Question

Prove that,
1/1 - 1/2 + 1/3 - 1/4 +... = ln(2)

This is a Harmonic Series with alternate signs.

Answer 1

$\large\underline{\sf{Solution-}}$

Let first derive the series of ln(1 + x)

Let assume that

$\rm \: f(x) = ln(1+x) = log_{e}(1 + x)$

Let evaluate differential coefficients of f(x).

So,

$\rm \: f'(x) = \dfrac{1}{1 + x}$

$\rm \: f''(x) = \dfrac{ - 1}{(1 + x)^{2} }$

$\rm \: f'''(x) = \dfrac{2}{(1 + x)^{3} }$

$\rm \: f''''(x) = \dfrac{ - 6}{(1 + x)^{4} }$

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Now, let evaluate the values of differential coefficients at x = 0.

So,

$\rm \: f(0) = log_{e}(1 + 0) = log_{e}(1) = 0$

$\rm \: f'(0) = \dfrac{1}{1 + 0} = 1$

$\rm \: f''(0) = \dfrac{ - 1}{(1 + 0)^{2} } = - 1$

$\rm \: f'''(x) = \dfrac{2}{(1 + 0)^{3} } = 2$

$\rm \: f''''(x) = \dfrac{ - 6}{(1 + 0)^{4} } = - 6$

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Now, By Maclaurin's Series, we have

$\rm \: f(x) = f(0) + xf'(0) + \dfrac{ {x}^{2} }{2!}f''(0) + \dfrac{ {x}^{3} }{3!}f'''(0) + \dfrac{ {x}^{4} }{4!}f''''(0) + - - -$

So, on substituting the values, we get

$\rm \: log_{e}(1 + x) = 0 + x + \dfrac{ {x}^{2} }{2!}( - 1) + \dfrac{ {x}^{3} }{3!}(2) + \dfrac{ {x}^{4} }{4!}( - 6) + - - -$

$\rm \: log_{e}(1 + x) = x - \dfrac{ {x}^{2} }{2} + \dfrac{ {x}^{3} }{3} - \dfrac{ {x}^{4} }{4} + - - -$

Thus, we derive the expression

$\rm \: log_{e}(1 + x) = x - \dfrac{ {x}^{2} }{2} + \dfrac{ {x}^{3} }{3} - \dfrac{ {x}^{4} }{4} + - - -$

Now, Substituting x = 1, we get

$\rm \: log_{e}(1 + 1) = 1 - \dfrac{ {1}^{2} }{2} + \dfrac{ {1}^{3} }{3} - \dfrac{ {1}^{4} }{4} - - -$

$\rm \: log_{e}(2) = 1 - \dfrac{ 1 }{2} + \dfrac{ 1 }{3} - \dfrac{1 }{4} - - -$

Hence,

$\rm\implies \:\rm \: log_{e}(2) = ln(2) = 1 - \dfrac{ 1 }{2} + \dfrac{ 1 }{3} - \dfrac{1 }{4} - - -$

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Additional Information :-

$\boxed{\sf{  \:\rm \: sinx = x - \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{5} }{5!} + - - - \: \: }}$

$\boxed{\sf{  \:\rm \: cosx = 1 - \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{4} }{4!} + - - - \: \: }}$

$\boxed{\sf{  \:\rm \: {e}^{x} = 1 + x + \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{3} }{3!} + - - - \: \: }}$

$\boxed{\sf{  \:\rm \: {e}^{ - x} = 1 - x + \dfrac{ {x}^{2} }{2!} - \dfrac{ {x}^{3} }{3!} + - - - \: \: }}$

$\boxed{\sf{  \:\rm \: {a}^{x} = 1 + loga + \dfrac{ {x}^{2} }{2!} {(loga)}^{2} + \dfrac{ {x}^{3} }{3!} {(loga)}^{3} + - - - \: \: }}$