$(x_n)$ is a positive sequence and $$\lim_{n\to\infty}\frac{x_n}{n}=0\hspace{6mm}\limsup_{n\to\infty} \frac{x_1+x_2+\ldots+x_n}{n}\in\mathbb{R}$$ and we need to find $$ \lim_{n\to\infty} \frac{x^2_1+x^2_2+\ldots+x^2_n}{n^2}$$ I have no idea how to approach this task. I was thinking that I could maybe  $$ \frac{x^2_1+x^2_2+\ldots+x^2_n}{n^2}=\frac{\frac{x^2_1}{2}+\frac{x^2}{2}+\ldots+\frac{x^2_n}{2}}{n^2}+\frac{\frac{x^2_1}{2}+\frac{x^2}{2}+\ldots+\frac{x^2_n}{2}}{n^2}=\frac{\frac{x^2_1}{3}+\frac{x^2}{3}+\ldots+\frac{x^2_n}{3}}{n^2}+\ldots$$ But that doesn't lead anywhere. How should I approach it? What does the second limit ($\limsup\ldots\in\mathbb{R}$) mean and how should I use it?
121280
Let's $x_n\ge0$ and $$\overline{\lim}_{n\rightarrow\infty}\frac{x_1+x_2+\cdots+x_n}{n}\lt+\infty,~~\lim_{n\rightarrow\infty}\frac{x_n}{n}=0.$$Prove that $$\lim_{n\rightarrow\infty}\frac{x_1^2+x_2^2+\cdots+x_n^2}{n^2}=0.$$ I have no idea how I can start the proof.  Thanks!
83859
We have a sequence of numbers $x_n$ determined by the equality   $$x_n = \frac{x_{n-1} + x_{n-2}}{2}$$  The first and zeroth term are $x_1$ and $x_0$.The following limit must be expressed in terms of $x_0$ and $x_1$ $$\lim_{n\rightarrow\infty} x_n $$  The options are:   A)$\frac{x_0 + 2x_1}{3}$ B)$\frac{2x_0 + 2x_1}{3}$  C)$\frac{2x_0 + 3x_1}{3}$ D)$\frac{2x_0 - 3x_1}{3}$  Since it was a multiple choice exam I plugged $x_0=1$ and $x_1=1$. Which means that all terms of this sequence is $1$,i.e, $$x_n=1, n\in \mathbb{N} $$  From this I concluded that option A was correct.I could not find any way to solve this one hence I resorted to this trick. What is the actual method to find the sequence's limit?
108876
Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$ If $P_n=\prod_1^n(1+a_i^{-1})$ then $$\lim_{n\to\infty}P_n $$is ?    I did: $$P_n=\prod_1^n\frac{(1+a_i)}{a_i} =\frac{a_{n+1}}{a_1}\prod1^{n-1}\frac{(1+a_i)}{a_{i+1}} =\frac{a_{n+1}}{a_1}\prod_1^{n-1}\frac1{i+1}=\frac{a_{n+1}}{n!}$$ Now how do I find the limit?
38451
"Examine whether the given progressions converge (eventually improperly) and determine their limits where applicable.  (a) $$(a_n)_{n\in\mathbb{N}}:=\frac{n^n}{n!}$$  [...]"  I am having problems getting this homework done as I have no clue about convergency at all. Through Mathematica I know that $\lim\limits_{n\to\infty}a_n=+\infty$ however I don't know how to proove it! During my research I found out that for sufficient big $n$ the inequality $x^n\leq n!\leq n^n$ for $x\in\mathbb{R}$ and $n\in\mathbb{N}$ is true. However this doesn't help me at all. Having a look at our lecture notes I developed the following statement: every $x^n$ with $|x| \geq 1$ diverges as $x^n$ is unbounded and therefore has the improper limit $\lim\limits_{n\to\infty}x^n=+\infty$. Furthermore it seems obvious to me that I need to express the terms in a simpler way, like fractions converging to 0 etc, but I don't know how. Neither some work with the $\varepsilon$-definition of limits in combination with the triangle inequality helped me out. Any suggestions or hints?  Information: This is "Analysis for computer scientist" and therefore we haven't "officially" learned any fancy tricks like L'Hôpital's rule etc. just basic properties of $\mathbb{R}$ and some inequalities as this is an introductory lecture.
