Does the sequence $t_n=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n}}} (a_k&gt;0,k=1,2,\cdots)$ converge? I need to prove that if $$ \limsup\limits_{n\to\infty} \dfrac{\ln\ln a_n}{n}&lt;\ln 2, $$ the sequence $\{t_n\}$ converge. (When $a_n\le 1$, $\ln\ln a_n$ is defined to be $-\infty$.) I have no idea about this question. But if I consider the sequence $$ p_n^{(r)} = \sqrt[r]{a_1+\sqrt[r]{a_2+\cdots+\sqrt[r]{a_n}}}, r\in\mathbb{Z}_+$$ and in particular, when $r=1$, I have that $$ p_n^{(1)} = \sum_{k=1}^n a_k, $$ thus, if $$ \limsup\limits_{n\to\infty}\dfrac{\ln a_n}{n}&lt;0(=\ln 1), $$ it's easy to obtain that $\{p_n^{(1)}\}$ converges. I guess there is a connection but I don't know how to  continue. Any hint?
44571
In my answer to the recent question , @GEdgar correctly raised the issue that the proof is incomplete unless I show that the intermediate expressions do converge to a (finite) limit. One such quantity was the nested radical  $$ \sqrt{1 + \sqrt{1+\sqrt{1 + \sqrt{1 + \cdots}}}} \tag{1} $$  To assign a value $Y$ to such an expression, I proposed the following definition. Define the sequence $\{ y_n \}$ by:  $$ y_1 = \sqrt{1}, y_{n+1} = \sqrt{1+y_n}. $$ Then we say that this expression evaluates to $Y$ if the sequence $y_n$ converges to $Y$.   For the expression (1), I could show that the $y_n$ converges to $\phi = (\sqrt{5}+1)/2$. (To give more details, I showed, by induction, that $y_n$ increases monotonically and is bounded by $\phi$, so that it has a limit $Y &lt; \infty$. Furthermore, this limit must satisfy $Y = \sqrt{1+Y}$.) Hence we could safely say (1) evaluates to $\phi$, and all seems to be good.  My trouble. Let us now test my proposed idea with a more general expression of the form $$\sqrt{a_1 + \sqrt{a_2 + \sqrt{a_3 + \sqrt{a_4+\cdots}}}} \tag{2}$$  (Note that the linked question involves one such expression, with $a_n = 5^{2^n}$.) How do we decide if this expression converges? Mimicking the above definition, we can write: $$ y_1 = \sqrt{a_1}, y_{n+1} = \sqrt{a_{n+1}+y_n}. $$ However, unrolling this definition, one get the sequence $$ \sqrt{a_1}, \sqrt{a_{2}+ \sqrt{a_1}}, \sqrt{a_3 + \sqrt{a_2 + \sqrt{a_1}}}, \sqrt{a_4+\sqrt{a_3 + \sqrt{a_2 + \sqrt{a_1}}}}, \ldots $$ but this seems little to do with the expression (2) that we started with.   I could not come up with any satisfactory ways to resolve the issue. So, my question is:     How do I rigorously define when an expression of the form (2) converges, and also assign a value to it when it does converge?   Thanks.
26873
I'm trying to re-learn my undergrad math, and I'm using Stephen Abbot's Understanding Analysis. In section 2.7, he has the following exercise:  Exercise 2.7.5 (a) Show that if $\sum{a_n}$ converges absolutely, then $\sum{a_n^2}$ also converges absolutely. Does this proposition hold without absolute convergence?  I'm posting about this here because my answer to that last question about whether the proposition holds without absolute convergence is "Yes", but my suspicions are raised by him merely asking the question. Usually questions like this are asked to point out that certain conditions are necessary in the statement of propositions, theorems, etc. I just want to see if I'm missing something here.  Anyway, here's how I prove the absolute convergence of $\sum{a_n^2}$, and note that I never use the fact that $\sum{a_n}$ is absolutely convergent:  Proof: Let $s_n = \sum_{i=1}^n{a_n^2}$. I want to show that $(s_n)$ is a Cauchy sequence. So, let $\epsilon &gt; 0$ and $n &gt; m$, and consider, \begin{equation}   \begin{aligned}     |s_n - s_m| &amp;= |(a_1^2 + a_2^2 + \cdots a_n^2) - (a_1^2 + a_2^2 + \cdots a_m^2)|\\                 &amp;= |(a_{m+1})^2 + (a_{m+2})^2 + \cdots + a_n^2|\\                 &amp;= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\   \end{aligned} \end{equation} Now, since $\sum{a_m}$ converges, the sequence $(a_m)$ has limit 0. Therefore I can choose an $N$ such that $|a_m| &lt; \sqrt{\epsilon/n}$ for all $m &gt; N$. So for all $n &gt; m&gt; N$,  we have, \begin{equation}   \begin{aligned}     |s_n - s_m| &amp;= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\                 &amp;&lt; \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n}}\right)^2\\                 &amp;&lt; \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2\\                 &amp;= \epsilon   \end{aligned} \end{equation} Therefore $(s_n)$ is Cauchy and $\sum{a_n^2}$ converges. And since $\sum{a_n^2} = \sum{|a_n^2|}$, the series $\sum{a_n^2}$ is absolutely convergent. &#x220E;  Am I doing something wrong here? Am I right in thinking that $\sum{a_n}$ need not be absolutely convergent for $\sum{a_n^2}$ to be absolutely convergent?
