How do we prove that $$\lim_{n \rightarrow \infty} \dfrac{\displaystyle\sum_{r=1}^{n} r^a}{n^{a+1}}=\frac{1}{a+1}$$?  This type of terms appear in problems on limits, but I am unable to prove this.  Please help me out.
5154
What is the result of the next limit:   $$ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $$  Why (theorem)?
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.
102122
How can we prove that  $\displaystyle \lim_{n\to\infty}  \sum_{k=0}^{n} \frac{{n \choose k}}{n^{k}(k+3)} = e-2$  ?  The only thing that came to my mind was writing out a few terms to see where that could get me, but I'm not able to make much progres. Any help in the form of hints or a solution would be appreciated.
70232
I'm asked to find $\displaystyle\lim_{n\to \infty}\sum_{k=1}^n \left(\frac{k}{n}\right)^n$.  It seems that the sum converges.  I managed to prove that $\forall n,1&lt;\sum_{k=1}^n \left(\frac{k}{n}\right)^n&lt;e$  But I don't know how to prove convergence, nor find the value of the limit.
98908
Let $(f_n)$- real sequence such that $$ \lim_{n\rightarrow \infty} \frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}=a+b, $$ and $$ \lim_{n\rightarrow \infty} \frac{f_{n}^2-f_{n-1}f_{n+1}}{f_{n+1}^2-f_nf_{n+2}}=ab   \quad    (|a|&lt;|b|). $$  Prove that:$$\lim_{n\rightarrow \infty} \frac{f_n}{f_{n+1}}=a $$  I think we must prove $\displaystyle\lim_{n\rightarrow \infty} \frac{f_n}{f_{n+1}} $ exists,and we prove this limit is $a$,But I can't prove this limit exists.  My idea: since $$\lim_{n\to\infty}\dfrac{\dfrac{f_{n}}{f_{n+1}}-\dfrac{f_{n-1}}{f_{n}}\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}{1-\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}=a+b$$ and  $$\lim_{n\to\infty}\dfrac{\left(\dfrac{f_{n}}{f_{n+1}}\right)^2-\dfrac{f_{n-1}}{f_{n}}\dfrac{f_{n}}{f_{n+1}}}{1-\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}=ab$$ But I felt this deal is not useful,  Other idea: I want to take the  Fibonacci sequence  to solve this problem, But I can't,   Thank you
116877
Let $a_n&gt;0$. Prove that $$\varlimsup_{n\to\infty}n\left(\frac{1+a_{n+1}}{a_n}-1\right)\geq 1.$$  I argue by contradiction. If it is not ture, then $$\exists\ N,\ \forall\ n\geq N, n\left(\frac{1+a_{n+1}}{a_n}-1\right)&lt;1\Rightarrow 1+a_{n+1}-a_n&lt;\frac{a_n}{n}.$$ What contradiction shall I get then....
118369
$$\sum_{i=0}^{n-1} ar^i = \frac{a(r^n-1)}{r-1}$$ for  $r \in \mathbb R, r  \ne 1, n \ge 1 $  i. Show true for $n=1$ $$\frac{a(r^1-1)}{r-1} = a$$  ii. I am stuck on how to do the second part of the proof, and the logic behind it.
21003
Prove or disprove: if $\sum\limits_{n=0}^\infty a_n$  converges and $\lim\limits_{n→\infty}⁡b_n =0$, then $\sum\limits_{n=0}^\infty a_n  b_n$ converges.  I tried this example $$a_n= b_n = \frac {(-1)^n} {\sqrt {n+1}},$$ this will show that $$\sum_{n=0}^\infty a_n  b_n = \sum_{n=0}^\infty \frac {1} { {n+1}},$$ which is the Harmonic series, diverges.   However, I tried every convergent test but I couldn't show that $$\sum_{n=0}^\infty \frac {(-1)^n} {\sqrt {n+1}}$$ converges.
50833
Let $\{a_n\}, {n\geq 1}$, be a sequence of real numbers satisfying $|a_n|\leq 1$ for all $n$. Define  $$A_n = \frac{1}{n}(a_1 + a_2 + \cdots + a_n),$$ for $n\geq 1$. Then find $\displaystyle\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)$ .    I proceed in this way  $$\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)=\lim_{n \rightarrow \infty}\sqrt{n}\left[\frac{1}{n+1}(a_1 + a_2 + \cdots + a_n+a_{n+1})-\frac{1}{n}(a_1 + a_2 + \cdots + a_{n})\right]=\lim_{n \rightarrow \infty}\left[{(na_{n+1}-a_1 - a_2 - \cdots - a_n})\frac{1}{\sqrt{n}(n+1)}\right]$$  Please help me to complete from here
22753
Prove that if one of $\displaystyle\lim_{x \rightarrow 0 }{f(x)}$ and $\displaystyle\lim_{x \rightarrow 0}{f(x^3)}$ exists, then the other one also exists. Can anyone guide me on this ? I have no idea on how to start .
376
Let $a_n$ be a positive real sequence such that the series $\sum a_n$ converges.  I was asked to prove that under such circumstances $\sum a_n^{\frac{n}{n+1}}$ converges.  The previous sum can be rewritten as $\sum \frac{a_n}{a_n^{1/(n+1)}}$.  I can't prove the convergence of this last series... I'm sure comparison test is enough here, but other than $a_n \rightarrow 0$ there is no information that helps.  Thanks for your help.
49091
I try to solve the following tricky limit:  $$\lim_{x\rightarrow\infty} \sum_{k=1}^{\infty} \frac{kx}{(k^2+x)^2} $$  For some large values, W|A shows that its limit tends to $\frac{1}{2}$ but not sure how to prove that.
122687
Let $f:[0, 1] \rightarrow \mathbb{R}$ a continuous function. If $a&gt;0$, show that:  $$ f(0)\ln(\frac{b}{a})=\lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx$$  Tried using Riemann sum, but did not succeed.
58830
I have to find the radius of convergence for $\displaystyle\sum_{n=0}^{\infty}\frac{n}{n+1}x^n$, for x$\in$R. I think it is $r^{-1}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{ \frac{n+1}{n+2} }{ \frac{n}{n+1}}=1$ so r=1. Then I have to find out if $\displaystyle\sum_{n=0}^{\infty}\frac{n}{n+1}x^n$ is convergent in  the end points of the convergence interval. I think I have to check for $|r|$. But we can't use ratio test (L=1) or integral test(not decreasing) for $\displaystyle\sum_{n=0}^{\infty}\frac{n}{n+1}$. How can I then check if it's convergent in the end points of the convergence interval?
103605
Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[  \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$  I have a question about the following solution:  We may write it in the form:  $$ \frac{1}{n} \left[ \frac{1}{1+(\frac{1}{n})^2} + \ldots + \frac{1}{1+(\frac{n}{n})^2} \right] $$  Somehow I need to figure out that the limit is actually the Riemann sum of $\frac{1}{1 + x^2}$ on $[0,1]$ for $\pi = 0 &lt; \frac{1}{n} &lt; \ldots &lt; \frac{n}{n}$.  Can you explain to me to reach this conclusion?
28023
Let $a_1,a_2,\ldots$ be a sequence of positive real numbers. Suppose that $a_{m+n}\leq a_m+a_n$ for all $n\geq 1$. Does $\lim_{n\rightarrow\infty}\dfrac{a_n}{n}$ always exist?  From $a_{m+n}\leq a_m+a_n$ we know that $a_n\leq na_1$, so that $\dfrac{a_n}{n}\leq a_1$, which means the sequence $\dfrac{a_n}{n}$ is bounded both from above and below. But this is not enough to conclude that the limit exists.
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