设 f ( x ) = ∑ n = 1 ∞ x n n 2 ，证明： f ( x ) + f ( 1 − x ) + ln x ln ( 1 − x ) = ∑ n = 1 ∞ 1 n 2 \text{设}f\left( x \right) =\sum_{n=1}^{\infty}{\frac{x^n}{n^2}}\text{，证明：}f\left( x \right) +f\left( 1-x \right) +\ln x\ln \left( 1-x \right) =\sum_{n=1}^{\infty}{\frac{1}{n^2}} 设f(x)=n=1∑∞n2xn，证明：f(x)+f(1−x)+lnxln(1−x)=n=1∑∞n21
f
′
(
x
)
=
∑
n
=
1
∞
x
n
−
1
n
=
1
x
∑
n
=
1
∞
x
n
n
=
1
x
∫
0
x
∑
n
=
1
∞
t
n
−
1
d
t
=
1
x
∫
0
x
∑
n
=
0
∞
t
n
d
t
=
1
x
∫
0
x
1
1
−
t
d
t
=
−
ln
(
1
−
x
)
x
f'\left( x \right) =\sum_{n=1}^{\infty}{\frac{x^{n-1}}{n}}=\frac{1}{x}\sum_{n=1}^{\infty}{\frac{x^n}{n}}=\frac{1}{x}\int_0^x{\sum_{n=1}^{\infty}{t^{n-1}}dt}=\frac{1}{x}\int_0^x{\sum_{n=0}^{\infty}{t^n}dt}=\frac{1}{x}\int_0^x{\frac{1}{1-t}dt}=-\frac{\ln \left( 1-x \right)}{x}
f′(x)=n=1∑∞nxn−1=x1n=1∑∞nxn=x1∫0xn=1∑∞tn−1dt=x1∫0xn=0∑∞tndt=x1∫0x1−t1dt=−xln(1−x)
f
′
(
x
)
−
f
′
(
1
−
x
)
=
−
ln
(
1
−
x
)
x
+
ln
x
1
−
x
f'\left( x \right) -f'\left( 1-x \right) =-\frac{\ln \left( 1-x \right)}{x}+\frac{\ln x}{1-x}
f′(x)−f′(1−x)=−xln(1−x)+1−xlnx
∵
[
ln
x
ln
(
1
−
x
)
]
′
=
ln
(
1
−
x
)
x
−
ln
x
1
−
x
\because \left[ \ln x\ln \left( 1-x \right) \right] '=\frac{\ln \left( 1-x \right)}{x}-\frac{\ln x}{1-x}
∵[lnxln(1−x)]′=xln(1−x)−1−xlnx
∴
f
′
(
x
)
−
f
′
(
1
−
x
)
+
[
ln
x
ln
(
1
−
x
)
]
′
=
0
\therefore f'\left( x \right) -f'\left( 1-x \right) +\left[ \ln x\ln \left( 1-x \right) \right] '=0
∴f′(x)−f′(1−x)+[lnxln(1−x)]′=0
令
g
(
x
)
=
f
(
x
)
+
f
(
1
−
x
)
+
ln
x
ln
(
1
−
x
)
x
∈
(
0
,
1
)
\text{令}g\left( x \right) =f\left( x \right) +f\left( 1-x \right) +\ln x\ln \left( 1-x \right) \ \ x\in \left( 0,1 \right)
令g(x)=f(x)+f(1−x)+lnxln(1−x) x∈(0,1)
∴
g
′
(
x
)
=
0
\therefore g'\left( x \right) =0
∴g′(x)=0
∵
∫
0
x
g
′
(
u
)
d
u
=
g
(
x
)
−
lim
t
→
0
+
g
(
t
)
=
0
\because \int_0^x{g'\left( u \right)}du=g\left( x \right) -\underset{t\rightarrow 0^+}{\lim}g\left( t \right) =0
∵∫0xg′(u)du=g(x)−t→0+limg(t)=0
∴
g
(
x
)
=
lim
t
→
0
+
g
(
t
)
\therefore g\left( x \right) =\underset{t\rightarrow 0^+}{\lim}g\left( t \right)
∴g(x)=t→0+limg(t)
∵
lim
t
→
0
+
g
(
t
)
=
lim
t
→
0
+
f
(
t
)
+
lim
t
→
0
+
f
(
1
−
t
)
+
lim
t
→
0
+
ln
t
⋅
ln
(
1
−
t
)
\because \underset{t\rightarrow 0^+}{\lim}g\left( t \right) =\underset{t\rightarrow 0^+}{\lim}f\left( t \right) +\underset{t\rightarrow 0^+}{\lim}f\left( 1-t \right) +\underset{t\rightarrow 0^+}{\lim}\ln t\cdot \ln \left( 1-t \right)
∵t→0+limg(t)=t→0+limf(t)+t→0+limf(1−t)+t→0+limlnt⋅ln(1−t)
其中
lim
t
→
0
+
f
(
t
)
=
0
,
lim
t
→
0
+
f
(
1
−
t
)
=
∑
n
=
1
∞
1
n
2
,
lim
t
→
0
+
ln
t
⋅
ln
(
1
−
t
)
=
lim
t
→
0
+
t
ln
t
=
lim
t
→
0
+
ln
t
1
t
=
lim
t
→
0
+
1
t
−
1
t
2
=
0
\text{其中}\underset{t\rightarrow 0^+}{\lim}f\left( t \right) =0,\ \underset{t\rightarrow 0^+}{\lim}f\left( 1-t \right) =\sum_{n=1}^{\infty}{\frac{1}{n^2}},\ \underset{t\rightarrow 0^+}{\lim}\ln t\cdot \ln \left( 1-t \right) =\underset{t\rightarrow 0^+}{\lim}t\ln t=\underset{t\rightarrow 0^+}{\lim}\frac{\ln t}{\frac{1}{t}}=\underset{t\rightarrow 0^+}{\lim}\frac{\frac{1}{t}}{-\frac{1}{t^2}}=0
其中t→0+limf(t)=0, t→0+limf(1−t)=n=1∑∞n21, t→0+limlnt⋅ln(1−t)=t→0+limtlnt=t→0+limt1lnt=t→0+lim−t21t1=0
∴
lim
t
→
0
+
g
(
t
)
=
∑
n
=
1
∞
1
n
2
\therefore \underset{t\rightarrow 0^+}{\lim}g\left( t \right) =\sum_{n=1}^{\infty}{\frac{1}{n^2}}
∴t→0+limg(t)=n=1∑∞n21
∴
g
(
x
)
=
∑
n
=
1
∞
1
n
2
\therefore g\left( x \right) =\sum_{n=1}^{\infty}{\frac{1}{n^2}}
∴g(x)=n=1∑∞n21
∴
f
(
x
)
+
f
(
1
−
x
)
+
ln
x
ln
(
1
−
x
)
=
∑
n
=
1
∞
1
n
2
\therefore f\left( x \right) +f\left( 1-x \right) +\ln x\ln \left( 1-x \right) =\sum_{n=1}^{\infty}{\frac{1}{n^2}}
∴f(x)+f(1−x)+lnxln(1−x)=n=1∑∞n21