# 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. Therefore the series does not converge.

## Exercise 1.4

$$\displaystyle{\sum_{n=1}^{\infty}}\dfrac{1}{\sqrt[n]{n}}$$

$$\displaystyle{\lim_{n\rightarrow\infty}}\dfrac{1}{\sqrt[n]{n}}=\displaystyle{\lim_{n\rightarrow\infty}}n^{-\frac{1}{n}}=\displaystyle{\lim_{n\rightarrow\infty}}e^{-\frac{1}{n}\log n}=1$$

$$\displaystyle{\lim_{n\rightarrow\infty}}\dfrac{\log n}{n}=0$$

$$\displaystyle{\lim_{n\rightarrow\infty}}a_n=1\neq 0$$ the series does not converge (that is, it diverges at $$+\infty$$, since it has positive terms).

## Exercise 1.5

$$\displaystyle{\sum_{n=1}^{\infty}}(-1)^n\dfrac{1}{\log \left(1+\dfrac{1}{n}\right)}$$

$$\displaystyle{\lim_{n\rightarrow\infty}}\dfrac{1}{\log \left(1+\dfrac{1}{n}\right)}=\dfrac{1}{\log (1)}=+\infty$$

the series does not converge but diverges to $$+\infty$$ because $$\forall n\in \mathbb{N},\;\log\left(1+\dfrac{1}{n}\right)>0$$

## Exercise 1.6

$$\displaystyle{\sum_{n=1}^{\infty}}\sin n$$

In this case since this limit does not exist: $$\displaystyle{\lim_{n\rightarrow\infty}}\sin n$$ the series does not converge.

## Exercise 1.7

$$\displaystyle{\sum_{n=0}^{\infty}}\dfrac{2^n+3^n}{5^n}=\displaystyle{\sum_{n=0}^{\infty}}\left[\left(\dfrac{2}{5}\right)^n+\left(\dfrac{3}{5}\right)^n\right]=\displaystyle{\sum_{n=0}^{\infty}}\left(\dfrac{2}{5}\right)^n+\displaystyle{\sum_{n=0}^{\infty}}\left(\dfrac{3}{5}\right)^n$$

It is the sum of two geometric series both converging (with reason less than 1). So the initial series converges to the sum of the two.

$$S_1=\displaystyle{\sum_{n=0}^{\infty}}\left(\dfrac{2}{5}\right)^n=\dfrac{1}{1-\dfrac{2}{5}}=\dfrac{5}{3}$$

$$S_2=\displaystyle{\sum_{n=0}^{\infty}}\left(\dfrac{3}{5}\right)^n=\dfrac{1}{1-\dfrac{3}{5}}=\dfrac{5}{2}$$

$$S=S_1+S_2=\dfrac{5}{3}+\dfrac{5}{2}=\dfrac{25}{6}$$

## Exercise 1.8

$$\displaystyle{\sum_{n=0}^{\infty}} \dfrac{n^2}{3^n}$$

Solve the series using the ratio criterion: $$\displaystyle{\lim_{n\rightarrow\infty}}\dfrac{a_{n+1}}{a_n}$$

$$\displaystyle{\lim_{n\rightarrow\infty}}\dfrac{(n+1)^2}{3^{n+1}}\dfrac{3^n}{n^2}=\lim_{n\rightarrow\infty}\dfrac{(n+1)^2}{n^2}\dfrac{1}{3}=\dfrac{1}{3}<1$$

So the series converges.