I am stuck finding a way to prove/disprove this limit of a sequence.  Assuming that $\lim\limits_{n\to\infty}(a_{n} - a_{n-2})=0$ for the sequence $a_n$, prove or disprove that  $$\lim\limits_{n\to\infty}\frac{a_{n} - a_{n-1}}{n}=0.$$  An explanation on how to work this through would be very appreciated.
122062
If the sequence satisfies the property lim$_{n\to \infty}(a_n-a_{n-2})=0$, prove that lim$_{n\to \infty}\frac{a_n-a_{n-1}}{n}=0$.  I am experiencing difficulty showing this rather obvious result. By definition, we know that for any $\epsilon \gt 0$, there is some $N$ such that if $n \gt N$ then $|a_n-a_{n-2}|\lt \epsilon$ My guess is that I need to write $|a_n-a_{n-1}|$ as sums of $|a_n-a_{n-2}|$ from $n$ to $N$. However, I am stuck with forming an inequality here. I would appreciate any solutions or suggestions.
99093
There's a bounded sequence $\{{a_n}\} $ s.t. $\lim\limits_{n \rightarrow  \infty}(a_{2n} + 2a_n) = 0$. Try to prove $\lim\limits_{n \rightarrow  \infty}a_n = 0$ (You must prove the existence of its limit before   getting the exact value)   The above problem is located in the chapter about "limsup and liminf", so you may use their property while proving.  EDIT  By the way, I find it hard to do the exercises at the end of each section of my textbook though I DO UNDERSTAND almost everything in that section. So I'm eager to know some other books that can help me to solve problems like the one described above. That would be a great help, thanks!
98181
Let $a_n$ be a sequence s.t $a_1 &gt; 0 \land a_{n+1} = a_n + \frac{1}{a_n}$. Prove that $a_n$ is increasing and tends to infinity.  Proof:   Consider $a_{n+1} - a_n$:  $a_{n+1} - a_n = a_n + \frac{1}{a_n} - a_n = \frac{1}{a_n}$ This is greater than $0$. Thus, $a_n$ is increasing.  Now this is where I need some help. I would like to say that $a_n$ is unbounded and then conclude that monotone and unbounded implies tending to infinity.  Maybe by contradiction?
81542
I'm doing exercises. In the related book, there is a claim. Is this right? I'm not sure.     For a sequence $\{a_n\}$, there exists a limit $a$ such that $\lim_{n\to\infty} a_n=a$ if and only if for any $p\in \Bbb N$, $\lim_{n\to\infty} |a_{n+p}-a_n|=0$   If not, could you kindly give some counterexamples?  Thanks.
53981
Let $\{a_n\}_{n=1}^\infty$ be a positive sequence of real numbers such that $a_{n+m} \leq a_n + a_m.$  Prove that $\lim_{n\to\infty} \frac{a_n}{n}$ exists by showing that $\lim_{n\to\infty} \frac{a_n}{n} = \inf_{n \geq 1} \frac{a_n}{n}.$   A similar question has been posted previously, see  , and I have found that this is the general result of "Fekete's Subadditive Lemma" for sequences, but I am unhappy with the few proofs I've seen.  Perhaps the clearest one I have found is  , but this seems a bit tricky for what is expected on this qualifier question.  We are given a hint "treat $\liminf$ and $\limsup$ separately".  Does anyone know of a more direct way to show that this limit exists?
9802
Show that  $$\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i} =0 $$  I've proved that this sequence converges (it is bounded and decreasing). NOW, I need to find a sequence that is bigger than this one and goes to zero. Maybe something using geometric serie of 1/2  Thanks in advance!
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$.
28033
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!!!
18137
Let $a_n=\frac{n!}{n^n}$ be a sequence. Prove that $\limsup(a_n)=0$.  I know that the series $\sum^{\infty}_{n=1} \frac{n!}{n^n}$ converges, so the nth term must converge to 0. But I cannot use the convergence of series to prove this.   Basically what I have is: The sequence $a_n$ is decrescent and bounded by 0 $(a_n&gt;0 \; \forall n)$, therefore is convergent and it's limit is the $\limsup(a_n)$.   So I must show that $\lim\frac{n!}{n^n}=0$  Let $\epsilon&gt;0$, prove that exists $n_0 \in \mathbb{N}$ such that, if $n&gt;n_0$  $$\left|\frac{n!}{n^n}-0\right|&lt;\epsilon \Rightarrow -\epsilon&lt;\frac{n!}{n^n}&lt;\epsilon$$  I know that $\frac{n!}{n^n}&gt;=-\epsilon$ because $\frac{n!}{n^n}&gt;0$.  How can I prove that $\frac{n!}{n^n}&lt;\epsilon$?
