For $x\in \left ( 0,1 \right )$ converges $\sum_{n=0}^{\infty}\left ( n+2 \right )\left ( x-1 \right )^n$. Is this true statment?

Question

For $x\in \left ( 0,1 \right )$ converges $\sum_{n=0}^{\infty}\left ( n+2 \right )\left ( x-1 \right )^n$. Is this true statment?

To examine convergence I usually use rules for convergence, but what to do when I have intervall?

Answer 1

Using the ratio test,

$$\lim_{n\rightarrow\infty} \left|\frac{[(n+1)+2](x-1)^{n+1}}{(n+2)(x-1)^n}\right|=\lim_{n\rightarrow\infty}\left|\frac{(n+3)}{(n+2)}(x-1)\right|=|x-1|$$

The series converges if $|x-1|<1$. Thus, $x\in(0,2)$.

However, we must check the boundaries: when $x=2$ and when $x=0$ as sometimes, depending on the series, it may be convergent.

For $x=2$ $$\sum_{n=0}^\infty (n+2)1^n$$

diverges.

For $x=0$

$$\sum_{n=0}^\infty (n+2)(-1)^n$$ diverges.

So, the interval of convergence is indeed $x\in(0,2)\supset(0,1)$

Answer 2

Using d'Alembert, $$\lim_{n\to \infty }\left|\frac{(n+1)+2}{n+1}\right|=1.$$ Therefore, the serie converge if $|x-1|<1$ and thus, for $x\in (0,2)$. Therefore, your statement is true.

Answer 3

It's a power series with center in $x=1$; you can calculate the radius of convergence: $$R=\dfrac{1}{\limsup_{n\to +\infty}\sqrt[n]{\lvert a_n\rvert}}$$

where $a_n=n+2$

You have: $$\limsup_{n\to +\infty}\sqrt[n]{\lvert a_n\rvert}=\limsup_{n\to +\infty}\sqrt[n]{n+2}=1$$ so $R=1$

The disc of convergence is: $$\lbrace x\in\mathbb{R} | \lvert x-1\rvert<R\rbrace=\lbrace x\in\mathbb{R} | \lvert x-1\rvert<1\rbrace=(0,2)$$

Is it true that for all $x\in(0,1)$ the series is convergent because it converges in $(0,2)$.

Answer 4

Why not sum the series directly?

Put $y=x-1$. Then $$\sum_{n=0}^{\infty}\left ( n+2 \right )\left ( x-1 \right )^n=\sum_{n=0}^{\infty}\left ( n+2 \right )y^n$$

Next, put $f(y)=\sum_{n=0}^{\infty}y^{n+1}$. Then $f´(y)= \sum_{n=0}^{\infty}(n+1)y^n$. Thus you have $$\sum_{n=0}^{\infty}\left ( n+2 \right )\left ( x-1 \right )^n=f´(y)+\sum_{n=0}^{\infty}y^n$$

For $|y|<1$, $f(y)$ converges to $\frac{y}{1-y}$ (which is differentiable in that range) and $\sum_{n=0}^{\infty}y^n$ converges to $\frac{1}{1-y}$. So you can not only deduce that your series converges for $x\in (0,2)$ but also give an explicit value for its sum. All without invoking any principle or theorem at all!