Let $$I_n=\int_0^1 \frac{x^n}{\sqrt{1+x^2}} dx$$ for $n\in \mathbb{N}$ Find $$\lim_{n\to \infty} nI_n$$ My approach: Let me explain my observations briefly. Note that $I_{n-1}-I_n=\int_0^1 \frac{x^{n-1}(1-x)}{\sqrt{1+x^2}} dx \ge 0$ Hence $I_n$ is decreasing.Also $I_n\ge 0$.So, $I_n$ is convergent. Let $\lim_{n\to \infty} I_n =l$ Now $I_n+I_{n-2} =\frac{\sqrt{2}}{n-1} -\frac{2}{n-1} I_n$ (By parts +little bit calculation) Applying limit to the above equation we get, $l=0$. Also from the above equation we have, $(n-1)I_n+(n-1)I_{n-2}=\sqrt{2}-2I_n$. If $J_n := nI_n$, then $\frac{n-1}{n} J_n +\frac{n-1}{n-2} J_{n-2}=\sqrt{2}-2I_n$ Yes I am only now one step away from getting $\lim_{n\to \infty} J_n=\frac{1}{\sqrt{2}}$ But I am unable to prove that $J_n$ is convergent 🙁 Any idea??
124121
Question: Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $$\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$$.    My approach: Given that $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx, \forall n\in\mathbb{N}.$ Let us make the substitution $x^n=t$, then $$nI_n=\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}}.$$  Now since $0\le t\le 1\implies \frac{1}{t}\ge 1\implies \left(\frac{1}{t}\right)^{2/n}\ge 1 \implies 1+\left(\frac{1}{t}\right)^{2/n}\ge 2\implies \sqrt{1+\left(\frac{1}{t}\right)^{2/n}}\ge \sqrt 2.$   This implies that $$\frac{1}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le\frac{1}{\sqrt 2}\\ \implies \int_0^1 \frac{dt}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le \int_0^1\frac{dt}{\sqrt 2}=\frac{1}{\sqrt 2}.$$  So, as you can see, I am trying to solve the question using Sandwich theorem.   Can someone help me to proceed after this?   Also, in $$\lim_{n\to\infty}nI_n=\lim_{n\to\infty}\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}},$$ the limit and integral interchangeable?
32193
I would like to find the following limit without using any Taylor expansion.  $$\lim _{n\to \infty }n \int_{-1}^0(x + e^x)^{n}dx = \: ?$$  I know the answer is $\frac{1}{2}$. I tried to write $-1 &lt; x &lt; 0$ and $0 &lt; e^x &lt; 1$, sum up the relations and then integrate, but I found nothing conclusive.
111865
Problem: Let $(x_i)$ be a sequence in (0,1) such that $\lim_{n \to \infty} (1/n)\sum_{i=1}^n x_i^k$ exists for every $k=0,1,2,...$ and $f\in C[0,1]$. Show that $\lim_{n \to \infty} (1/n)\sum_{i=1}^n f(x_i)$ exists.  I've used the fact that $B_n(f)$ converges uniformly to $f$ to get   $\lim_{n \to \infty} (1/n)\sum_{i=1}^n f(x_i) = \lim_{n \to \infty} (1/n)\sum_{i=1}^n \sum_{k=0}^n f(\frac{k}{n}) {n \choose_k }x_i^k (1-x_i)^{n-k}$  And then you can arrange to get   $\lim_{n \to \infty} (1/n) \sum_{k=0}^n \left[ f(\frac{k}{n}) {n \choose_k }\sum_{i=1}^nx_i^k (1-x_i)^{n-k}\right]$  I want to use the fact that $(1-x_i) &lt; 1$ to consolidate some of the equation but I get nowhere.
69930
Prove that $\lim_{n \to \infty} a^{\frac{1}{n}} = 1$ if $a &gt;0$.  In my textbook, we are given a suggestion to let $a^{\frac{1}{n}} = (1+h_n)$ and then show that the $h_n$ term goes to zero using a Theorem that states the following conditions:  $\lim_{n \to \infty} = \infty,$ if $a &gt;1, 1,$ if $a = 1,$ and 0, if $\vert a \vert &lt; 1$. I apologize for not formatting the above cases appropriately; I could not figure out how to use a giant left brace to group them altogether.  If I rewrite it as $a^{n^{-1}}$, then I thought that this might help, but I don't think the Binomial Theorem would then apply since the exponent is negative. Furthermore, the problem is only concerned with the positive, real $n$-th roots. I am quite stuck on this problem and any suggestions or advice would be greatly appreciated.   I am using the textbook Introduction to Analysis by Arthur Mattuck.
47565
I get that the limit is  $$\lim\limits_{n\to\infty}f(x)=1$$ because $$\lim\limits_{n\to\infty}(1+x^{2n})=1,\,\,\lvert x\rvert&lt;1.$$  Is this right?
106090
I have tried $f(n) = \int_1^n \ln x dx$, but it fails. The difference is  $$\lim_{n\to \infty}(\ln n! - \int_1^n \ln x dx)$$  $\int_1^n \ln x dx = n\ln n -n +1 \\ \ln n! - \int_1^n \ln x dx = \ln n! - n\ln n +n -1 = \ln(\frac{n!e^n}{n^n})-1$  where $\frac{n!e^n}{n^n}\to \infty$, so the limit diverges.  Do we have some function $f(n)$that converges to the series $\ln n!$ ?
