let-a-gt-0-find-the-sum-of-the-infinite-series-1-loga-2-2-loga-4-4-loga-6-6-

Question Number 177957 by infinityaction last updated on 11/Oct/22

let a > 0 find the sum of the infinite

series

1 + \frac{{(loga)}^{2}}{2!} + \frac{{(loga)}^{4}}{4!} + \frac{{(loga)}^{6}}{6!} \dots

Answered by mr W last updated on 11/Oct/22

e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots

e^{- x} = 1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \dots

e^{x} + e^{- x} = 2 (1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \dots)

\frac{e^{x} + e^{- x}}{2} = 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \dots

w i t h x = \log a

\frac{a + \frac{1}{a}}{2} = 1 + \frac{{(\log a)}^{2}}{2!} + \frac{{(\log a)}^{4}}{4!} + \dots

\Rightarrow 1 + \frac{{(\log a)}^{2}}{2!} + \frac{{(\log a)}^{4}}{4!} + \dots = \frac{a^{2} + 1}{2 a}

Commented by infinityaction last updated on 11/Oct/22

t h a n k y o u s i r

Commented by Tawa11 last updated on 11/Oct/22

Great sir

Answered by Ar Brandon last updated on 11/Oct/22

1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + \cdot \cdot \cdot = \cosh (x)

\Rightarrow 1 + \frac{{(\log a)}^{2}}{2!} + \frac{{(\log a)}^{4}}{4!} + \frac{{(\log a)}^{6}}{6!} + \cdot \cdot \cdot = \cosh (\log a)

= \frac{1}{2} (e^{\log a} + e^{- \log a}) = \frac{1}{2} (a + \frac{1}{a}) = \frac{a^{2} + 1}{2 a}

Commented by Tawa11 last updated on 11/Oct/22

Great sir

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