = \frac{\lfloor a_{n+1} \rfloor \{ a_{n+1} \}}{ a_{n+1}} &lt; 1$$  since $\{ a_{n+1} \} \le 1$ (we can exclude the case where it's equal to one since then it becomes the null sequence that satisfies the thesis).  Moreover this sequence is bounded between $0$ and $a_0$ so by the monotone convergence theorem it must have a limit.  Taking the limit I obtain $$\ell = \lim \lfloor a_n \rfloor (\ell - \lim \lfloor  a_n \rfloor) \iff \ell = - \lim \lfloor a_n \rfloor^2 / (1 - \lim \lfloor  a_n \rfloor) $$  And so if I prove that there exists $n$ s.t. $ \lfloor a_n \rfloor &lt; 1$ I am done. Because then the limit could only be $0$ and this implies that there exist infinite $a_n = a_{n+2} = 0$.  But I can't seem to prove it. Is my reasoning up to now ok? How could I proceed?  Edit:  Add original text of the problem.