34275
Find the limit $$ \lim_{x \to \infty} \frac{e^{x}}{(1+\frac{1}{x})^{x^{2}}}. $$  I know that the limit is of indeterminate form type $\frac{\infty}{\infty}$, but it seems using L'Hopital's rule directly does not help here.  Do I somehow use the fact that $\lim_{x \to \infty} (1+\frac{1}{x})^{x}=e$?
110785
Introduction:   An exercise from "Principles of mathematical Analysis, third edition" by Rudin, page 78.     Exercise:  Calculate $\lim_{n\to\infty}(\sqrt{n^2+n}-n)$.    Explanation:  I have a hard time to grasp how to handle limits like this. I don't know how to start and what to look for. I've checked with mathematica and the answer should be $\frac{1}{2}$, and of course i've got the wrong answer. I find limits unintuitive. In the book, they proved "the limits of some sequences which occur frequently". The limits they proved were:  (a) If $p&gt;0$ then $\lim_{n\to\infty}\frac{1}{n^p}=0$  (b) If $p&gt;0$ then $\lim_{n\to\infty}\sqrt[n]{p}=1$  (c) $\lim_{n\to\infty}\sqrt[n]{n}=1$  (d) If $p&gt;0$ and $\alpha$ is real, then $\lim_{n\to\infty}\frac{n^\alpha}{(1+p)^n}=0$  (e) If $|x|&lt;1$, then $\lim_{n\to\infty}x^n=0$.  When they proved all of the above theorems it felt like they used the fact that they knew the limits. For example:    Proof of (b): If $p&gt;1$, put $x_n=\sqrt[n]{p}-1$. Then, $x_n&gt;0$ and by the binomial theorem,  $$1+nx_n\leq(1+x_n)^n=p$$  so that $$0&lt;x_n\leq\frac{p-1}{n}.$$  Hence $x_n\to 0$. And so on...    That is, I think they used the fact that the limit were 1 when they did put $x_n=\sqrt[n]{p}-1$. Before I compute a limit do I have to guess one? How can I do that when I don't think this is intuitive? Have you any tips how you do when you shall tackle a problem like this? How do you start when you want to compute a limit?    Solution:  This is how I did it:  $\sqrt{n^2+n}-n=\sqrt{n}\sqrt{n-1}-n=\sqrt{n}(\sqrt{n-1}-\sqrt{n})$.  Since $$(\sqrt{n-1}-\sqrt{n})\to 0\text{ when }n\to\infty.$$  The product approaches $0$. Which is obviously not true. I did realise this after a while. Since one of the factor grows really big while the other gets really small and I guess they tend to take each other out, so it's pretty clear it shouldn't approach 0, but I don't think it's clear that it should approach $\frac{1}{2}$ either. Thanks for your help.
545
Let $s_n$ be a sequence defined as given below for $n \geq 1$. Then find out $\lim\limits_{n \to \infty} s_n$.   \begin{align}   s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx  \end{align}  I have written a solution of my own, but I would like to know it is completely correct, and also I would like if people post more alternative solutions.
3821
Can you help me with this limit? What do I have to do? I'm lost.  $$\lim_{n\to\infty}n\left(\sum_{i=1}^{n}\dfrac{1}{(n+i)^2}\right)$$  The solution given is $\dfrac{1}{2}$.
64672
Wonder how to determine this limit by the use of Riemann integral. The limit is as follows:  $$\lim_{n \to +\infty} \frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}} $$  My instructor told me that the usage of Riemann integral gives spectacular result. Checked Rudin, but did not find any valuable references.  I am very interested in seeing how this "spectacular result" emanates.  Help/advices/solutions very, very appreciated!
74620
Okay, maybe I'm missing something obvious but consider $h_n(x) = x^{ 1+ \frac{1}{2n-1}}$ on the set $[-1,1]$. Now consider a fixed $x \in [-1,1]$, we have:  $$\lim_{n\rightarrow \infty} h_n(x) = x \lim_{n \rightarrow \infty} x^\frac{1}{2n-1} = x$$  But my book says it's $|x|$? Why?
24010
Evaluate the limit   $$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$   My approach :   If I divide numerator and denominator by $n^2$ I get :   $$\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0$$   but the answer is $\dfrac{1}{2}$ please suggest how to solve this.. thanks.
35981
$$\lim_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}.$$  With a first look this must give $1$ as a result but have a problem to explain it.  How can I do it?  Edit  I noticed that it is $\frac{\infty}{\infty}$.  $$\lim_{n \to \infty}{n^{n}\frac{(\frac{1^1}{n^{n}}+\frac{2^2}{n^{n}}+\frac{3^3}{n^{n}}+\cdots+1)}{n^n}}= \lim_{n \to \infty}{\frac{1^1}{n^{n}}+\frac{2^2}{n^{n}}+\frac{3^3}{n^{n}}+\cdots+1}=1$$  Is this correct?
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 $\left(\frac{1}{2},1\right)$.
53733
Find $$\lim_{n\to\infty} \left(1+\frac{x^2}{n^2}\right)^n$$   I particular, I am hoping to find the above to be $1+x^kg(x/n)$ where $g(\cdot)$ is a function with uniformly bounded derivatives.  Edit: I'm trying to prove a result that would be true if  $(1+x^2/n^2)^n -1 = x^kO(x/n).$
47463
It is quite easy to show that (by using residue theorem) $$\int_0^{\infty} \frac{1}{1+x^n}dx = \frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)} $$ for $$n \ge 2$$  Is there any possibility to simplify $$\frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)}$$ or it is best result?  Thanks in advance!