I know, when n is even, $$\int_{0}^{\frac {\pi}{2}} \sin^{n} x dx = \frac {(2m-1)!!}{(2m)!!}\frac {\pi}{2} $$ and n is odd, $$\int_{0}^{\frac {\pi}{2}} \sin^{n} x dx = \frac {(2m)!!}{(2m+1)!!} $$ But when let limit n to infinite, how can I get the limit, or if the limit exists. -- I find a good way to solve the limit:$$\lim_{x \to +\infty} \frac{(2n-1)!!}{(2n)!!} $$ as we know, $2 =\frac{1+3}{2} &gt; \sqrt{1 \cdot 3}$ , so $2n = \frac{(2n-1)+(2n+1)}{2} &gt; \sqrt{(2n-1)(2n+1)}$ so, $$\lim_{x \to +\infty} \frac{(2n-1)!!}{(2n)!!} = \frac{\sqrt{1 \cdot 3}\cdots \sqrt{(2n-3)(2n-1)}\sqrt{2n-1}}{\sqrt{1\cdot 3} \cdots \sqrt{(2n-3)(2n-1)} \sqrt{(2n-1)(2n+1)}} = \frac{1}{\sqrt{2n+1}} \to 0 (n \to \infty)$$
120543
Show:       $$ \lim_{n\to \infty} \int_{0}^{\frac{\pi }{2}} \sin^n(x) = 0. $$   I showed this using the . I want other methods to prove it. Thank you.
99321
Limit to evaluate: $$\lim_{x \rightarrow 0} \cfrac{\sin{(x)}+\sin{(3x)}+\sin{(5x)}}{\sin{(2x)}+\sin{(4x)}+\sin{(6x)}}$$  Proposed solution: $$ \cfrac{\sin(x)+\sin(3x)+\sin(5x)}{\sin(2x)+\sin(4x)+\sin(6x)} \Bigg/ \cdot\ \cfrac{1/x}{1/x}\Bigg/= \frac{\cfrac{\sin(x)}x + \cfrac{\sin(3x)}{3x} \cdot 3 + \cfrac{\sin(5x)}{5x} \cdot 5} {\cfrac{\sin(2x)}{2x} \cdot 2 + \cfrac{\sin(4x)}{4x} \cdot 4 + \cfrac{\sin(6x)}{6x} \cdot 6} $$  Using $\lim_{x \rightarrow 0} \frac{\sin x}x=1$, we get $$\frac{1+1\cdot 3+1\cdot 5}{1\cdot 2+1\cdot 4+1\cdot 6} = \frac 9{12} = \frac 3 4$$  Please tell me if I am correct.
25244
I want to compute $$\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}.$$  Is it OK how I want to do?    $$\sin{\sqrt{x+1}}-\sin{\sqrt{x}}=2\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}=2\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}$$  I'm not surem but I want to say that $$|\cos \frac{\sqrt{x+1}-\sqrt{x}}{2}| \leq 1$$ $$\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}} \to 0 \mbox{ when } x \to \infty$$ So the limit is $0$ ?  I thinks is not because  if I say that $ \displaystyle|\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}| \leq 1$ I will obtain that another limit than $0$.   Thanks :)
77099
Evaluate the limit  $$\displaystyle\lim_{n \to \infty} \int_{0}^{\pi} \frac{\sin x}{1+\cos ^2(nx)} dx$$  Using property of definite integral $\int_{0}^{2a} f(x).dx=2\int_{0}^{a} f(x)dx$,when $f(2a-x)=f(x)$ I got  $$\displaystyle\lim_{n \to \infty} \int_{o}^{\pi} \frac{\sin x}{1+\cos ^2(nx)} dx=2\displaystyle\lim_{n \to \infty} \int_{o}^{\pi/2} \frac{\sin x}{1+\cos ^2(nx)} dx$$ but I cannot proceed after that. Could someone provide me with some hint? Till now I have only done integration in terms of elementary functions. Any hint would be appreciated.
13847
I am trying to compute this integral/limit, I don't feel like I have any good insight...   $$\lim_{n \to \infty} \int_0^{n^2} e^{-x^2}n\sin\frac{x}{n} \, dx.$$  I have tried to make a change of variable to get rid of the $n^2$, I changed to $X=\frac{x}{n^2}$ but got something even worse, I've tried to reach a situation where I could use a convergence theorem for Lebesgue Integrals,... I'm not sure I'm even on the right track!  Could you give me a hint on how to start this?  Thank you very much!
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!
