技术问询：求解幂级数∑_{k=1}^∞x^k/(k(k+1))的和

Hey there! I totally get how frustrating it can be to stare at those stubborn ( k ) terms and feel stuck—let’s break this down with straightforward power series tricks that will make those denominators disappear.

Step 1: Partial Fraction Decomposition #

First, split the rational term to simplify the series. We can rewrite ( \frac{1}{k(k+1)} ) using partial fractions:
[
\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}
]
This lets us split the original series into two separate sums:
[
\sum_{k=1}^\infty \frac{x^k}{k(k+1)} = \sum_{k=1}^\infty \frac{x^k}{k} - \sum_{k=1}^\infty \frac{x^k}{k+1}
]

Step 2: Use the Known Logarithmic Power Series #

The first sum is a standard result. For ( |x| < 1 ), the power series for the natural logarithm is:
[
\sum_{k=1}^\infty \frac{x^k}{k} = -\ln(1 - x)
]
We’ll call this ( S_1(x) = -\ln(1 - x) ).

Step 3: Rewrite the Second Sum to Match Known Series #

The second sum ( \sum_{k=1}^\infty \frac{x^k}{k+1} ) needs a small tweak to use the same logarithmic series. Multiply and divide by ( x ) to shift the index:
[
\sum_{k=1}^\infty \frac{x^k}{k+1} = \frac{1}{x} \sum_{k=1}^\infty \frac{x^{k+1}}{k+1}
]
Let ( n = k+1 ), so when ( k=1 ), ( n=2 ). The sum becomes:
[
\frac{1}{x} \left( \sum_{n=1}^\infty \frac{x^n}{n} - \frac{x^1}{1} \right)
]
Substitute ( \sum_{n=1}^\infty \frac{x^n}{n} = -\ln(1 - x) ):
[
\frac{1}{x} \left( -\ln(1 - x) - x \right) = -\frac{\ln(1 - x)}{x} - 1
]
We’ll call this ( S_2(x) = -\frac{\ln(1 - x)}{x} - 1 ).

Step 4: Combine the Two Sums #

Subtract ( S_2(x) ) from ( S_1(x) ) to get the original series sum ( S(x) ):
[
S(x) = S_1(x) - S_2(x) = -\ln(1 - x) - \left( -\frac{\ln(1 - x)}{x} - 1 \right)
]
Simplify the expression by combining terms:
[
S(x) = 1 + \frac{(1 - x)\ln(1 - x)}{x}
]

Step 5: Check Endpoints of Convergence #

We should verify the sum at the endpoints of the convergence interval (( x = 1 ) and ( x = -1 )):

• At ( x = 1 ): The original series becomes ( \sum_{k=1}^\infty \frac{1}{k(k+1)} ). The partial sum is ( 1 - \frac{1}{n+1} ), which approaches 1 as ( n \to \infty ). Using the limit as ( x \to 1^- ) in our formula gives ( 1 + 0 = 1 ), which matches.
• At ( x = -1 ): The series becomes ( \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1)} ). Plugging ( x=-1 ) into our formula gives ( 1 + \frac{(1 - (-1))\ln(2)}{-1} = 1 - 2\ln2 ), which is the correct sum.

Final Result #

For ( |x| \leq 1 ) (note: at ( x=0 ), the series sum is 0, which aligns with the limit of our formula as ( x \to 0 )), the sum is:
[
\boxed{1 + \frac{(1 - x)\ln(1 - x)}{x}}
]

内容的提问来源于stack exchange，提问作者Quinn