Show that e2-1e2-1 - 1-1-2-1-4-1-6- - - 1-1-1-3-1-5- - -

e^x=1+x+x^22!+x^33!+...... Hence e=1+1+12!+13!+14!..... ...(i) e^ 1=1 1+12! 13!+14!.... ...(ii)Adding i and ii, we get #160; e+e^ 1=2[1+12! ...

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Sum of n Terms

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Show that $\frac{e^{2} + 1}{e^{2} - 1} = \frac{1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + . .}{\frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + . .}$

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\frac{4}{1 +} \frac{6}{2 +} \frac{8}{3 +} \dots \frac{2 n + 2}{n +} \dots = \frac{2 (e^{2} - 1)}{e^{2} + 1}

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

E_{1}

E_{2}

P (E_{1}) = \frac{1}{4}

P (\frac{E_{2}}{E_{1}}) = \frac{1}{2}

P (\frac{E_{1}}{E_{2}}) = \frac{1}{4}

e^{2} - 1 = \frac{1 + \frac{a^{2}}{2!} + \frac{a^{4}}{3!} + \frac{a^{6}}{4!} + . . .}{1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^{2}}{4!} + . . .}

a

E_{1}

E_{2}

P (E_{1}) = 1 / 4, P (E_{2} / E_{1}) = 1 / 2

P (E_{1} / E_{2}) = 1 / 4

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