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 \( \displaystyle \sum_{n=1}^{\infty} t_n \)
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- diverges
- converges to \( 3S - a_1 - a_2 \)
- converges to \( 3S - a_1 - 2a_2 \)
- converges to \( 3S - 2a_1 - a_2 \)
The Correct Option is D
Solution and Explanation
Step 1: Write out the series $\sum_{n=1}^{\infty} t_n$:
$$\sum_{n=1}^{\infty} t_n = t_1 + t_2 + t_3 + t_4 + \cdots$$
Step 2: Substitute the definition of $t_n$:
$$= (a_1 + a_2 + a_3) + (a_2 + a_3 + a_4) + (a_3 + a_4 + a_5) + \cdots$$
Step 3: Rearrange by grouping terms with the same subscript:
$$= a_1 + (a_2 + a_2) + (a_3 + a_3 + a_3) + (a_4 + a_4 + a_4) + (a_5 + a_5 + a_5) + \cdots$$
Step 4: Count the coefficient of each term:
- $a_1$ appears 1 time
- $a_2$ appears 2 times
- $a_n$ appears 3 times for $n \geq 3$
Therefore: $$\sum_{n=1}^{\infty} t_n = a_1 + 2a_2 + 3\sum_{n=3}^{\infty} a_n$$
$$= a_1 + 2a_2 + 3\left(\sum_{n=1}^{\infty} a_n - a_1 - a_2\right)$$
$$= a_1 + 2a_2 + 3S - 3a_1 - 3a_2$$
$$= 3S - 2a_1 - a_2$$
Answer: (D) converges to $3S - 2a_1 - a_2$
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