Let S k = 1 + 1/2 + ⋯ + 1/ k for positive k. For n ≥ 2 , what equals S 1 + S 2 + ⋯ + S n − 1 ?

Reverse the finite double sum: ∑ k = 1 n − 1 S k = ∑ j = 1 n − 1 ( n − j ) / j = n S n − 1 − ( n − 1 ) = n S n − n .

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MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minQB

MathematicsMediummultiple choice

Let $S_{k} = 1 + 1/2 + \dots + 1/ k$ for positive k. For $n \geq 2$ , what equals $S_{1} + S_{2} + \dots + S_{n - 1}$ ?

Answer:

A, D

Solution:

Reverse the finite double sum: $\sum_{k = 1}^{n - 1} S_{k} = \sum_{j = 1}^{n - 1} (n - j) / j = n S_{n - 1} - (n - 1) = n S_{n} - n$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: QB

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