65279
How can I calculate this limit? $$\lim_{n\rightarrow\infty} \frac{7^{\sqrt{n+1}-\sqrt{n}}\cdot(\frac{n+1}{2})!\cdot(\frac{n+1}{2})!}{(n+1)\cdot(\frac{n}{2})!\cdot(\frac{n}{2})!}$$ I don't have idea and I will be happy for help.
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?
1614
Evaluate   $$\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}.$$   $\underline{\bf{My\;\;Try}}::$ Let $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n} = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \left(\frac{(n+1)\cdot (n+2)\cdot (n+3)\cdots (n+n)}{(1)\cdot (2)\cdot (3)\cdots (n)}\right)$  $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \left\{\ln \left(\frac{n+1}{1}\right)+\ln \left(\frac{n+2}{2}\right)+\ln \left(\frac{n+3}{3}\right)+\cdots+\ln \left(\frac{n+n}{n}\right)\right\}$  $\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{n+r}{r}\right) = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{1+\frac{r}{n}}{\frac{r}{n}}\right)$  Now Using Reinman Sum  $\displaystyle y = \frac{1}{2}\int_{0}^{1}\ln \left(\frac{x+1}{x}\right)dx = \frac{1}{2}\int_{0}^{1}\ln (x+1)dx-\frac{1}{2}\int_{0}^{1}\ln (x)dx =  \ln (2)$  My Question is , Is there is any method other then that like Striling Approximation OR Stolz–Cesàro theorem OR Ratio Test  If yes then please explain here  Thanks
108257
Find the limit of the sequence $a_n=\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}.$  The limit is supposed to be $e^{-1}$ and there are a couple of posts in MSE proving it. But here is a proof I encountered showing the limit is $1.$ Please tell me what's going wrong here.      Proof:  We can write $a_n=\sqrt[n]{n!}(b_n-1),$ where $b_n=\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}.$ Hence $$a_n=n\cdot\frac{\sqrt[n]{n!}}{n}\cdot\frac{b_n-1}{\ln{b_n}}=\frac{\sqrt[n]{n!}}{n}\cdot\frac{b_n-1}{\ln{b_n}}\cdot\ln{b_n^n.}$$   But $\lim \limits_{n\to \infty}\frac{\sqrt[n]{n!}}{n}=e^{-1},$ so $b_n\to1$ as $n\to \infty.$ On the other hand, $$\lim \limits_{n\to \infty}b_n^n=\lim \limits_{n\to \infty}\frac{(n+1)!}{n!}.\frac{1}{\sqrt[n+1]{(n+1)!}}=e.$$   Also $$\lim \limits_{n\to \infty}\frac{b_n-1}{\ln{b_n}}=\lim \limits_{n\to \infty}\frac{1}{\ln{[1+(b_n-1)]^{1/(b_n-1)}}}=\frac{1}{\ln e}=1$$         And the author goes on to say that the limit is $e^{-1}.$ But doesn't it say that the limit is $1?$ I must have overlooked something so obvious.
95514
I'm trying to find some nice proofs for the following limit  $$\lim_{n\to\infty} \int_{0}^{\pi/4}\tan^n x \ dx$$  One way is to use the integration by parts. What else can we do here? Are there some fast ways? All answers will have my upvote. Thanks!
44551
Let $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. Show that $a_n&gt;0\  \forall\  n\ge1$.   Prove or disprove: $\sum\limits_{n=1}^\infty a_n$ is convergent.     I can't show that $a_n &gt; 0\  \forall n\ge1$. I tried using induction but it wouldn't work.  Attempt: $$ \begin{align} 2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}&amp;=(2\sqrt{n}-(\sqrt{n-1}+\sqrt{n+1}))\cdot {2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\\&amp;={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\end{align}$$ (The calculations are true for sure. No check is desired.) Denote $$a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\text{ and }b_n={1\over n^{2}}.$$   Then  $$\begin{align*} \lim_{n\to \infty}{a_n\over b_n} &amp; =\lim_{n\to \infty}n^{2}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\ &amp;=\lim_{n\to \infty}n^{2}\lim_{n\to \infty}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\ &amp;=0 \end{align*}$$ By the comparison test for series convergence, since $\lim_{n\to \infty}{a_n\over b_n}=0$, then if $b_n$ converges, which it does, so does $a_n$.
