Using the Method of Differences to Find the Sum of a Series

[converting1]

find \displaystyle\sum_{r=1}^n (r-1 + \dfrac{1}{r} - \dfrac{1}{r+1})

using the method of differences

if I split it up (i.e do sum of r and sum of 1 and sum of (1/r - 1/(r+1)) to n I get the right answer of \dfrac{n(n^2 + 1)}{2(n+1)}

however if I do the method of differences right away of the expression I get:

r= 1: 0 + 1 - 1/2
r= 2: 1 + 1/2 - 1/3
r= 3: 2 + 1/3 - 1/4
...

r= n-1: n-2 + 1/(n-1) - 1/n
r = n: n-1 + 1/n - 1/(n+1)

and adding all of this I get \displaystyle\sum_{r=0}^{n-1} r - \dfrac{1}{n+1}

= \dfrac{n(n-1)}{2} - \dfrac{1}{n+1}

which doesn't give me the right answer...

where have I gone wrong using the second method?

 

[Ray Vickson]

find \displaystyle\sum_{r=1}^n (r-1 + \dfrac{1}{r} - \dfrac{1}{r+1})

using the method of differences

if I split it up (i.e do sum of r and sum of 1 and sum of (1/r - 1/(r+1)) to n I get the right answer of \dfrac{n(n^2 + 1)}{2(n+1)}

however if I do the method of differences right away of the expression I get:

r= 1: 0 + 1 - 1/2
r= 2: 1 + 1/2 - 1/3
r= 3: 2 + 1/3 - 1/4
...

r= n-1: n-2 + 1/(n-1) - 1/n
r = n: n-1 + 1/n - 1/(n+1)

and adding all of this I get \displaystyle\sum_{r=0}^{n-1} r - \dfrac{1}{n+1}

= \dfrac{n(n-1)}{2} - \dfrac{1}{n+1}

which doesn't give me the right answer...

where have I gone wrong using the second method?

You dropped a "+1". You should have gotten
1 + \frac{n(n-1)}{2} - \frac{1}{n+1}

 

[kevinj888]

This is an telescoping Sequence