![\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k+1-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{(k+1)^2-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\ ](/tpl/images/0419/7356/7a627.png)
![=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\dfrac{\sqrt{(k+1)^2-1}}{\sqrt{(k+1)^2}}-\dfrac{1}{k+1}\cdot\dfrac{\sqrt{k^2-1}}{\sqrt{k^2}}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{\dfrac{(k+1)&^2-1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{\dfrac{k^2-1}{k^2}}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\dfrac{1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{1-\dfrac{1}{k^2}}\right]=\\\\\\ ](/tpl/images/0419/7356/f2f2b.png)
![=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\left(\dfrac{1}{k+1}\right)^2}-\dfrac{1}{k+1}\cdot\sqrt{1-\left(\dfrac{1}{k}\right)^2}\right]=\\\\\\= \sum\limits_{k=1}^n\left[\arcsin\left(\dfrac{1}{k}\right)-\arcsin\left(\dfrac{1}{k+1}\right)\right]=\\\\\\](/tpl/images/0419/7356/cf3c5.png)
![=\bigg[\arcsin(1)-\arcsin\left(\frac{1}{2}\right)\bigg]+\bigg[\arcsin\left(\frac{1}{2}\right)-\arcsin\left(\frac{1}{3}\right)\bigg]+\\\\\\+ \bigg[\arcsin\left(\frac{1}{3}\right)-\arcsin\left(\frac{1}{4}\right)\bigg]+\dots+ \bigg[\arcsin\left(\frac{1}{n}\right)-\arcsin\left(\frac{1}{n+1}\right)\bigg]=\\\\\\](/tpl/images/0419/7356/3a9a1.png)
![\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)}](/tpl/images/0419/7356/a7793.png)
![\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \lim\limits_{n\to\infty}\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \lim\limits_{n\to\infty}\Bigg(\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)\Bigg)=\dfrac{\pi}{2}-\lim\limits_{n\to\infty}\arcsin\left(\dfrac{1}{n+1}\right)=\\\\\\= \Bigg\{\dfrac{1}{n+1}\xrightarrow{n\to\infty}0\Bigg\}=\dfrac{\pi}{2}-\arcsin(0)=\dfrac{\pi}{2}-0=\dfrac{\pi}{2}](/tpl/images/0419/7356/94cf4.png)
![\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}](/tpl/images/0419/7356/823e9.png)
A) compute the sum
arcsin
and express your answer in terms of n.
...
Mathematics, 15.12.2019 18:31 ayoismeisjuam
A) compute the sum
arcsin
and express your answer in terms of n.
b) prove that the series
arcsin
converges to π/2.
Answers: 2
Another question on Mathematics
Mathematics, 21.06.2019 15:30
Will used 20 colored tiles to make a design in art class. 5/20 of his tiles were red. what is an equivalent fraction for the red tiles?
Answers: 1
Mathematics, 21.06.2019 22:00
You buy 4 items that cost 2.69 14.45 8.05 13.95 what is the estimated total round to the nearest dollar
Answers: 1
Mathematics, 22.06.2019 03:40
(25 ! ) what does it mean for an equation to have no solution or infinitely many solutions?
Answers: 1
You know the right answer?
Questions
Social Studies, 11.02.2021 22:50
Mathematics, 11.02.2021 22:50
Biology, 11.02.2021 22:50
Mathematics, 11.02.2021 22:50
Biology, 11.02.2021 22:50
Mathematics, 11.02.2021 22:50
Computers and Technology, 11.02.2021 22:50
Mathematics, 11.02.2021 22:50
English, 11.02.2021 22:50
Mathematics, 11.02.2021 22:50
History, 11.02.2021 22:50
Chemistry, 11.02.2021 22:50
