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Today's Calculation Of Integral
2006 Today's Calculation Of Integral
166
Today's calculation of Integral 166
Today's calculation of Integral 166
Source: Toyama Medical and Pharmaceutical University entrance exam 1989
November 11, 2006
calculus
integration
limit
logarithms
calculus computations
Problem Statement
Express the following the limit values in terms of a definite integral and find them. (1)
I
=
lim
n
→
∞
1
n
ln
(
1
+
1
n
)
(
1
+
2
n
)
⋯
⋯
(
1
+
n
n
)
.
I=\lim_{n\to\infty}\frac{1}{n}\ln \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots\cdots \left(1+\frac{n}{n}\right).
I
=
lim
n
→
∞
n
1
ln
(
1
+
n
1
)
(
1
+
n
2
)
⋯⋯
(
1
+
n
n
)
.
(2)
J
=
lim
n
→
∞
1
n
2
(
n
2
−
1
+
n
2
−
2
2
+
⋯
⋯
+
n
2
−
n
2
)
.
J=\lim_{n\to\infty}\frac{1}{n^{2}}(\sqrt{n^{2}-1}+\sqrt{n^{2}-2^{2}}+\cdots\cdots+\sqrt{n^{2}-n^{2}}).
J
=
lim
n
→
∞
n
2
1
(
n
2
−
1
+
n
2
−
2
2
+
⋯⋯
+
n
2
−
n
2
)
.
(3)
K
=
lim
n
→
∞
1
n
3
(
n
2
+
1
+
2
n
2
+
2
2
+
⋯
+
n
n
2
+
n
2
)
.
K=\lim_{n\to\infty}\frac{1}{n^{3}}(\sqrt{n^{2}+1}+2\sqrt{n^{2}+2^{2}}+\cdots+n\sqrt{n^{2}+n^{2}}).
K
=
lim
n
→
∞
n
3
1
(
n
2
+
1
+
2
n
2
+
2
2
+
⋯
+
n
n
2
+
n
2
)
.
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