The sum in problem 6.4a
The sum we need to complete is,
S
=
1
n
2
−
m
2
⎛
⎝
⎜
⎞
⎠
⎟
m
≠
n
∑
.
This can be simplified using partial fractions,
S
=
1
2
n
1
n
−
m
+
1
n
+
m
⎛
⎝
⎜
⎞
⎠
⎟
m
≠
n
∑
=
1
2
n
1
n
−
m
m
≠
n
∑
+
1
n
+
m
m
≠
n
∑
⎛
⎝
⎜
⎞
⎠
⎟
.
The sums can each be broken into the part up to n1 and the part from n+1 upward,
S
=
1
2
n
1
n
−
m
m
=
0
n
−
1
∑
+
1
n
−
m
m
=
n
+
1
k
∑
+
1
n
+
m
m
=
0
n
−
1
∑
+
1
n
+
m
m
=
n
+
1
k
∑
⎛
⎝
⎜
⎞
⎠
⎟
,
where k will be allow to approach infinity shortly. Reordering the sums,
S
=
1
2
n
1
n
−
m
m
=
0
n
−
1
∑
+
1
n
+
m
m
=
0
n
−
1
∑
+
1
n
+
m
m
=
n
+
1
k
∑
+
1
n
−
m
m
=
n
+
1
k
∑
⎛
⎝
⎜
⎞
⎠
⎟
.
Replacing m with –m in the first sum then flipping the limits,
S
=
1
2
n
1
n
+
m
m
=
1
−
n
0
∑
+
1
n
+
m
m
=
0
n
−
1
∑
+
1
n
+
m
m
=
n
+
1
k
∑
+
1
n
−
m
m
=
n
+
1
k
∑
⎛
⎝
⎜
⎞
⎠
⎟
Now, the first three sums can be turned into a single sum if we pull out one of the redundant m=0
terms (
1
n
) from the first two sums and subtract off the m=n term
(
1
2
n
) that needs to be added write a single sum,
S
=
1
2
n
1
n
−
1
2
n
+
1
n
+
m
m
=
1
−
n
k
∑
+
1
n
−
m
m
=
n
+
1
k
∑
⎛
⎝
⎜
⎞
⎠
⎟
.
Now, replace m with jn in the first sum,
S
=
1
2
n
1
n
−
1
2
n
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 Spring '14
 D.Kagan

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