A log of wood

You can rewrite it as a:

$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...$

And it's equal to:

$\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n}$

Now our sum can be written as

$S=1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{2n-1}-\frac{1}{2n}$

$S=H_{2n}-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2n})$

$S=H_{2n}-H_{n}$

Now there is theorem that harmonic series can be represented as

$H_n=ln(n)+\gamma+\alpha_n$

Where $\gamma=0.577$ called Euler constant, and $\alpha_n$ is some zero-sequence. (Try to prove this.)

And in limits where $n$ goes to infinity its equal to:

$S=ln(2n)-ln(n)$

$S=ln(2)$

The given series is:- $\dfrac{1}{2} + \dfrac{1}{12}+\dfrac{1}{30} + \ldots$ On observation we can write it as:- $\dfrac{1}{1.2}+\dfrac{1}{3.4} + \dfrac{1}{5.6}+\ldots$ $\sum_{r=0}^{\infty} \dfrac{1}{(2r+1)(2r+2)}$ $\sum_{r=0}^{\infty} \dfrac{1}{2r+1}-\dfrac{1}{2r+2}$ Now comes the tricky part. Here we will strategically introduce an integral to make summation possible. $\sum_{r=0}^{\infty} \int_{0}^{1} (x^{2r}-x^{2r+1})dx$ $\int_{0}^{1}\Bigg( \sum_{r=0}^{\infty} x^{2r} - \sum_{r=0}^{\infty} x^{2r+1} \Bigg) dx$ $\int_{0}^{1} \Bigg( \dfrac{1}{1-x^{2}} - \dfrac{x}{1-x^{2}}\Bigg) dx$ $\int_{0} ^{1} \dfrac{1}{1+x}dx$ $=\ln 2$

Using mc lawrence series ,

F(x)=f(0)+f'(0)x/1! +f''(0)x^2/2! ..................

Using this we get ln(1+x)=x/1 - x^2/2 + x^3/3 -x^4/4 ...............

Put x=1 to get the answer as ln2=0.693