Telescoping Series/Example 2
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Theorem
Let $\left \langle {b_n} \right \rangle$ be a sequence in $\R$.
Let $\left \langle {a_n} \right \rangle$ be a sequence whose terms are defined as:
- $a_k = b_k - b_{k - 1}$
Then:
- $\displaystyle \sum_{k \mathop = m}^n a_k = b_n - b_{m - 1}$
Proof
\(\ds \displaystyle \sum_{k \mathop = m}^n a_k\) | \(=\) | \(\ds \sum_{k \mathop = m}^n \left({b_k - b_{k - 1} }\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^n b_k - \sum_{k \mathop = m}^n b_{k - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = m}^n b_k - \sum_{k \mathop = m - 1}^{n - 1} b_k\) | Translation of Index Variable of Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({\sum_{k \mathop = m}^{n - 1} b_k + b_n}\right) - \left({b_{m - 1} + \sum_{k \mathop = m}^{n - 1} b_k}\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b_n - b_{m - 1}\) |
$\blacksquare$
Linguistic Note
The term telescoping series arises from the obvious physical analogy with the folding up of a telescope.
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $19$