Question

A) Compute the sum


`\mathsf{\displaystyle\sum_(k=1)^n}` arcsin
`\mathsf{\left[(√(k^2+2k)-√(k^2-1))/(k(k+1))\right]}`

and express your answer in terms of n.

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b) Prove that the series


`\mathsf{\displaystyle\sum_(k=1)^(\infty)}` arcsin
`\mathsf{\left[(√(k^2+2k)-√(k^2-1))/(k(k+1))\right]}`

converges to π/2.

Answer 1

A)

To calculate this sum, we could use trigonometric identity:


`\arcsin(x)-\arcsin(y)=\arcsin\left(x√(1-y^2)-y√(1-x^2)\right)`

We have:


`\sum\limits_(k=1)^n\arcsin\left[(√(k^2+2k)-√(k^2-1))/(k(k+1))\right]=\\\\\\= \sum\limits_(k=1)^n\arcsin\left[(√(k^2+2k))/(k(k+1))-(√(k^2-1))/(k(k+1))\right]=\\\\\\= \sum\limits_(k=1)^n\arcsin\left[(√(k^2+2k+1-1))/(k(k+1))-(√(k^2-1))/(k(k+1))\right]=\\\\\\= \sum\limits_(k=1)^n\arcsin\left[(√((k+1)^2-1))/(k(k+1))-(√(k^2-1))/(k(k+1))\right]=\\\\\\`


`=\sum\limits_(k=1)^n\arcsin\left[(1)/(k)\cdot(√((k+1)^2-1))/(√((k+1)^2))-(1)/(k+1)\cdot(√(k^2-1))/(√(k^2))\right]=\\\\\\= \sum\limits_(k=1)^n\arcsin\left[(1)/(k)\cdot\sqrt{((k+1)&^2-1)/((k+1)^2)}-(1)/(k+1)\cdot\sqrt{(k^2-1)/(k^2)}\right]=\\\\\\= \sum\limits_(k=1)^n\arcsin\left[(1)/(k)\cdot\sqrt{1-(1)/((k+1)^2)}-(1)/(k+1)\cdot\sqrt{1-(1)/(k^2)}\right]=\\\\\\`


`=\sum\limits_(k=1)^n\arcsin\left[(1)/(k)\cdot\sqrt{1-\left((1)/(k+1)\right)^2}-(1)/(k+1)\cdot\sqrt{1-\left((1)/(k)\right)^2}\right]=\\\\\\= \sum\limits_(k=1)^n\left[\arcsin\left((1)/(k)\right)-\arcsin\left((1)/(k+1)\right)\right]=\\\\\\`


`=\bigg[\arcsin(1)-\arcsin\left((1)/(2)\right)\bigg]+\bigg[\arcsin\left((1)/(2)\right)-\arcsin\left((1)/(3)\right)\bigg]+\\\\\\+ \bigg[\arcsin\left((1)/(3)\right)-\arcsin\left((1)/(4)\right)\bigg]+\dots+ \bigg[\arcsin\left((1)/(n)\right)-\arcsin\left((1)/(n+1)\right)\bigg]=\\\\\\`


`=\arcsin(1)-\arcsin\left((1)/(2)\right)+\arcsin\left((1)/(2)\right)-\arcsin\left((1)/(3)\right)+\arcsin\left((1)/(3)\right)-\\\\\\-\arcsin\left((1)/(4)\right)+\dots+\arcsin\left((1)/(n)\right)-\arcsin\left((1)/(n+1)\right)=\\\\\\= \arcsin(1)-\arcsin\left((1)/(n+1)\right)=(\pi)/(2)-\arcsin\left((1)/(n+1)\right)`

So the answer is:


`\sum\limits_(k=1)^n\arcsin\left[(√(k^2+2k)-√(k^2-1))/(k(k+1))\right]=(\pi)/(2)-\arcsin\left((1)/(n+1)\right)}`

B)


`\sum\limits_(k=1)^\infty\arcsin\left[(√(k^2+2k)-√(k^2-1))/(k(k+1))\right]=\\\\\\= \lim\limits_(n\to\infty)\sum\limits_(k=1)^n\arcsin\left[(√(k^2+2k)-√(k^2-1))/(k(k+1))\right]=\\\\\\= \lim\limits_(n\to\infty)\Bigg((\pi)/(2)-\arcsin\left((1)/(n+1)\right)\Bigg)=(\pi)/(2)-\lim\limits_(n\to\infty)\arcsin\left((1)/(n+1)\right)=\\\\\\= \Bigg\{(1)/(n+1)\xrightarrow{n\to\infty}0\Bigg\}=(\pi)/(2)-\arcsin(0)=(\pi)/(2)-0=(\pi)/(2)`

So we prove that:


`\sum\limits_(k=1)^\infty\arcsin\left[(√(k^2+2k)-√(k^2-1))/(k(k+1))\right]=(\pi)/(2)`