Evaluate This Infinite Series!!!

ohayme

New member
Joined
Dec 13, 2015
Messages
7
Hi, I'm doing my calculus homework over the break, and I got stuck on this series problem.

70. Evaluate the series:

. . . . .k=1\displaystyle \displaystyle \sum_{k\, =\, 1}^{\infty}\, [Sin1(1k)Sin1(1k+1)]\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.
 
Last edited by a moderator:
Hi, I'm doing my calculus homework over the break, and I got stuck on this series problem.

70. Evaluate the series:

. . . . .k=1\displaystyle \displaystyle \sum_{k\, =\, 1}^{\infty}\, [Sin1(1k)Sin1(1k+1)]\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.

k=1sin1(1k)sin1(1k+1)\displaystyle \displaystyle{\sum_{k=1}^{\infty}\sin^{-1}\left (\frac{1}{k}\right ) - \sin^{-1}\left (\frac{1}{k+1}\right )}

= [sin1(11)sin1(12)] + [sin1(12)sin1(13)] + [sin1(13)sin1(14)]\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....
 
Last edited by a moderator:
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.
limn(arcsin(1)arcsin[(1+n)1])=?\displaystyle \displaystyle{\lim _{n \to \infty }}\left( {\arcsin (1) - \arcsin \left[ {{{(1 + n)}^{ - 1}}} \right]} \right) = ?
 
Ohhh

70. Evaluate the series:

. . . . .k=1\displaystyle \displaystyle \sum_{k\, =\, 1}^{\infty}\, [Sin1(1k)Sin1(1k+1)]\displaystyle \bigg[\, Sin^{-1}\left(\dfrac{1}{k}\right)\, -\, Sin^{-1}\left(\dfrac{1}{k\, +\, 1}\right)\, \bigg]
k=1sin1(1k)sin1(1k+1)\displaystyle \displaystyle{\sum_{k=1}^{\infty}\sin^{-1}\left (\frac{1}{k}\right ) - \sin^{-1}\left (\frac{1}{k+1}\right )}

= [sin1(11)sin1(12)] + [sin1(12)sin1(13)] + [sin1(13)sin1(14)]\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!!
 
Last edited by a moderator:
Top