23242
Let $a_1, \ldots, a_n &gt; 0$. How to compute the limit $\lim_{n \to \infty} \sqrt[n]{a_1^n + \cdots + a_m^n}$?  My solution:  $$ \lim_{n \to \infty} \sqrt[n]{a_1^n + \cdots + a_m^n} \\ = \lim_{n \to \infty} (1 + a_1^n + \cdots + a_m^n - 1)^{1/n} \\ = \lim_{n \to \infty} \left( (1 + a_1^n + \cdots + a_m^n - 1)^{\frac{1}{a_1^n + \cdots + a_m^n - 1}} \right)^{ \frac{a_1^n + \cdots a_m^n-1}{n}}. $$ But then we need to have the condition $a_1^n + \cdots + a_m^n - 1 \to 0$ ($n \to \infty$).   Thank you very much.
104458
Suppose that $$Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} $$and $$P(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}.$$ How do I find $$\lim_{x\rightarrow\infty}\frac{Q(x)}{P(x)}$$ and what does the sequence $$\frac{Q(k)}{P(k)}$$ converge to?  For example, how would I find what the sequence $$\frac{8k^2+2k-100}{3k^2+2k+1}$$ converges to?  Or what is $$\lim_{x\rightarrow\infty}\frac{3x+5}{-2x+9}?$$   This is being asked in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: .
119969
Suppose $\sum_{n=2}^{\infty} a_n$ is a convergent series. Is the following series convergent as well? $$\sum_{n=2}^{\infty}{\sqrt{a_n} \over \ln{n}}\left( n^{a_n}-1 \right)$$ The part with $n^{a_n}-1 $ looks quite similar to $\sqrt[n]{n}-1$ which behaves like $\ln{n}$, however, I'm not able to tie this fact to the behavior of the whole expression. Any suggestions?
70203
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.
8737
Suppose $a_n \ge 0$ for all natural numbers $\Bbb{N}$ and $\sum{a_n}$ converges. Is it necessarily true that $\sum{(a_n)^2}$ converges?  I think yes.  Clearly as $a_n \ge 0$ it follows $(a_1+a_2+\dots+a_n)^2 \ge a_1^2+a_2^2+\dots+a_n^2 \ge 0$ and so $(\sum_{n=1}^{\infty}{a_n})^2 \ge \sum_{n=1}^{\infty}{a_n^2} \ge 0$.  As we are told $\sum{a_n}$ converges to say $L$ where $L$ is a real number it follows $L^2 \ge \sum_{n=1}^{\infty}{a_n^2} \ge 0$ and so therefore $\sum{(a_n)^2}$ converges.  Is this correct?
112089
Let $(a_n)$ be a sequence such that $(a_n+a_{n+1})$ is convergent. Prove that $(a_n/n)$ converges to $0$.   $(a_n+a_{n+1})$ convergent to $L$ means that for all $\epsilon$, there exists $N$ such that for all $n\geq N$, $|L-a_n-a_{n+1}|&lt;\epsilon$, or in other words $L-\epsilon&lt;a_n+a_{n+1}&lt;L+\epsilon$. How to proceed from here?
23637
I am trying to find out if the following sequence $(a_n)_{n\geq 1}$ will either converge or diverge. $$ a_n = (-1)^n + \frac1n $$  If the limit converges, I need to find its limit.   I tried plugging in various numbers for n besides 1, but I am not entirely sure if it really does converge.
