Here is a question: Show that the limit exists and find it $$\lim_{n\to\infty}\left(\frac n{n^2+1}+\frac n{n^2+2}+\frac n{n^2+3}+\dots+\frac n{n^2+n}\right)$$ I'm finding it a bit tricky to evaluate the limit of this series. I tried finding the limit of each term, but I think that's not the correct approach. How should I go about solving it? Can I use squeeze theorem here? If so, then how?
120997
$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$  Can we write it as following  $E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$  Let's see what happens:-  $$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$ $$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$  In the same way for further terms, we will get $0$  Let's also confirm for general term  $$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$ $$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$  So the whole expression $E$ will be zero  But actual answer is $1$  Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$  $OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$  Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?  I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.  But I got the following counter thought:-  $\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$  As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|&lt;1$  $\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$  $\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$  Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.   But surprisingly $\dfrac{1}{3}$ is the correct answer.  I am feeling very confused in these two things. Please help me.
96253
Just out of curiosity, I was wondering if anybody knows any methods (other than the integral test) of proving the infinite series where the nth term is given by $\frac{1}{n^2}$ converges.
66365
I am facing difficulty with the following limit. $$ \lim_{n\to\infty}\left(\binom{n}{0}\binom{n}{1}\dots\binom{n}{n}\right)^{\frac{1}{n(n+1)}} $$ I tried to take log both sides but I could not simplify the resulting expression. Please help in this regard. Thanks.
82879
I have to find $$\lim_{n\to\infty}{(\sqrt[3]{n+1}-\sqrt[3]{n})}$$ here what I did $$\lim_{n\to\infty}{(\sqrt[3]{n+1}-\sqrt[3]{n})}$$=$$\lim_{n\to\infty}{\left(\sqrt[3]{n\left(1+\frac{1}{n}\right)}-\sqrt[3]{n}\right)}$$ How can I proceed?
113546
Find the value of infinite series   $$ \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{2}{n^2} \right) $$  I tried to find sequence of partial sums and tried to find the limit of that sequence. But I didn't get the answer.
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.
48433
Hi I have a question regarding of limits to infinity please help! Thank You!  The question states the user to find the following limit:  $  \lim_{n\to\infty} n^2 ({\sqrt[n]{x}-\sqrt[n+1]{x}}) $      Where $(x &gt; 0) $  Thank You!!!
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!
90576
$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$  is equal to :  $\frac{9}{e^2}$ $3 \log3−2$ $\frac{18}{e^4}$ $\frac{27}{e^2}$    My attempt : $\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$ $=\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(n+2n)}{n^{2n}})^{\frac{1}{n}}$ $=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(1+2n/n)\}}{n^{2n}})^{\frac{1}{n}}$ $=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(1+2)\}}{n^{2n}})^{\frac{1}{n}}$ $=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(3)\}}{n^{2n}})^{\frac{1}{n}}$ $=\lim_{n \to \infty} (\{(1+1/n)(1+2/n)\dots(3)\})^{\frac{1}{n}}$ I'm stuck here.  Can you please explain?
24190
I got this question in my maths paper     Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$   and find the sum if it exists.   I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.
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.
5928
$$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k+n}$$  Could you please give me a hint how to find this limit?
96869
How to find $\lim\limits_{n\to\infty}\left({2\sqrt n}-\sum\limits_{k=1}^n\frac1{\sqrt k}\right)$ ?  And generally does the limit of the integral of $f(x)$ minus the sum of $f(x)$ exist? How to prove that and find the limit?
95156
Determine the following limit $$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)\ldots (n+n)} $$ Idea #1 the nth root could be rewritten as $$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{\frac{(n+n)!}{n!}} $$ Noticing that $n^{-1}$ is an infinitesimal, if the n-th root were bounded, the limit would be $0$.  But hoping that $\sqrt[n]{\frac{(n+n)!}{n!}}$ is bounded is a pipe dream, this clearly tends to $\infty$.  Idea #2 $$\lim_{n\to\infty}\ln\left (\frac{1}{n}[(n+1)\ldots (n+n)]^\frac{1}{n}\right ) = \lim_{n\to\infty}\left [-\ln n+\frac{1}{n}\left (\ln (n+1)+\ln (n+2)+\ldots +\ln (n+n) \right )\right ]$$ but trying to bring it to a common denominator doesn't help either, what I would have is $$\lim_{n\to\infty}\frac{1}{n} \left [-n\ln n+\ln (n+1)+\ldots +\ln 2+\ln n\right ]$$ Which doesn't bring me any closer to simplification. The solution of this must something entirely different. I would like a hint/point me in the direction I should go.
65629
I came across this strange limit whilst showing convergence of a series: $$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$$ How can I calculate this limit?
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!
19745
Prove that : $\displaystyle \lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$    the only thing I could think of is that it can be written like this :  $$ \lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k+n} =\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\frac{k}{n}+1}=\int_0^1 \frac{1}{x+1} \ \mathrm{d}x=\ln 2$$  is my answer right ? and are there any other method ?(I'm sure there are)