Series - Problem 3

We know,

$\\ \dfrac{e^x+e^{-x}}{2} = 1 + \dfrac{x^2}{2!} + \dfrac{x^4}{4!}+ \dfrac{x^6}{6!} + ............ \\ \Rightarrow \dfrac{e^{\ln 2}+e^{-\ln 2}}{2} = 1 + \dfrac{(\ln 2)^2}{2!} + \dfrac{(\ln 2)^4}{4!}+ \dfrac{(\ln 2)^6}{6!} +......... \\ \Rightarrow 1 + \dfrac{(\ln 2)^2}{2!} + \dfrac{(\ln 2)^4}{4!}+ \dfrac{(\ln 2)^6}{6!} +......... = \dfrac{(2+\dfrac{1}{2})}{2} = \boxed{\dfrac{5}{4}}$

First of all let's denote

$1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots = S(x)$

Then observe that

$S(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} = \sum_{n=0}^{\infty} \frac{x^n}{n!} - \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} = e^{x} - \sinh(x) = \frac{e^{x}+e^{-x}}{2}$

And if we plug $x = \ln(2)$ we get

$S(\ln(2)) = 1 + \frac{(\ln(2))^{2}}{2!} + \frac{(\ln(2))^{4}}{4!} + \cdots = \frac{e^{\ln(2)} + e^{-\ln(2)}}{2} = \frac{2 + \frac{1}{2}}{2} = \frac{5}{4}$

I used a calculator and haha, I got 1.24999. Your solution makes me feel guilty too.