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Japan Contests
Today's Calculation Of Integral
2010 Today's Calculation Of Integral
621
621
Part of
2010 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 621
Source:
7/1/2010
Find the limit
lim
n
→
∞
1
n
∑
k
=
1
n
k
ln
(
n
2
+
(
k
−
1
)
2
n
2
+
k
2
)
.
\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n k\ln \left(\frac{n^2+(k-1)^2}{n^2+k^2}\right).
lim
n
→
∞
n
1
∑
k
=
1
n
k
ln
(
n
2
+
k
2
n
2
+
(
k
−
1
)
2
)
.
2010 Yokohama National University entrance exam/Engineering, 2nd exam
calculus
integration
limit
logarithms
real analysis
calculus computations