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Unformatted text preview: n n = −1 to see that B = −1. This works
becasue
−1
n + 1 − (n − 1)
2
1
+
=
=2
.
2−1
n−1 n+1
n
n −1
(a) We see that
M sM = 1
2−1 − ···+ = = 1
1 + 1
2+1 1
3−1 + 1
(M −2)−1
1
1 − ···+
1
2 M 2
1
1
=
=
−
2−1
n
n−1 n+1
n=2
n=2 − 1
3 1
3+1 + 1
4−1 1
(M −2)+1 + 1
(M −1)−1 +
1
M −3 + ···+ − 1
2 1
1
+ 3−1
4
5
1
1
M −1 + M −2 − − +
+ − 1
4+1 +
− + 1
5−1 − 1
(M −1)+1 1
4
1
M − 1
6 + − + − + 1
M −1 1
M +1 − = (b) The sum of the series is
3
3
1
1
.
−
−
=
M →∞ 2
M
M +1
2 lim sM = lim M →∞ 1 1
5+1 +... 1
M −1 − 1
M +1 +... +...
1
M − 1
M +1 1
1
3
−
−
.
2M
M +1...
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This document was uploaded on 03/23/2014 for the course MATH 142 at South Carolina.
 Fall '11
 KUSTIN
 Calculus, Fractions

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