## How to solve series #1

Verify that the following series converges. Exercise 1.0 $$\displaystyle{\sum_{n=1}^{\infty}}\dfrac{2}{n^2+2n}$$ $$\dfrac{2}{n^2+2n}=\dfrac{2}{n(n+2)}=\dfrac{1}{n}-\dfrac{1}{n+2}$$ $$S_1=a_1=1-\dfrac{1}{3}$$ $$S_2=a_1+a_2=\left(1-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)=1+\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}$$ $$S_3=a_1+a_2+a_3=\left(1-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)=1+\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{5}$$ $$S_4=a_1+a_2+a_3+a_4=\left(1-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\left(\dfrac{1}{4}-\dfrac{1}{6}\right)=1+\dfrac{1}{2}-\dfrac{1}{5}-\dfrac{1}{6}$$ $$S_n=a_1+a_2+…+a_n=1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}$$ $$\displaystyle{\lim_{n\rightarrow\infty}}S_n=1+\dfrac{1}{2}=\dfrac{3}{2}$$ the series converge and the sum is: $$\dfrac{3}{2}$$ Exercise 1.1 $$\displaystyle{\sum_{n=1}^{\infty}}\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n^2+n}}$$ $$a_n=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n^2+n}}=\dfrac{\sqrt{n+1}}{\sqrt{n}\sqrt{n+1}}-\dfrac{\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}$$ $$S_1=a_1=1-\dfrac{1}{\sqrt{2}}$$ $$S_2=a_1+a_2=\left(1-\dfrac{1}{\sqrt{2}}\right)+\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)=1+\dfrac{1}{\sqrt{3}}$$ $$S_3=a_1+a_2+a_3=\left(1-\dfrac{1}{\sqrt{2}}\right)+\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)+\left(\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{4}}\right)=1+\dfrac{1}{\sqrt{4}}$$ $$S_n=a_1+a_2+…+a_n=1-\dfrac{1}{\sqrt{n}}$$ the series converge and the sum is: $$\displaystyle{\lim_{n\rightarrow\infty}}S_n=1-\dfrac{1}{\sqrt{n}}=1$$ Exercise 1.3 $$\displaystyle{\sum_{n=1}^{\infty}}(-1)^n\dfrac{n}{n+1}$$ $$\displaystyle{\lim_{n\rightarrow\infty}}(-1)^n\dfrac{n}{n+1}$$ Unfortunately this limit does not exist, so it is not zero. …