$$\lim_{n\to\infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$$  I have found that  the sequence is increasing as $n$ increases. I guess that the answer should be $\frac{1}{1-e}$.
39151
I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks!
46401
What would you suggest here?  $$\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} n^{1/k} $$
69669
For $a,b,c&gt;0$, Find $$ \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n$$  how can I find the limit of sequence above?  Provide me a hint or full solution. thanks ^^
115748
The sequence is defined: $$u_1=1, u_{n+1}=1+\dfrac n {u_n}$$  The question asks to find an asymptotic development of 2 terms for $n\to+\infty$. I have got $u_n\sim_\infty\sqrt{n} $, but how to derive the 2nd term?
99636
Here is an exercise:  Suppose that $\{x_n\}$ is a sequence such that $\lim \limits_{n\to\infty}(x_n-x_{n-2})=0$.  Show that:   $$\lim \limits_{n\to\infty}\frac{x_n-x_{n-1}}{n}=0 $$  Thanks.
91034
I just learned the definition of limits, and I learned that if $\{a_n\}, \{b_n\} $ converges, then $$\lim_{n\to \infty} (a_n+b_n)=\lim_{n \to \infty} a_n+\lim_{n \to \infty}b_n$$ holds.  And my teacher said that $\lim_{n \to \infty}\frac{1+2+3+ \cdots +n}{n^2}=\lim_{n\to \infty}\frac{\frac{n(n+1)}{2}}{n^2}=\frac{1}{2}$.  But can't we compute like  $$\lim_{n \to \infty}\frac{1+2+3+ \cdots +n}{n^2}=\lim_{n\to\infty}\frac{1}{n^2}+\lim_{n\to\infty}\frac{2}{n^2}+\lim_{n\to\infty}\frac{3}{n^2}+\cdots +\lim_{n\to\infty}\frac{n}{n^2}=0+0+\cdots+0=0$$?
14445
$ \lim_{n\to \infty} ( \lim_{x\to0} (1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} $  The answer should be $ {e}^\frac{1}{3} $  I haven't encountered problems like this before and I'm pretty confused, thank you.  I guess we must use the remarkable limit of $ \frac{\tan(x)}{x} $ when $ x $ approaches $ 0 $ by dividing and then multiplaying with $x$.
53954
How can we compute the following limit:     $$\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$$   Mathematica gives the answer $\sqrt{e}$. However, I do not know to do it.
121550
There is a couple of limits that I failed to find:  $$\lim_{n\to\infty}\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}$$  and  $$\lim_{n\to\infty}1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots + \frac {(-1)^{n-1}}{3^{n-1}}$$  There is no problem to calculate a limit using Wolfram Alfa or something like that. But what I am interested in is the method, not just a concrete result.  So my questions are:   What should I do when I need a limit of infinite sum? (are there any rules of thumb?) What theorems or topics from calculus should I know to solve these problems better?   I am new to math and will appreciate any help. Thank you!
121419
How to prove the following: $$ \lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 \frac{x_1^2+x_2^2+ \cdots +x_n^2}{x_1+x_2+ \cdots +x_n} dx_1 dx_2 \cdots dx_n = \frac23 $$  I would really appreciate if you could help me!
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 $\dfrac{1}{2}$.
34560
$ \lim_{n\to\infty} {\sqrt{1+{\sqrt{2+{\sqrt{\cdots +\sqrt{n}\ }\ }\ }\ }\ \ }\ } = ? $ Either closed answer or an upper bound would help.
10600
I have the following exercise: Let $S_1$ = 1  and for $n \geq 1$  let $S_{n+1} = \sqrt{S_n + 1}$. Prove that the limit of this sequence is $\frac{1}{2}(1 + \sqrt{5})$
59359
Does $\displaystyle\lim_{n\to \infty}\ \frac{1+\sqrt[2]{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}=1$?   Thanks a lot for your time and help.
32427
Evaluate  $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$
65397
Let $x&gt;0$. Find $$\lim_{n\to\infty}\frac{n^{x}}{(1 + x)^{n}}.$$   What I was thinking involved an inequality, by reason of considering the fact that  $$(1 + x)^{n} &gt; C(n, k) x^{k}$$ for some $k$ such that $n&gt;k$. Expanding the binomial coefficient and taking the $n$ common from each bracket, we have something like $$(1 + x)^{n} &gt; C(n, k) x^{n} &gt; \frac {x^{k}n^{k}}{k! 2^{k}}.$$ But now, I am totally clueless. By intuition, I know that the limit has to be zero, so I just want to prove using the concept of inequalities that what I thought was the right approach (being the sandwich theorem). Any help would be appreciated!
52587
Find $$\lim_{n \to \infty} \frac {1}{2^n} \sum_{k=1}^n \frac{1}{\sqrt{k}} \binom{n}{k}.$$  First time I thought about Stirling's approximation but didn't get anything by applying it. I would also think about a Riemann Sum, but no idea how to rewrite...  The answer is $0$.