Problem: Find  $$\lim_{n\to \infty} \left[\frac{1}{n^k}(1^k+2^k+...+n^k)-\frac{n}{k+1}\right].$$  We solve this problem by taking $x_n=(k+1)(1^k+2^k+...+n^k)-n^{k+1}$ and $y_n=n^k(k+1)$ then by Stolz's theorem we have that $$\lim_{n\to \infty}\frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}=\frac{(k+1)n^{k}-(n)^{k+1}+(n-1)^{k+1}}{(k+1)[n^{k}-(n-1)^k]}$$  $$=\frac{(k+1)/n-1+(1-1/n)^{k+1}}{\frac{k+1}{n}[1-(1-1/n)^k]}$$ $$=\frac{(k+1)/n-1+(1-1/n)^{k+1}}{\frac{k+1}{n^2}[(1-1/n)^{k-1}+(1-1/n)^{k-2}+...+1]}$$ $$=\frac{nk+n-n^2+n^2(1-1/n)^{k+1}}{(k+1)[(1-1/n)^{k-1}+(1-1/n)^{k-2}+...+1]}$$  I am not sure how to proceed from here and thus any hints will be much appreciated.
122981
Suppose $a \in \mathbb{R} ,a&gt;1$ Is there an idea to compute the limit below ? $$\lim_{n \to \infty}\left( \frac{1^a+2^a+3^a+...+n^a}{n^a}-\frac{n}{a+1} \right)$$ I tried it for $a=1,2,3$ but I get stuck in general form . Can someone help me ? Thanks in advance.
18292
I'm just beginning my journey with calculus, and this problem is giving me hard time.  $$\lim_{n \to \infty} \sum_{k=1}^n \frac{k}{{n^2+k}}$$  I calculated first three sums: $$\frac{1}{2}, \frac{16}{30}, \frac{117}{220}$$ From that I guessed that the limit is $$\frac{1}{2}$$. The hard part, though, is proving it.    I reckon that the squeeze theorem should be used here, but I can't seem to find anything useful. (The only sequences I was able to figure out give me limits of 1 and 0, so it's not something I can use)  Maybe you guys can give me some ideas or theorems that can be of use here? I know that this problem could be probably solved using some more advanced calculus techniques, but I'm not familiar with derivatives and integrals yet, and this problem is in the limits section of my problem book, so I'm not supposed to use anything more advanced.
8638
How would I go about finding $\prod_{n=1}^\infty (1+\frac{1}{n^2})$ ? Setting $a_k = \prod_{n=1}^k (1+\frac{1}{n^2})$, the first few partial products are $a_1 = 2$, $a_1 = 2.5$, $a_3 \approx 2.78$, $a_4 \approx 2.591$, $a_{20} \approx 3.50$, and they eventually appear to converge at around $3.639$. (These values were found with a quick python program. So much for my initial conjecture that the product converged to $e$.)   I tried writing $1+\frac{1}{n^2}$ as $\frac{n^2 + 1}{n^2}$, but it wasn't immediately obvious how to prove convergence / find the limit because the product doesn't really factor or telescope. Is there a way to calculate the limit?
110152
I need to prove that $$\lim_{n\rightarrow \infty} \frac{n^k}{2^n}=0$$ where $k\in \mathbb{N}$. All I can think of is to use something like L'Hopital's rule but I suppose there must be a another simpler way. I would much appreciate if someone could give me a hint. Thanks
4837
Let $$a_n=\left(1-\dfrac{1}{\sqrt2}\right)\dots \left(1-\dfrac{1}{\sqrt{n+1}}\right),n\ge1$$ Then find $\lim_{n\to \infty} a_n$.   How can I proceed? I am stuck at the first step. Please help.
46401
What would you suggest here?  $$\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} n^{1/k} $$
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?  $\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| &lt; \epsilon$  $\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} &lt; \epsilon$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} &gt; \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow \sqrt{n} &gt; \frac{1-2 \epsilon}{2}$  $\Rightarrow n &gt; \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
10877
How can I prove $\lim_{n \to \infty} \frac{n^2}{2^n}=0$  I tried to use $\mid \frac{n^2}{2^n} - 0 \mid &lt;\epsilon $ However, because of $n^2$ I cannot use it. Also, I tried to use ratio test and I got $\lim_{n-&gt;\infty}\frac{(n+1)^2}{2}\frac{1}{n^2}$, but after that one, I dont know how to get limit &lt; 0  I have no idea to solve this problem.   I can also use a squeez lemma, but dont know how to apply to this question.
84235
I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:  \begin{align*} \lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&amp;=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\ &amp;=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4} \end{align*}  where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?
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.
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.
109362
Evaluate the limit      $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$   My Attempt:  $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} = \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$  How can I solve the problem from this point?
49267
The limit to solve is $$\lim_{n\to\infty}\left(\frac{n^2+2}{2n^2+1}\right)^{n^2}$$ I tried to use L'Hôpital's rule but the derivatives are quite complex.
28088
I have the following problem.   I have to find the limit of $b_n = \frac{a_n}{n}$, where $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=l$  My approach:  I express $a_n$ in terms of $b_n$, i.e. $a_n=nb_n$ and $a_{n+1}=(n+1)b_n$  We look at the difference: $a_{n+1}-a_n=(n+1)b_{n+1}-nb_n$  Assuming that $b_n$ converges to a real number m, we see that:  $l=(n+1)m-nm$, from where I conclude that $m=l$.  What I'm left with is proving that $b_n$ is convergent which I'm not sure how to do.  Thanks in advance!
44975
How may one calculate  $$\lim_{n\to\infty} \ \left(\left(\sum_{k=1}^{n} \frac{1}{3k-1}\right) - \frac{\ln n}{3}\right) \ ?$$
77486
How can I find the following limit :$$\lim_{n\to\infty}\left(1-\frac{1}{3}\right)^2\left(1-\frac{1}{6}\right)^2\left(1-\frac{1}{10}\right)^2\cdots\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2$$There are four options $(a)\frac1{3}(b)\frac1{9}(c)\frac1{81}(d)0$ Thanks!
120017
$$\lim_{n \to \infty} \frac{n!}{n^{n}}$$    Attempt:  $$n!=n\left( n-1 \right)\left( n-2 \right)...\left( n-\left( n-1 \right) \right)=n^{n}+...$$  $$\lim_{n \to \infty} \frac{n!}{n^{n}}=\lim_{n \to \infty} \frac{n^{n}+...}{n^{n}}=\lim_{n \to \infty} \frac{1+...}{1}=1+\lim_{n \to \infty} \ ...=1$$    I based this on the fact that $n$ is the highest power. The answer's meant to be $0$ though.