Find-the-value-of-S-S-1-1-1-2-1-2-3-1-3-4-

Question Number 182875 by HeferH last updated on 15/Dec/22

F i n d t h e v a l u e o f S :

S = 1 + \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} \dots

Answered by TheSupreme last updated on 16/Dec/22

S = 1 + Σ \frac{1}{i (i + 1)} = 1 + Σ \frac{A}{i} + Σ \frac{B}{i + 1}

1 + Σ \frac{A i + A + B i}{i (i + 1)}

A + B = 0

A = 1

S = 1 + Σ \frac{1}{i} - Σ \frac{1}{i + 1}

S_{n} = 1 + \frac{1}{2} - \frac{1}{n + 1}

S = \frac{3}{2}

Commented by Frix last updated on 16/Dec/22

S_{n} = 1 + \underset{i = 1}{\sum^{n}} \frac{1}{i} - \underset{i = 1}{\sum^{n}} \frac{1}{i + 1} = 1 + 1 - \frac{1}{n + 1} \Rightarrow

S = 2

Answered by manxsol last updated on 16/Dec/22

S = 1 + (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) \dots

\dots (\frac{1}{n - 2} - \frac{1}{n - 1}) + (\frac{1}{n - 1} - \frac{1}{n})

S = 1 + 1 - \frac{1}{n}

n \Rightarrow \infty \Rightarrow \frac{1}{n} = 0

S = 2

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