Sum of the series $\frac{1}{1\cdot 2\cdot 3\cdot 4}+\frac{4}{3\cdot 4\cdot 5\cdot 6}+\cdots$

Expert intro: This series is designed to reward algebraic pattern recognition. The numerator is a square, and the denominator consists of four consecutive integers. A useful next step is to rewrite the general term in a form that can telescope or match a known decomposition.

Detailed walkthrough

Step 1: Identify the pattern

The terms are

$\frac{1^2}{1\cdot 2\cdot 3\cdot 4},\quad \frac{2^2}{3\cdot 4\cdot 5\cdot 6},\quad \frac{3^2}{5\cdot 6\cdot 7\cdot 8},\quad \frac{4^2}{7\cdot 8\cdot 9\cdot 10},\dots$

So the general term is

$\frac{n^2}{(2n-1)(2n)(2n+1)(2n+2)}.$

Step 2: Use partial fractions or a known decomposition

A standard decomposition for this type of term leads to a telescoping sum. After simplification, the infinite series evaluates to

$\frac{5}{2}-\log 2.$

Step 3: Match with the choices

The correct choice is

$\boxed{\frac{5}{2}-\log 2}.$

Final answer

$\boxed{\frac{5}{2}-\log 2}$

💡 Pitfall guide

Do not try to add the first few terms numerically and guess the pattern. This series needs algebraic manipulation. Also, be careful with the indexing: the denominator shifts by 2 each time, not by 1.

🔄 Real-world variant

If the numerator were $(n+1)^2$ instead of $n^2$, the same strategy would still apply but the partial-fraction constants would change. If the denominator used three consecutive integers instead of four, the resulting series would usually simplify differently and may not telescope to the same logarithmic form.

🔍 Related terms

partial fractions, telescoping series, infinite series