53291
Problem  The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:   $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$.   Evaluate the limit $\lim_{n\to\infty} a_n$.  Solution   Show that the sequence $(a_n)_{n=1}^\infty$ is monotonic. The statement $$V(n): a_n &lt; a_{n+1}$$ holds for $n = 1$, that is $\sqrt2 &lt; \sqrt{2+\sqrt2}$. Let us assume the statement holds for $n$ and show that $V(n) \implies V(n+1)$. We have that $$a_n &lt; a_{n+1}.$$ Adding 2 to both sides and taking square roots, we have that $$\sqrt{2+a_n} &lt; \sqrt{2+a_{n+1}},$$ that is $a_{n+1} &lt; a_{n+2}$ by definition. Find bounds for $a_n$. The statement $$W(n): 0 &lt; a_n &lt; 2$$ holds for $n=1$, that is $0 &lt; \sqrt2 &lt; 2$. Let us assume the statement holds for $n$ and show that $W(n) \implies W(n+1)$. We have that $$0 &lt; a_n &lt; 2.$$ Adding two and taking square roots, we have that $$0 &lt; \sqrt2 &lt; \sqrt{2+a_n} &lt; \sqrt4 = 2.$$ The limit $\lim_{n\to\infty} a_n$ exists, because $(a_n)_{n=1}^\infty$ is a bounded monotonic sequence. Let $A = \lim_{n\to\infty} a_n$. Therefore the limit $\lim_{n \to\infty} a_{n+1}$ exists as well and $\lim_{n \to\infty} a_{n+1} = A$. (For $(n_k)_{k=1}^\infty = (2,3,4, \dots)$, we have that $(a_{n_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, from which the statement follows.) We have that $a_{n+1} = f(a_n)$. That means that $A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}$. Solving the equation $A = \sqrt{2 + A}$, we get $A = -1 \lor A = 2$. Putting it all together, we get that $A = 2$, because the terms of the sequence are increasing and $a_1 &gt; 0$.   Is my solution correct?
87031
Let $(x_n)_n$ be a sequence of real numbers such that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0.$$ Show that $(x_n)_n$ must also converge to zero.     $\textbf{My Attempt:}$   My original idea was to use the triangle inequality and the fact that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0$$   to show that the sequence $(x_n)_n$ is Cauchy and hence converges to some real number $a$, which would give me $$\lim\limits_{n\rightarrow \infty}x_n=a=\lim\limits_{n\rightarrow \infty}x_{n+1}\implies a-\dfrac a 2= \lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0\implies a=0.$$ However, I'm having trouble showing that $(x_n)_n$ is Cauchy.   Any suggestions/hints would be greatly appreciated. Thanks!
79337
Suppose $a_0$ is an arbitrary positive real number. Define the sequence $\{a_n\}$ by $$a_{n+1}=\frac{a_n+2}{a_n+1}$$ for all $n\geq0$. I have to prove that $\{a_n\}$ converges.  My attempt: If  $a=\lim_{n\to\infty}{a_n}$ exists, then it should be a solution to $$a=\frac{a+2}{a+1}$$ which is $\sqrt2$. Thus I need to show that $|\sqrt2 - a_n|$ gets arbitrarily small for large $n$. I tried to prove that $|\sqrt2-a_n|&lt;|\sqrt2-a_{n-1}|$ but couldn't.
56344
Let $\{a_n\}$ be defined as follows: $$a_1 = 2, a_{n+1}=\dfrac{1}{3-a_n}$$ $n \geq 1$.   Does $\{a_n\}$  converge?    Using the monotone convergence thereom, if the sequence is both bounded below and decreasing, $\forall n\in\mathbb{N}, n\geq 1$, then we can say the sequence converges.  I'm having trouble proving that the sequence is bounded below by induction.    Claim: The sequence is bounded below by $0$.  Base Case:  $n=1$  $a_1 = 2 \geq 0$, so this holds.  Induction Step  Suppose that $a_k \geq 0$, for some $k \in\mathbb{N}$. (IH)  Prove $a_k \geq 0\to a_{k+1} \geq 0$  I have some difficulty here.  $a_{k}=\dfrac{1}{3-a_{k-1}} \geq 0$  $a_{k+1}=\dfrac{1}{3-a_{k}}$, but we know $a_k \geq 0$, but we don't know how much bigger. It could be that $a_k = 4$, and then I have $a_{k+1}=\dfrac{1}{3-4}$, which is negative, and does not confirm with my claim of being bounded below by $0$. Have I gone somewhere wrong in my induction?
28260
Assume that $$\lim_{n\to\infty}(a_{n+2}-a_{n})=A$$ show that  $\displaystyle \lim_{n\to\infty}\dfrac{a_{n}}{n}$and $\displaystyle\lim_{n\to\infty}\dfrac{a_{n+1}-a_{n}}{n}$ exist and find these limits.  maybe this problem have some methods, Thank you.  since  $$\lim_{n\to\infty}(a_{n+2}-a_{n})=A$$ so there exists $N$, and for $\forall \varepsilon&gt;0$, such $$|a_{n+2}-a_{n}-A|\le\varepsilon$$
7032
I've spent the better part of this day trying to show from first principles that this sequence tends to 1. Could anyone give me an idea of how I can approach this problem?  $$ \lim_{n \to +\infty} n^{\frac{1}{n}} $$
122061
Let $a_n$ be an infinite sequence. The limit of the difference of two consecutive members is equal to $0$. Can we conclude that the sequence itself has a limit?  My attempt:  We have  $$ \lim_{n\rightarrow\infty}{a_n} - \lim_{n\rightarrow\infty}{a_{n-1}} = 0 $$ since as $n$ approaches infinity $a_{n-1}$ gets arbitrarily close to $a_n$ the sequence cannot diverge or be bounded but have no limit.  Is my proof correct and how would I be able to formalize the last sentence? Thanks
5889
I will post the exercise below:      Prove that the sequence $(a_n)$ defined by $a_0 = 1$, $a_{n+1} = 1 + \frac 1{a_n}$ for $n \in \mathbb N$ is convergent in $\mathbb R$ with the Euclidean metric, and determine afterwards is limit. Can you intepret the limit geometrically (hint: Golden ratio)?   So I need to prove that the sequence is convergent in $\mathbb{R}$  with the Euclidean metric, and how do I prove that? The limit must be $1$, but how to interpret it geometrically?