Infinite Series (Problem 1)

Consider the infinite series for all $\lvert x \rvert <1$ :

$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$

Integrating both sides from $0$ to $y$ :

$-\ln(1-y) = \sum_{n=0}^{\infty} \frac{y^{n+1}}{n+1}$

Multiplying both sides by $y$ :

$-y\ln(1-y) = \sum_{n=0}^{\infty} \frac{y^{n+2}}{n+1}$

Integrating both sides from $0$ to $1$ gives:

$-\int_{0}^{1}y\ln(1-y) \ dy = \sum_{n=0}^{\infty} \frac{1}{(n+1)(n+3)}$

Multiplying both sides by $\frac{4 \pi}{3}$ :

$-\frac{4 \pi}{3} \int_{0}^{1}y\ln(1-y) \ dy = \sum_{n=0}^{\infty} \frac{4 \pi}{3(n+1)(n+3)}$

The steps required to evaluate the integral are provided below. The required answer is:

$\boxed{\sum_{n=0}^{\infty} \frac{4 \pi}{3(n+1)(n+3)}=-\frac{4 \pi}{3} \int_{0}^{1}y\ln(1-y) \ dy = \pi}$

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Consider the indefinite integral:

$I=-\int y\ln(1-y) \ dy$

Integrating by parts:

$I = -\frac{y^2 \ln(1-y)}{2} -\int \frac{y^2}{2(1-y)} \ dy$ $I = -\frac{y^2 \ln(1-y)}{2} + \int \frac{1-y^2 - 1}{2(1-y)} \ dy$ $I = -\frac{y^2 \ln(1-y)}{2} + \int \frac{1+y}{2} \ dy - \int \frac{1}{2(1-y)} \ dy$ $I =-\frac{y^2 \ln(1-y)}{2} + \frac{y}{2}+\frac{y^2}{4} + \frac{\ln(1-y)}{2}$ $I = \frac{(1-y^2) \ln(1-y)}{2}+ \frac{y}{2}+\frac{y^2}{4}$

Now consider the definite integral:

$-\int_{0}^{a} y\ln(1-y) \ dy = \frac{(1-a^2) \ln(1-a)}{2}+ \frac{a}{2}+\frac{a^2}{4}$

$-\lim_{a \to 1^{-}} \int_{0}^{a} y\ln(1-y) \ dy = \lim_{a \to 1^{-}} \left( \frac{(1-a^2) \ln(1-a)}{2}+ \frac{a}{2}+\frac{a^2}{4} \right)$ $-\int_{0}^{1} y\ln(1-y) \ dy = \lim_{a \to 1^{-}} \left( \frac{(1-a^2) \ln(1-a)}{2}\right) + \frac{3}{4}$

The only limit on the right-hand side can be evaluated as such:

$\lim_{a \to 1^{-}} \frac{(1-a^2) \ln(1-a)}{2}=\lim_{a \to 1^{-}} \frac{ \ln(1-a)}{\frac{2}{1-a^2}}$

Beyond this, point L'hopital's rule can be applied to prove that the above limit evaluates to zero.

Therefore: $\boxed{-\int_{0}^{1} y\ln(1-y) \ dy = \frac{3}{4}}$

The given expression is equal to

$\dfrac{2π}{3}\left (\displaystyle \sum_{n=0}^\infty \left (\frac{1}{n+1}-\frac{1}{n+3}\right )\right )$ .

This telescopic sum reduces to $\dfrac{2π}{3}\times \dfrac{3}{2}=π\approx \boxed {3.14159}$ .

$\begin{aligned} S & = \sum_\blue{n=0}^\infty \frac {4\pi}{3(n+1)(n+3)} \\ & = \frac {4\pi}3 \sum_\red{n=1}^\infty \frac 1{n(n+2)} \\ & = \frac {2\pi}3 \sum_\red{n=1}^\infty \left(\frac 1n - \frac 1{n+2} \right) \\ & = \frac {2\pi}3 \left(\frac 11 + \frac 12 \right) \\ & = \pi \approx \boxed{3.142} \end{aligned}$

The first $5$ digits of $\pi$ only are needed

Therefore, the answer is $3.1415 \approx \pi$ .