First slide

Series of natural numbers

Question

Let $m$ be a positive integer, then $S = \sum_{k = 1}^{m} k (\frac{1}{k} + \frac{1}{k + 1} + \frac{1}{k + 2} + \dots + \frac{1}{m})$ is equal to :

Moderate

A

$\frac{1}{4} m (m + 2)$
B

$\frac{1}{4} m (m + 3)$
C

$\frac{1}{4} m (m + 4)$
D

$\frac{1}{4} m (m + 6)$

Solution

$\begin{array}{l} S = (1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{m}) \\ + 2 (\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{m}) \\ + 3 (\frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{m}) \\ + \dots \\ + (m - 1) (\frac{1}{m - 1} + \frac{1}{m}) + m (\frac{1}{m}) \\ = (1) (1) + \frac{1}{2} (1 + 2) + \frac{1}{3} (1 + 2 + 3) \\ + \frac{1}{4} (1 + 2 + 3 + 4) \\ + \dots + \frac{1}{m} (1 + 2 + 3 + \dots + m) \end{array}$
$\begin{array}{l} = \sum_{k = 1}^{m} \frac{1}{k} (1 + 2 + \dots + k) \\ = \sum_{k = 1}^{m} \frac{1}{k} \cdot \frac{k (k + 1)}{2} = \frac{1}{2} \sum_{k = 1}^{m} (k + 1) \\ = \frac{m}{4} [2 + (m + 1)] \\ = \frac{1}{4} m (m + 3) \end{array}$

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