\(\lim_{x\to 2} \sum_{n=1}^{9} \frac{x}{n(n+1)x^2 + 2(2n+1)x + 4}\) is equal to :
Show Hint
When you see a summation with terms involving \(n\) and \(n+1\) in the denominator, always suspect a telescoping series. The main task is to express the general term as a difference, \(f(n) - f(n+1)\), using partial fractions.
Step 1: Understanding the Question
We need to evaluate the limit of a summation. The general term of the summation depends on both \(n\) and \(x\). Since the limit is a finite value (\(x \to 2\)), we can first simplify the summation and then apply the limit. Step 2: Key Formula or Approach
The key is to simplify the general term of the series, likely using factorization and partial fractions, to see if it forms a telescoping series. Step 3: Detailed Explanation Factorize the Denominator:
Let's analyze the denominator \(D = n(n+1)x^2 + 2(2n+1)x + 4 = n(n+1)x^2 + (4n+2)x + 4\).
Let's try to factor it in the form \((ax+b)(cx+d)\). A good guess would be to involve the terms \(n\) and \(n+1\).
Consider \((nx+2)((n+1)x+2)\). Expanding this gives:
\[ (nx+2)((n+1)x+2) = n(n+1)x^2 + 2nx + 2(n+1)x + 4\] \[ = n(n+1)x^2 + (2n+2n+2)x + 4 = n(n+1)x^2 + (4n+2)x + 4 \]
This matches the denominator. So, the general term is \(\frac{x}{(nx+2)((n+1)x+2)}\). Apply Partial Fractions:
We decompose the term using partial fractions.
\[ \frac{x}{(nx+2)((n+1)x+2)} = \frac{A}{nx+2} + \frac{B}{(n+1)x+2} \]
Let's try a quicker method for this specific form:
\[ \frac{1}{nx+2} - \frac{1}{(n+1)x+2} = \frac{((n+1)x+2) - (nx+2)}{(nx+2)((n+1)x+2)} = \frac{x}{(nx+2)((n+1)x+2)} \]
This works perfectly. So the general term \(T_n = \frac{1}{nx+2} - \frac{1}{(n+1)x+2}\). Evaluate the Sum (Telescoping Series):
The summation is \(S = \sum_{n=1}^{9} T_n = \sum_{n=1}^{9} \left(\frac{1}{nx+2} - \frac{1}{(n+1)x+2}\right)\).
\[ S = \left(\frac{1}{x+2} - \frac{1}{2x+2}\right) + \left(\frac{1}{2x+2} - \frac{1}{3x+2}\right) + \dots + \left(\frac{1}{9x+2} - \frac{1}{10x+2}\right) \]
All intermediate terms cancel out.
\[ S = \frac{1}{x+2} - \frac{1}{10x+2} \]
Calculate the Limit:
Now we take the limit as \(x \to 2\).
\[ \lim_{x\to 2} S = \lim_{x\to 2} \left(\frac{1}{x+2} - \frac{1}{10x+2}\right) = \frac{1}{2+2} - \frac{1}{10(2)+2} = \frac{1}{4} - \frac{1}{22} \]
\[ = \frac{11 - 2}{44} = \frac{9}{44} \]
Step 4: Final Answer
The value of the limit is \(\frac{9}{44}\).