59390
Let $x&gt;0$, find the limit $$\lim_\limits{n\to{\infty}}{n^2\left(\sqrt[n]{x}-\sqrt[{n + 1}]{x}\right)}$$  I use Maclaurin series and find out that the limit is $\ln x$. And this is the answer I get from a math forum:  "Let $x_n=\sqrt[n]{x}-1$, $y_n=\dfrac{1}{n}$, then $(x_n)\to0,(y_n)\to0,(y_n)\downarrow$ when $n\to\infty$.  Use Stolz-Cesaro theorem, we have:  $$\mathop {\lim }\limits_{n \to \infty } \dfrac{{\sqrt[n]{x} - \sqrt[{n + 1}]{x}}}{{\frac{1}{n} - \frac{1}{{n + 1}}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\sqrt[n]{x} - 1}}{{\frac{1}{n}}} = \ln x$$ Then $\mathop {\lim }\limits_{n \to \infty } {n^2}\left( {\sqrt[n]{x} - \sqrt[{n + 1}]{x}} \right) = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{n^2}}}{{n\left( {n + 1} \right)}}\cdot\dfrac{{\sqrt[n]{x} - \sqrt[{n + 1}]{x}}}{{\frac{1}{n} - \frac{1}{{n + 1}}}} = \ln x$.  /END"  I just wonder what form Stolz-Cesaro theorem that he used, because I just learn Stolz-Cesaro theorem when $(y_n)\to\infty$.  Could anyone help me to get this? Thank you in advance.     This is "another form of Stolz theorem" in the forum that I mentioned: Let $(x_n),(y_n)$     be two sequences of real numbers. Assume $\mathop {\lim }\limits_{n \to \infty } {x_n} = \mathop {\lim }\limits_{n \to \infty } {y_n} = 0$; $y_n&gt;0$, $(y_n)$ is strictly decreasing  and the following limit exist $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x_{n + 1}} - {x_n}}}{{{y_{n + 1}} - {y_n}}} = L$. Then $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x_n}}}{{{y_n}}} = L$. Is this correct?
97153
I want to evaluate $$\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right).$$  I have computed  \begin{align*} a_{n+1}-a_n &amp; = \frac{1}{\sqrt{(2n+1)(2n+2)}}-\frac{1}{\sqrt{n(n+1)}}\\             &amp; = \frac{1}{\sqrt{n+1}}\left( \frac{1}{\sqrt{2(2n+1)}}-\frac{1}{\sqrt{n}} \right). \end{align*} Now I'm stuck.
79877
I have this difficult integral to solve.  $$ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} $$  Now my approach is this: split $(\sin x)^{\sqrt 2 + 1}$ and $(\sin x)^{\sqrt 2 - 1}$ as $(\sin x)^{\sqrt 2}.(\sin x)$ and $(\sin x)^{\sqrt 2 - 2}.(\sin x)$ respectively, and then apply parts. But that doesn't seem to lead anywhere. Hints please!  Edit:  This is what I did (showing just for the numerator)  $$ \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx $$ $$ = \int_0^{\pi/2} (\sin x)^{\sqrt 2}.(\sin x) dx  $$ $$ = (-\cos x)(\sin x)^{\sqrt 2}\Bigg|_0^{\pi/2} + \int_0^{\pi/2}(\sin x)^{ \sqrt 2 - 1 }(\cos^2 x) dx  $$  (taking $ v = \sin x $ and $ u = (\sin x)^{\sqrt 2} $ in the $ \int uv $ formula)  $$ = \int_0^{\pi/2}\left( (\sin x)^{ \sqrt 2 - 1 } - (\sin x)^{ \sqrt 2 + 1 } \right) dx  $$  Similarly for the denominator. This does give a reduction formula but then I don't see how to really use it for finding the answer.
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 $x$.
23277
I am working on trying to solve this problem:     Prove: $\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$   Here are the steps that I follow in the example that I am reading:     $u = \sin^{n - 1}{x}$      $du = (n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx$      $v = -\cos{x}$      $dv = \sin{x} \ dx$            $\int \sin^n{x} \ dx =  \sin^{n - 1}{x} \cdot \sin{x} \ dx$      $\int \sin^n{x} \ dx = \underbrace{\sin^{n - 1}{x}}_{u} \cdot \underbrace{-\cos{x}}_{v} - \int \underbrace{-\cos{x}}_{v} \cdot \underbrace{(n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx}_{du}$      $\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \cdot \cos^{2}{x} \ dx$      $\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \cdot \left(1 - \sin^{2}{x}\right) \ dx$   Here is where I get lost. How did we go from $\int \sin^{n - 2}{x} \cdot \left(1 - \sin^{2}{x}\right) \ dx$ to $\int \sin^{n - 2}{x} \ dx - (n - 1) \int \sin^{n}{x} \ dx$? Even more specifically, where did $\sin^{n}{x}$ come from?     $\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \ dx - (n - 1) \int \sin^{n}{x} \ dx$   I get this part.     $n\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \ dx$      $\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot x \ \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$   Could someone please explain what I am missing?  Thank you for your time.