How to calculate $$\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt[n]{{{3^n} + {4^n}}}} \right)?$$    I saw on the suggestions that you can use the sandwich theorem or compression. And the fact that $\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt[n]{2}} \right) = 1$  So I must first find upper bounds and lower.  someone help me with this please?
31851
How can I prove the convergence of the sequence $b_n=\sqrt[n]{x^n+y^n}$ where $x, y &gt; 0$?  Can I divide it in two cases?  Case 1: $x &gt; y$.  $$ b_n=\sqrt[n]{x^n+y^n} &lt; \sqrt[n]{x^n+x^n} = \sqrt[n]{2 \cdot x^n}=x \cdot \sqrt[n]{2} $$  Case 2: $x &lt; y$.  $$ b_n=\sqrt[n]{x^n+y^n} &lt; \sqrt[n]{y^n+y^n} = \sqrt[n]{2 \cdot y^n}=y \cdot \sqrt[n]{2}$$  Result: Does the sequence converges to $\max(x,y)$?
24947
When $\lim_{n\to\infty} x_n = a$, and $\lim_{n\to\infty} y_n = b$, find the limit, $$\lim_{n\to\infty} \frac{x_1 y_n + x_2 y_{n-1} + \cdots + x_n y_1}{n}.$$ Thank you for your help in advance.
17619
I have two limit problems which are quite similar, so I've put them both into this post.   $((4^{10}+2^{n})^{\frac{1}{n}})$ $((3n^{2}+n)^{\frac{1}{n}})$   Attempt:  For 1, I'm not sure if this is permissible. I know that $\lim_{n \to \infty}(x^{n}+y^{n})^{\frac{1}{n}} = \max\{x,y\}$. Obviously $4^{10}$ is a constant so it's not quite in the same form, but I believe it is still okay to use this result, and so the limit is $2$.  For 2, I've got $((3n^{2}+n)^{\frac{1}{n}}) = (n(3n+1)^{\frac{1}{n}})$, but then I'm not too sure how to proceed.   I also thought of taking the log and proceeding that way, however in the notes I'm learning from they haven't covered l'hopital yet, so I'm trying to not use it.  Thanks.
52779
How does one prove the following limit? $$   \lim_{n \to \infty}   \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots \sqrt{1 + (n - 1) \sqrt{1 + n}}}}} = 3. $$
61993
What is the limit of:  $$\lim_{n\to \infty} \frac{n^1+\dots+n^n}{1^n+\dots+n^n} = ?$$  I did some computation with big numbers, I guess it is in the interval $\left(\frac{1}{2},1\right)$.
18978
Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$  Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?  $\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} &lt; \epsilon$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} &gt; \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \sqrt{n} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow n &gt; \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
19929
I have a problem with the calculation of the following limit. \begin{equation} \lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n} \end{equation} I do not know where to start! Thank you very much
17446
I've seen some limit problems where you can do this: $$ \lim_{x \to \infty} \exp\left({g(x)}\right) = \exp \left( \lim_{x \to \infty }g(x)\right) . $$ So, I've tried to generalize the result as: $$ \lim_{x \to \infty} f(g(x)) = f \left( \lim_{x \to \infty }g(x) \right) $$ and I was just wondering if the result above is actually true, and on what conditions? Perhaps someone can point me to a theorem.
