Question

Show that the interval of convergence of the series$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{x}}^{\mathbf{n}}}{{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{n}}$is $\left|x\right|\mathbf{\le }\mathbf{1}\mathbf{}\mathbf{(}\mathbf{For}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{1}$. (For , this is the series of Problem 9.) Using theorem$\left(14.4\right)$, show that for$\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$, four terms will give two decimal place accuracy.

Short answer

The series is convergent for $\left|x\right|\le 1$ , and as $R_4\le \left|\frac{1}{960}\right|$for $x=\frac{1}{2}$, four terms will give two decimal place accuracy.

Step-by-step solution

Given Information

$\mathrm{The}\mathrm{function}\mathrm{is}\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^2+\mathrm{n}}.$

Definition of the Error of the Series

The calculated series is Underestimate if the Error is less than zero. The calculated series is Overestimate if the Error is more than zero.

Verify the statement

Test the convergence as:

$$\begin{gathered} \underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{x^{n+1}}{{\left(n+1\right)}^2+n+1}\times \frac{n^2+n}{x^n}\right| \\ =\left|x\right|\underset{n\to \infty }{\lim }\left|\frac{n^2+n}{{\left(n+1\right)}^2+n+1}\right| \\ =\left|x\right|\underset{n\to \infty }{\lim }\left|\frac{n\left(n+1\right)}{\left(n+1\right)+\left(n+2\right)}\right| \\ =\left|x\right|\underset{n\to \infty }{\lim }\left|\frac{n}{\left(n+2\right)}\right| \\ =\left|x\right| \end{gathered}$$

Series is convergence if $\left|x\right|<1$.

Check the boundary conditions as:

For x= 1:

$$\begin{gathered} \sum _{n=1}^{\infty }\frac{x^n}{n^2+n}=\sum _{n=1}^{\infty }\frac{{\left(1\right)}^n}{n^2+n} \\ =\sum _{n=1}^{\infty }\frac{1}{n\left(n+1\right)} \end{gathered}$$

The above series is convergent.

Hence, the series is convergent for $\left|x\right|\le 1$

For $x<\frac{1}{2}$, Error of remainder is given below:

$$\begin{gathered} R_4\le \left|a_5\right| \\ {R}_4\le \left|\frac{1}{2^5\left(5^2+5\right)}\right| \\ {R}_4\le \left|\frac{1}{\left(30\right)\left(30\right)}\right| \\ {R}_4\le \left|\frac{1}{960}\right| \end{gathered}$$

Hence, the series is convergent for $\left|x\right|\le 1$ , and as $R_4\le \left|\frac{1}{960}\right|$for $x=\frac{1}{2}$, four terms will give two decimal place accuracy.

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