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Putnam
2017 Putnam
B4
B4
Part of
2017 Putnam
Problems
(1)
Putnam 2017 B4
Source:
12/3/2017
Evaluate the sum
∑
k
=
0
∞
(
3
⋅
ln
(
4
k
+
2
)
4
k
+
2
−
ln
(
4
k
+
3
)
4
k
+
3
−
ln
(
4
k
+
4
)
4
k
+
4
−
ln
(
4
k
+
5
)
4
k
+
5
)
\sum_{k=0}^{\infty}\left(3\cdot\frac{\ln(4k+2)}{4k+2}-\frac{\ln(4k+3)}{4k+3}-\frac{\ln(4k+4)}{4k+4}-\frac{\ln(4k+5)}{4k+5}\right)
k
=
0
∑
∞
(
3
⋅
4
k
+
2
ln
(
4
k
+
2
)
−
4
k
+
3
ln
(
4
k
+
3
)
−
4
k
+
4
ln
(
4
k
+
4
)
−
4
k
+
5
ln
(
4
k
+
5
)
)
=
3
⋅
ln
2
2
−
ln
3
3
−
ln
4
4
−
ln
5
5
+
3
⋅
ln
6
6
−
ln
7
7
−
ln
8
8
−
ln
9
9
+
3
⋅
ln
10
10
−
⋯
.
=3\cdot\frac{\ln 2}2-\frac{\ln 3}3-\frac{\ln 4}4-\frac{\ln 5}5+3\cdot\frac{\ln 6}6-\frac{\ln 7}7-\frac{\ln 8}8-\frac{\ln 9}9+3\cdot\frac{\ln 10}{10}-\cdots.
=
3
⋅
2
ln
2
−
3
ln
3
−
4
ln
4
−
5
ln
5
+
3
⋅
6
ln
6
−
7
ln
7
−
8
ln
8
−
9
ln
9
+
3
⋅
10
ln
10
−
⋯
.
(As usual,
ln
x
\ln x
ln
x
denotes the natural logarithm of
x
.
x.
x
.
)
Putnam
Putnam 2017
Putnam calculus