59390
Let $x&gt;0$, find the limit $$\lim_\limits{n\to{\infty}}{n^2\left(\sqrt[n]{x}-\sqrt[{n + 1}]{x}\right)}$$  I use Maclaurin series and find out that the limit is $\ln x$. And this is the answer I get from a math forum:  "Let $x_n=\sqrt[n]{x}-1$, $y_n=\dfrac{1}{n}$, then $(x_n)\to0,(y_n)\to0,(y_n)\downarrow$ when $n\to\infty$.  Use Stolz-Cesaro theorem, we have:  $$\mathop {\lim }\limits_{n \to \infty } \dfrac{{\sqrt[n]{x} - \sqrt[{n + 1}]{x}}}{{\frac{1}{n} - \frac{1}{{n + 1}}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\sqrt[n]{x} - 1}}{{\frac{1}{n}}} = \ln x$$ Then $\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {\sqrt[n]{x} - \sqrt[{n + 1}]{x}} \right) = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^2}}}{{n\left( {n + 1} \right)}}\cdot\dfrac{{\sqrt[n]{x} - \sqrt[{n + 1}]{x}}}{{\frac{1}{n} - \frac{1}{{n + 1}}}} = \ln x$.  /END"  I just wonder what form Stolz-Cesaro theorem that he used, because I just learn Stolz-Cesaro theorem when $(y_n)\to\infty$.  Could anyone help me to get this? Thank you in advance.     This is "another form of Stolz theorem" in the forum that I mentioned: Let $(x_n),(y_n)$     be two sequences of real numbers. Assume $\mathop {\lim }\limits_{n \to \infty } {x_n} = \mathop {\lim }\limits_{n \to \infty } {y_n} = 0$; $y_n&gt;0$, $(y_n)$ is strictly decreasing  and the following limit exist $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x_{n + 1}} - {x_n}}}{{{y_{n + 1}} - {y_n}}} = L$. Then $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x_n}}}{{{y_n}}} = L$. Is this correct?
72741
this it the limit to evaluate: $$\mathop {\lim }\limits_{n \to \infty } {1 \over {{n^{k + 1}}}}(k! + {{(k + 1)!} \over {1!}} + ... + {{(k + n)!} \over {n!}})$$  I've given an hint which is: $$(1 - q)(1 + q + ... + {q^N}) = (1 - {q^{N + 1}})$$  I tried to use Cesaro theorem but got to sort of a dead end.. I also noticed a similarity to the , is that going to help me?  Any help will be appreciated
20697
I am trying to find the limit of:  $$\lim_{n \to \infty} n(\sqrt[n]{2} - 1)$$  I know it should be very simple but I don't seem to get it. Thanks in advance for any help!
93160
I need to find the limit, not sure what to do. $\lim_{x \to \infty} \sqrt{x^2 +ax} - \sqrt{x^2 +bx}$  I am pretty sure I have to divide by the largest degree which is x^2 but that gets me some weird numbers that don't seem to help.
51963
Evaluate the limit: $$\mathop {\lim }\limits_{x \to \infty } x\left( {{{\left( {1 + {1 \over x}} \right)}^x} - e} \right)$$  My attempts didn't yield a result. I'd be glad for a guidance. Thanks!
61557
A problem I saw recently is to compute $$\lim_{n\to\infty} n^2 \left(\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\right)$$   I thought it would be fun to give this a go but it completely stumped me. I am not aware of any of the standard techniques used to compute $\infty\times (e-e)$ - type limits, bar possibly L'Hôpital's Rule. But I doubt that would be a good idea here, at best it would be ugly. I would appreciate solutions using any methods, the more elegant the better of course. Though I welcome any relevant hints as well if full solutions are discouraged.
91510
Determine value of the limit:  $$L=\lim_{n\to \infty}\sqrt[n]{1^2+2^2+\cdots+n^2}$$  My try:  $$1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$$  Hence: $$L=\lim_{n\to \infty} \sqrt[n]{\frac{n(n+1)(2n+1)}{6}}=?$$  But, come here, i do not know how, because please help me.
65279
How can I calculate this limit? $$\lim_{n\rightarrow\infty} \frac{7^{\sqrt{n+1}-\sqrt{n}}\cdot(\frac{n+1}{2})!\cdot(\frac{n+1}{2})!}{(n+1)\cdot(\frac{n}{2})!\cdot(\frac{n}{2})!}$$ I don't have idea and I will be happy for help.
74371
I want to show that $$\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n}-1) = 0$$ and my assistant teacher gave me the hint to find a proper estimate for $\sqrt[n]{n}-1$ in order to do this. I know how one shows that $\lim_{n \to \infty} \sqrt[n]{n} = 1$, to do this we can write $\sqrt[n]{n} = 1+x_n$, raise both sides to the n-th power and then use the binomial theorem (or to be more specific: the term to the second power). However, I don't see how this or any other trivial term (i.e. the first or the n-th) could be used here.   What estimate am I supposed to find or is there even a simpler way to show this limit?  Thanks for any answers in advance.