Chapter 11.4, Problem 25WE

To determine

To calculate:

The following series with sigma notation:

  $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\mathrm{....}$

Answer to Problem 25WE

The sigma notation for $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\mathrm{....}$ is ${\displaystyle \sum _{n=1}^6{n}^2+1}$ .

Explanation of Solution

Given information:

  $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\mathrm{....}$

Calculation:

In the given series:

The first term is $1$ . Here, each term appears as $1$ divided by square of the power of the index. Therefore the summand includes $\frac{1}{n^2}$

The value of the lower limit is calculated by using the last term as:

  $\begin{array}{l}\frac{1}{n^2}=1\\ \frac{1}{n^2}=\frac{1}{1^2}\\ n=1\end{array}$

So, the lower and upper limit is $1$ and $\infty$ . Now, the sigma notation for the given series is expressed as:

  $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\mathrm{....}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}}$