Consider the sequence $\displaystyle a_{n} = \sum_{i = 1}^{n}i^{-1/3} $ and $b_{n} = n^{2/3}$.  No we want to find $\displaystyle \lim_{n\rightarrow \infty}a_{n}-\frac{3}{2}b_{n}$?   Actually I find it lower bound , but I used Abel summation, have no idea how to find it easier.
122723
I'm noticing some things:  $$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/3}-\frac{3}{2}n^{2/3} \right)=\zeta(1/3)$$  Note $\int n^{-1/3} dn=\frac{3}{2}n^{2/3}+c$  $$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/2}-\frac{2}{1}n^{1/2} \right)=\zeta(1/2)$$  And  $$\int n^{-1/2}dn=2n^{1/2}+c$$  It seems as though  $$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/s}-\frac{s}{s-1}n^{(s-1)/s} \right)=\zeta(1/s)$$  If $s \neq 1$, may someone please explain why.
5928
$$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k+n}$$  Could you please give me a hint how to find this limit?
72225
I have to compute the limit of this sequence ${\textstyle\sum_{k=1}^n}\frac1{\sqrt{n^2+k}}$ as $n\rightarrow\infty$. First I was thinking about some Riemann sum and and forced the $n^{2}$ outside the square root but the function was not so pleasant.
113842
Given the series $A=\sum\limits_{n=1}^{+\infty}\frac{p_n}{p_{n+1}}$ and $B=\sum\limits_{n=1}^{+\infty}\frac{p_{2n}}{p_{2n+1}}$, where $p_n$ is the sequence where the nth number are the nth prime number, is it possible to determine whether they are converging or diverging?   I think there are some ways using the fact that $\sum\limits_{p\in\mathbb{P}}\frac{1}{p}$ diverges, but I'm not sure of how I can use it.
123600
I have recently come across a problem concerning the limit of a sequence given as, $$\lim_{n\rightarrow \infty} (a^n + b^n)^\frac{1}{n}$$  Where $0&lt;a&lt;b$. Solving the limit is not a concern. But applying L'Hospital's rule to this limit is one to me.   On rearranging we get, $b \lim_{n\rightarrow \infty} ((\frac{a}{b}) ^n + 1)^\frac{1}{n}$. The solver had then applied the L'Hospital's rule to evaluate the limit. Here, as $n\rightarrow \infty$, the limit approaches the form $1^0$. I have tried to think about this limit approaching an indeterminate form. It does not. From my knowledge of Cauchy's Mean Value theorem, I say that we cannot apply the rule to determinate forms. But it seems to me that we can because it worked for the solver. I think I am not clear on this. Can somebody help?
28565
Find the limit $\displaystyle\lim_{n\rightarrow\infty}{(1+1/n)^{n^2}e^{-n}}$?  I found the limit as $e^{-1/2}$ using l'Hospital rule. I guess I made a mistake. Because the limit seems to be 1. Also, can we find the limit without L'Hospital rule?
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.
31748
How to find the value of $\lim_{n\to\infty}S(n)$, where $S(n)$ is given by $$S(n)=\displaystyle\sum_{k=1}^{n} \dfrac{k}{n^2+k^2}$$   is unable to calculate it.  This is a question from a questions booklet, and the options for the answer are--  $\begin{align} &amp;A) \dfrac{\pi}{2} \\ &amp;B) \log 2 \\ &amp;C) \dfrac{\pi}{4} \\ &amp;D) \dfrac{1}{2} \log 2 \end{align}$
5210
Suppose $x_{1}>0$, and consider the sequence, $\{x_{n}\}$ defined as follows: $$x_{n+1}=\log(1+x_{n}) \quad n\geq 1 $$ Find the value of $\displaystyle \lim_{n \to \infty} nx_{n}$  I am having trouble solving it. One thing is clear, that since $x_{n}>0$ and $x_{n+1} &lt; x_{n}$, we can have a sequence which converges an $f$ which satisfies $f=\log(1+f)$, so that $f=0$. Any way as to how we can proceed from here.
103599
can anyone help me with this, I got stuck on it.  let $a_n$ and $b_n$ (for $n$ a non-negative integer) be two sequences of real numbers such that $\lim_{n\to \infty}a_n=a$ and $\lim_{n\to \infty}b_n=b$. prove that $\lim_{n\to \infty} \frac{a_0b_n+a_1b_{n-1}+....+a_nb_0}{n}=ab$.
30372
I have a sequence defined as follow:  $a_0 = 1, a_n=\cos\left(a_{n-1}\right)$.  I want to count $\lim_{n\rightarrow\infty} a_n$ - it definitely does have limit by looking at the graph, the first few numbers of the limit are 0.7390851, but I have no idea, if that number is related to some other real number ($\pi$ or something like that).  The sequence is from , but they don't provide actual result for their own sequence.
91830
Suppose you have a double sequence $\displaystyle a_{nm}$.  What are sufficient conditions for you to be able to say that $\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}} = \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}$?  Bonus points for necessary and sufficient conditions.  For an example of a sequence where this is not the case, consider $\displaystyle a_{nm}=\left(\frac{1}{n}\right)^{\frac{1}{m}}$.  $\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}}=\lim_{n\to \infty}{\left(\frac{1}{n}\right)^0}=\lim_{n\to \infty}{1}=1$, but $\displaystyle \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}=\lim_{m\to \infty}{0^{\frac{1}{m}}}=\lim_{m\to \infty}{0}=0$.
51385
It seems that I'm missing something about this.  First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)  What I tried to do is to find a limit of a partial sum $\lim_{n\rightarrow\infty}S_n$ as follows: $S_n=\frac{\frac{1}{6} n (n+1) (2 n+1)}{\frac{1-\left(\frac{1}{2}\right)^n}{2 \left(1-\frac{1}{2}\right)}}$. Still, the limit is $\infty$ and I'm clearly doing something wrong.
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.
42955
Given a sequence $a_{n}$, if I know that the sequence of even terms converges to the same limit as the subsequence of odd terms:    $$\lim_{n\rightarrow\infty} a_{2n}=\lim_{n\to\infty} a_{2n-1}=L$$  Is this sufficient to prove that the $\lim_{n\to\infty}a_{n}=L$?    If so, how can I make this more rigorous?  Is there a theorem I can state that covers this case?
21873
Computing $\displaystyle \lim_{n\rightarrow \infty}\left(\frac{1^n+2^n+\cdots \cdots +n^n}{n^n}\right)$  Attempt: $\displaystyle \lim_{n\rightarrow \infty}\bigg[\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots \cdots +\left(\frac{n-1}{n}\right)^n+\left(\frac{n}{n}\right)^n\bigg] = 1$  Is my answer right? If not, then could someone help me? Thanks.
116751
Assume $\{a_n\}$is a sequence which  satisfies the following recursion formula: $$a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}~(n\geqslant 1).$$ and $a_0=\pi,a_1=\pi^2 $. How to compute $\displaystyle\lim_{n\to \infty}\frac{a_n}{n^2}$?