We have a sequence:  $$a_n=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$  We need to show that it is rising and above unbounded.  So I did the following:  $$a_{n+1}=a_n+\frac{1}{n} $$  $$a_{n+1}&gt;a_n $$  $$a_n+\frac{1}{n}&gt;a_n $$  $$\frac{1}{n}&gt;0$$  $$n&gt;0$$   Which holds.   First question: Is the upper proof enough to satisfy that the sequence is rising  As far as I understand, I need to also now show that supremum doesn't exist.  So I have done the following:  $$ \sup{a_n}=\lim{a_n} = M $$  $$M&gt;0$$  $$M&gt;a_n$$ for every $n$  $$a_n+\frac{1}{n}&lt;M$$  $$1&lt;(M-a_n)n $$  Because $M&gt;a_n$ we can divide without sign direction change  So we get:  $$\frac{1}{M-a_n}&lt;n$$  Which I think isn't the end of the proof. I do not know how to continue.
2411
Can someone give a simple explanation as to why the       $$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$   doesn't converge, on the other hand it grows very slowly?   I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.
80485
$(a_n)$ is a bounded sequence with the following condition  $a_{n+1}\geq a_n-\frac{1}{2^n}$   The sequence converges, but how do we show it?
48112
This question came from the prelim exam I took last month. I have a proof that seems a bit unwieldy to me (posted as an answer), so I'm opening it up to ask if there are other ways of showing this.     Let $x$ be any positive real number, and define a sequence $\{a_n\}$ by   $$ a_n = \frac{ [x] + [2x] + [3x] + \dotsb + [nx] }{n^2} $$   where $[x]$ is the largest integer less than or equal to $x$. Prove that $\displaystyle{\lim_{n \to \infty} a_n = x/2}$.
104218
I am triyng to solve this interesting problem, but I have no idea. I tried induction, but it didn't work. Let $(x_n)_{n \ge 1}$ be a sequence, where $x_1=1$ and defined as follows: $$x_{n+1}=x_n+ \frac {1} {3x_n^2},$$ for every $n \ge 1$. Show that there exists a real number $c \gt 0$ so as $x_n \lt \sqrt[3] {n+c}.$
42323
Let $S\subset\mathbb{R}$ be a non-empty bounded above set. Then there exists a monotone increasing sequence $\{x_n\}\subset S$ such that    $$\lim_{n\to\infty}x_n=\sup S.$$   I'm struggling with creating a formal proof for this problem. I understand that this has to be true, but I'm unsure of how to go about proving it.  I said that $\sup S = C$ and $x_n= C - y/n$ where $y$ is dependent on the size of $S$, but this is just a bunch of jumbled ideas and I can't really think of a way to formalize a proof. Any idea, even totally different from mine would be appreciated.
57008
Suppose that $(u_n)$ is unbounded above. Then, we pick any arbitrary monotone increasing subsequence $(v_n)$ of $(u_n)$. But by hypothesis, we can find a subsequence of $(v_n)$ that converges to $0$. Hence, $\lim v_n =0$. Therefore, $(u_n)$ must be bounded above.  Again, let $(u_n)$ be unbounded below. Then we pick any monotone decreasing subsequence $(w_n)$. By similar arguments, $(u_n)$ must be bounded below.   Being bounded, it has a finite $\limsup =l$ and $\liminf =m$.  Hence, there must be convergent subsequences $(a_n)$ and $(b_n)$ that converge to $l$ and $m$ respectively. But every subsequences of the above two must converge to $l$ and $m$ respectively (treating the two as convergent sequences). Again, by hypothesis, they both have subsequences that converge to $0$. Therefore, both $(a_n)$ and $(b_n)$ converge to $0$.   In conclusion, $\limsup u_n = \liminf u_n = 0 =\lim u_n$  Is this alright?
