First slide

Special Series in Sequences and Series

Question

$The sum to infinity of the series$ $\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + \dots ., is equal to -$

Moderate

A

3
B

1
C

2
D

$\frac{3}{2}$

Solution

We have S= $\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + \dots . .$
$Let t_{n} = \frac{1}{1 + 2 + 3 + \dots . n} = \frac{2}{n (n + 1)} = 2 [\frac{1}{n} - \frac{1}{n + 1}]$
$∴ Sum = S = 2 [(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots \dots .] = 2$

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