64589
How to prove that $$ \frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots+\frac{x_n}{1+x_1^2+\cdots+x_n^2}&lt;\sqrt{n} $$ knowing that $(x_n)$ is a positive sequence ? I looked up all kinds of inequalities such AM-GM, Chebyshev, Cauchy-Schwarz, but I couldn't manage to obtain anything useful.. Can anyone help ?
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.]
90012
I am studying for a test and I am given this problem:  $$\lim_{n\to\infty}\frac{n^n}{n!}.$$  How do I go about solving this limit? Intuitively I see how the numerator is growing much faster, but how do I express this precisely? Thanks!
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
96657
What I've done so far: I have proved that this sequence is bounded below by 0, which is a very rough estimate. I know that the infimum is $\sqrt2$.  Anyway, the question first asks me to prove that the sequence decreases monotonically. And I've tried the following:  Suppose $a_{n+1} \le a_n$, then we would get $\frac{1}{a_n} \le \frac{a_n}{2}$. After some algebra, I ended up with $\frac{1}{a_n^2 +2}\le\frac{a_n^2 +2}{2}$. Now, $a_{n+2} - a_{n+1}=\frac{2a_n}{a_n^2 +2}-\frac{a_n^2 +2}{4a_n}$. I was looking for some similarities between these two expressions but it doesn't seem to work out.  A more general question is, if the sequence is defined recursively, what are some common strategies to find the limit? I just started studying analysis and this kind of questions kind of troubles me.  Thank you guys.
92775
Let $\alpha$ be a positive number. Calculate $$\lim_{n\to\infty} \sum_{k=1}^{n} \Big(\frac{k}{n}\Big)^{\alpha k}$$  Edit: I have deleted my attempt, it didn't seem to lead me anywhere and I discovered a couple of mistakes. :)  Please give me a hint!
99383
Find the limit,  $$L=\lim_{n\to \infty}\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$  My try:  $$ \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}xdx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}(1-x)dx&lt; \int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx&lt; \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}(1-x)dx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}xdx$$  Now taking the limit I get that,  $$\frac{1}{4}&lt;L&lt;\frac{3}{4}$$  But, how can I get the exact answer!!  This is  from the American Mathematical Monthly.
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.
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.
119125
Evaluate the value of $$\lim_{n \to \infty}\frac{x_n}{\prod_{i=1}^{n-1} x_i}$$ where $x_n=x_{n-1}^2-2, (x_1=5).$  I've found that $x_n$ is an increasing sequence so that the limit becomes an indeterminate form of $(\frac{\infty}{\infty})$. I can't think of any other workable ideas though. Some hints are appreciated.
64033
I am trying to calculate the limit of  $$\lim_{n \to \infty} \frac{(2n-1)!}{(2n+1)!}$$  and I am stuck. I tried n division but that didn't work. Any help is appreciated
4262
fine the limit :  $$\lim_{ n \to \infty }\frac{1}{n}\int_{0}^{n}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=?$$  My Try: in the  $$I=\int_{}^{}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=\left(x+n\right)\ln\left(\left|x+n\right|\right)+\left(\ln\left(n\right)-\ln\left(n-1\right)\right)\ln\left(\left|x+1\right|\right)+\operatorname{Li}_2\left(-\dfrac{x+1}{n-1}\right)+\left(-\ln\left(n\right)-1\right)+c$$  now?
99697
I want to prove that the sequence defined by $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ has a limit.  By evaluating the sequence I notice that the sequence is strictly monotonically decreasing starting from $x_2=1.5$.  It seems to suggest itself to prove that the sequenced is bounded by $1\le\big( x_n \big)_{n\ge1} \le 1.5$ and to prove that it is strictly monotonically decreasing starting at $x_2$ which would imply convergence.  How would I proceed and could one prove the existence of the limit without first evaluating the values of the sequence to see how the sequence behaves?