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Today's Calculation Of Integral
2011 Today's Calculation Of Integral
695
Today's calculation of Integral 695
Today's calculation of Integral 695
Source: 2011 Tokyo Medical and Dental University entrance exam
May 7, 2011
calculus
integration
limit
logarithms
calculus computations
Problem Statement
For a positive integer
n
n
n
, let
S
n
=
∫
0
1
1
−
(
−
x
)
n
1
+
x
d
x
,
T
n
=
∑
k
=
1
n
(
−
1
)
k
−
1
k
(
k
+
1
)
S_n=\int_0^1 \frac{1-(-x)^n}{1+x}dx,\ \ T_n=\sum_{k=1}^n \frac{(-1)^{k-1}}{k(k+1)}
S
n
=
∫
0
1
1
+
x
1
−
(
−
x
)
n
d
x
,
T
n
=
k
=
1
∑
n
k
(
k
+
1
)
(
−
1
)
k
−
1
Answer the following questions:(1) Show the following inequality.
∣
S
n
−
∫
0
1
1
1
+
x
d
x
∣
≤
1
n
+
1
\left|S_n-\int_0^1 \frac{1}{1+x}dx\right|\leq \frac{1}{n+1}
S
n
−
∫
0
1
1
+
x
1
d
x
≤
n
+
1
1
(2) Express
T
n
−
2
S
n
T_n-2S_n
T
n
−
2
S
n
in terms of
n
n
n
.(3) Find the limit
lim
n
→
∞
T
n
.
\lim_{n\to\infty} T_n.
lim
n
→
∞
T
n
.
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