97754
So My professor assigned this question and I am really stuck on part B of the question.  For $n \in \mathbb{N}$ let {$T_n$} = {$1 + \frac{1}{1!} + \frac {1}{2!} + \cdots + \frac{1}{n!}$}.  (a) Prove that {$T_n$} is non-decreasing.  (b) Use {$S_n$} = {$(1+\frac{1}{n})^n$} to prove that {$T_n$} is bounded above and prove that $lim_n(T_n)\ge lim_n (S_n)$   Like I mentioned I can prove part (a) simply by supposing that $T_{n+1} \lt T_n$ then after some algebra concluding the statement is false and than with one other example proving that its not strictly increasing either.  Now for part (b) I have not a slightest clue about how to solve, I assume I will have to use the expanded form of $S_n$ but thats the best I have. So please give me some hints or nudges in the right direction.
87463
Assume that $(a_n)$ is a sequence of positive real numbers satisfying $a_{n+1}=\frac{2}{a_n+a_{n-1}}$ for $n=2,3,\dots$.  Prove that $(a_n)$ is convergent and find the limit.    I have no idea how to prove convergence. I tried to prove that $(a_n)$ is bounded and monotone, but both trials failed.  If convergence is proven, the limit is easy: if $g=\lim a_n$, then $g\ge 0$ because all $a_n$'s are positive, the equivalent relation $a_{n+1}(a_n+a_{n-1})=2$ gives $g(g+g)=2$, so finally $g=1$.
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!
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?
88487
Let we have the following sequence $(a_n)$ such that $$a_n=\sum_{k=1}^n\frac{1}{n+k}$$ How can I prove that $(a_n)$ is increasing bounded sequence, then prove it is convergent and find its limit?
84065
I would appreciate a little help in finalizing a proof for the following:  Let $a_n$ be the sequence defined as $a_1 = 1$, $a_2 = 8$, and $a_n = a_{n-1} + 2*a_{n-2}$ when $n \geq 3$. Prove that $a_n = 3*2^{n-1} + 2(-1)^n$.  I decided to use strong induction and show that if the statement is true for $1,...,n$ , then it is true for $n+1$ (going off the fact that the statement is true for $n=3$). I have: $$a_{n+1} = a_n + 2*a_{n-1} = 3*2^{n-1} + 2(-1)^n + 2(3*2^{n-2} + 2(-1)^{n-1}) = 3*2^n + 2(-1)^n + 4(-1)^{n-1}$$  Which is close to the result I want $(3*2^n + 2(-1)^{n+1})$ but the signs are switched for the $(-1)$ terms; therefore I suspect I missed a $-1$ somewhere, but I could not see where the error is.  I appreciate all and any help. Thank you kindly!
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!
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.
112564
$|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.  let $|a_2 - a_1 |  =x$.   $|a_3-a_2| \leq \frac{1}{2}x$  $|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$  In general, for every $n$:  $|a_{n+1}-a_n| \le \frac {1}{2^{n-1}}x$  Let $\epsilon &gt;0$  for every $m,n &gt; 0 $, say $m&gt;n$ the following holds:  $$|a_m-a_n| = |a_m-a_{m-1} + a_{m-1} - a_{m-2} \dots a_{n+1}- a_n| \le$$  $$|a_m-a_{m-1}| + |a_{m-1} - a_{m-2}| + \dots |a_{n+1}-a_n| \leq \frac{x}{2^{m-2}} + \frac{x}{2^{m-3}} + \dots + \frac{x }{2^{n-1}}$$.  I can say that for big enough $n,m$, $|a_m-a_n|&lt;\epsilon$ and therefore, this sequence is Cauchy? It sounds not good enough to me yet I don't know what else to do here.
68710
I am trying to prove that the sequence $$a_n = (n!)^{\frac2n}$$ tends to infinity as $ n \to \infty $.  I've tried different methods but I haven't really got anywhere. Any solutions/hints?
104768
Show the sequence isn't bounded: $a_1=1$, $a_{n+1}=a_n+\frac 1 {a_n}$.   Proof by contradiction: Let $M&gt;0$ such that $\forall n: |a_n|&lt; M$.   Let $\epsilon &gt;0 $ and for some $n=N, \epsilon: a_N=M-\epsilon&lt;M $ pluging that in the recursion: $a_{N+1}=M-\epsilon+\frac 1 {M-\epsilon}&gt;M&gt;M-\epsilon$. Contradiction.  I wondered if I could suppose about the boundary that $\forall n: |a_n|\le M$ ? The proof would basically be the same only I could drop the epsilon.