Evaluate This Infinite Series!!!

[ohayme]

Hi, I'm doing my calculus homework over the break, and I got stuck on this series problem.

70. Evaluate the series:

. . . . .\(\displaystyle \displaystyle \sum_{k\, =\, 1}^{\infty}\, \)\(\displaystyle \bigg[\, Sin^{-1}\left(\dfrac{1}{k}\right)\, -\, Sin^{-1}\left(\dfrac{1}{k\, +\, 1}\right)\, \bigg]\)

I know that first you would probably split the series, but I don't know what to do next.

 

[Deleted member 4993]

Hi, I'm doing my calculus homework over the break, and I got stuck on this series problem.

70. Evaluate the series:

. . . . .\(\displaystyle \displaystyle \sum_{k\, =\, 1}^{\infty}\, \)\(\displaystyle \bigg[\, Sin^{-1}\left(\dfrac{1}{k}\right)\, -\, Sin^{-1}\left(\dfrac{1}{k\, +\, 1}\right)\, \bigg]\)

I know that first you would probably split the series, but I don't know what to do next.

\(\displaystyle \displaystyle{\sum_{k=1}^{\infty}\sin^{-1}\left (\frac{1}{k}\right ) - \sin^{-1}\left (\frac{1}{k+1}\right )}\)

\(\displaystyle = \ \displaystyle{\left [\sin^{-1}\left (\frac{1}{1}\right ) - \sin^{-1}\left (\frac{1}{2}\right )\right ] \ + \ \left [\sin^{-1}\left (\frac{1}{2}\right ) - \sin^{-1}\left (\frac{1}{3}\right )\right ] \ + \ \left [\sin^{-1}\left (\frac{1}{3}\right ) - \sin^{-1}\left (\frac{1}{4}\right )\right ]}\)......

This is a telescoping series....

 

[pka]

The problem is attached in the image below. I know that first you would probably split the series, but I don't know what to do next.

\(\displaystyle \displaystyle{\lim _{n \to \infty }}\left( {\arcsin (1) - \arcsin \left[ {{{(1 + n)}^{ - 1}}} \right]} \right) = ?\)

 

[ohayme]

Ohhh

70. Evaluate the series:

. . . . .\(\displaystyle \displaystyle \sum_{k\, =\, 1}^{\infty}\, \)\(\displaystyle \bigg[\, Sin^{-1}\left(\dfrac{1}{k}\right)\, -\, Sin^{-1}\left(\dfrac{1}{k\, +\, 1}\right)\, \bigg]\)

\(\displaystyle \displaystyle{\sum_{k=1}^{\infty}\sin^{-1}\left (\frac{1}{k}\right ) - \sin^{-1}\left (\frac{1}{k+1}\right )}\)

\(\displaystyle = \ \displaystyle{\left [\sin^{-1}\left (\frac{1}{1}\right ) - \sin^{-1}\left (\frac{1}{2}\right )\right ] \ + \ \left [\sin^{-1}\left (\frac{1}{2}\right ) - \sin^{-1}\left (\frac{1}{3}\right )\right ] \ + \ \left [\sin^{-1}\left (\frac{1}{3}\right ) - \sin^{-1}\left (\frac{1}{4}\right )\right ]}\)......

This is a telescoping series....

Ohhh i didnt realize. ..I see it now!! Thank you!!