Question

Suppose that \( S \) is the sum of a convergent series \( \displaystyle \sum_{n=1}^{\infty} a_n. \) Define \( t_n = a_n + a_{n+1} + a_{n+2}. \) Then the series \( \displaystyle \sum_{n=1}^{\infty} t_n \)

• A. diverges
• B. converges to \( 3S - a_1 - a_2 \)
• C. converges to \( 3S - a_1 - 2a_2 \)
• D. converges to \( 3S - 2a_1 - a_2 \)

Hint:

When summing shifted terms of a convergent series, the new series' sum is a linear combination of the original sum minus initial terms.

Answer 1

The Correct Option is D

Step 1: Write the expression for \( t_n. \)
\[ t_n = a_n + a_{n+1} + a_{n+2}. \]

Step 2: Consider the partial sum of the new series.
\[ T_N = \sum_{n=1}^{N} t_n = \sum_{n=1}^{N} (a_n + a_{n+1} + a_{n+2}). \]

Step 3: Expand the terms.
\[ T_N = (a_1 + a_2 + \cdots + a_N) + (a_2 + a_3 + \cdots + a_{N+1}) + (a_3 + a_4 + \cdots + a_{N+2}). \]

Step 4: Combine overlapping terms.
For large \( N \), \[ T_N = 3S - (a_{N+1} + a_{N+2}) - (a_1 + a_2). \]

Step 5: Take the limit as \( N \to \infty. \)
Since the given series \( \sum a_n \) is convergent, \( a_n \to 0 \). Hence, \[ \lim_{N \to \infty} T_N = 3S - a_1 - a_2. \]

Final Answer: The series converges to \( 3S - a_1 - a_2. \)