3816
I have a sequence $(a_n)$ where for each natural number $n$,   $$a_n = (1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})$$ and I want to find its limit as $n\to\infty$.   I obviously couldn't prove it and after several futile attempts decided to post it here.   Here is a list of a few observations which I got from those attempts:   The sequence $(a_n)$ is a strictly increasing sequence. To prove this, I rewrote each element as $$a_n = (1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})= \frac{(n+1)\cdots(n+n)}{n^n}= \frac{(2n)!}{n!n^n}.$$ Then $$\frac{a_{n+1}}{a_n}= \frac{2(n+1)!}{(n+1)!(n+1)^{n+1}}\frac{n!n^n}{(2n)!}=\frac{(2n+1)(2n+2)}{(n+1)^2(1+\frac{1}{n})^n} \to \frac{4}{e}$$ as $n\to\infty$. Since $\frac{4}{e}&gt;1$ we have $a_{n+1}&gt;a_n$ eventually. The limit of this sequence is bounded below by $e$. By replacing $1,2, \ldots, n$ with $1$ in the expression of $a_n$, we get $a_n \geq (1+\frac{1}{n})^n$. And thus $\lim{(a_n)}\geq e$.  $\lim{(a_n)}\geq e^2$ and $\lim{(a_n)}\geq e^3$. The first assertion follows from the fact that $$a_n\geq(1+\frac{1}{n})(1+\frac{2}{n})^{n-1}= \frac{(1+\frac{1}{n})(1+\frac{2}{n})^{n}}{(1+\frac{2}{n})} \to e^2.$$ And the last one follows the same way because $$a_n\geq (1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})^{n-2}.$$   Now I have a gut feeling that for any natural number $k$, one can show that for all sufficiently large natural number $n$, $$a_n\geq (1+\frac{1}{n})\cdots(1+\frac{k-1}{n})(1+\frac{k}{n})^{n-(k-1)}.$$ And therefore for all $k \in \mathbb{N}$, $\lim{(a_n)}\geq e^k$ making the sequence divergent. But I'm really not sure about this approach and I'll appreciate any help towards this end. Thank you.  [Note: As this sequence is quite common, there may be other posts on math.SE asking the same question. I didn't search for them because I just don't know how to search for an expression this big. Though a link related to any previous question concerning this particular sequence will be good enough, I will greatly appreciate if someone takes the trouble to look into my approach/observations and point out where  I'm going wrong.]
81275
Question:  Let ${a_n}$ be a sequence in $\mathbb{R}$ with $\lim_{n\to\infty}a_n=a$  Prove that  $\lim_{n\to\infty}a_n^2=a^2$  Attempt: Without using any properties of limits (this is a question in a section of the book that came before any list of any limit theorems)  By definition:  $|a_n^2-a^2|&lt;\epsilon$   $|a_n^2-a^2|$ $\leq$ $|(a_n+a)(a_n-a)|$ $\leq$ $|(a_n+a)| |(a_n-a)|$ $\leq$   $(|a_n | + |a|)(|a_n -a|)$ $&lt;$ $\epsilon$  And since I know that $\lim_{n\to\infty}a_n=a$, I'll subsitute that in for $|a_n|^2$ as $a^2$   $\therefore$ $a^2-|a|^2 &lt; \epsilon$ which is true since $a^2-|a|^2$ in my case.  I wonder if this is a possible solution?
15864
A sequence $a_n$ is defined as $a_1=1$ and $a_{n+1}=\sqrt{a_1+a_2+\cdots+a_n}$ .Prove that $ \lim\limits_{n \rightarrow \infty} \frac{a_n}{n}=\frac{1}{2}$  I have no idea how to approach this. But I have a feeling that Cesaro's lemma may come in handy
44461
I would appreciate your help!  How can we show that if a sequence of positive real numbers $a_n$ converges to $a\in\mathbb{R}$ with $a&gt;0$, then $(a_0 a_1\cdots a_n)^{\frac{1}{n}}$ converges to $a$.
64951
Let $(a_n)_{n\in\mathbb{N}}$ be a sequence such that lim inf $|a_n|=0$. Prove that there is a subsequence $(a_{n_{k}})_{k\in\mathbb{N}}$ such that $\sum_{k=1}^\infty a_{n_k}$ converges.  I'm thought about showing that you can make a subsequence that is smaller then $1/n^2$. Is this the right strategy ? Any hints ? Should I use the Cauchy criterion ?
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.
43031
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?
61480
We define $a_1=1$ and $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)$.  I want to prove that $a_n$ converges. But first I'd like to show that $a_n&lt;\sqrt{2}$ for every $n\in\mathbb{N}$.  I try to use induction. For a fixed $n$, I suppose that $a_n&lt;\sqrt{2}$.  Any hint to prove that $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)&lt;\sqrt{2}$?  Thanks.