# 1-1/9+1/25-1/49+1/81 in sigma notation?

May 1, 2018

${\sum}_{n = 1}^{5} \frac{{\left(- 1\right)}^{n - 1}}{2 n - 1} ^ 2$

#### Explanation:

Notice that each of our terms seems to be divided by a perfect square. That is, $1 - \frac{1}{9} + \frac{1}{25} - \frac{1}{49} + \frac{1}{81} = \frac{1}{1} ^ 2 - \frac{1}{3} ^ 2 + \frac{1}{5} ^ 2 - \frac{1}{7} ^ 2 + \frac{1}{9} ^ 2$.

We also note that we initially start with a 1, then we subtract our next term. We add the term after that and subtract the term following that. That is, we alternate adding and subtracting terms.

Using these two facts, we wish to devise a formula $f \left(n\right)$ such that $1$ corresponds with $n = 1$, $- \frac{1}{9}$ corresponds with $n = 2$, $\frac{1}{25}$ corresponds with $n = 3$, and so on.

We will first figure out the denominator. We create a small table in which we have our value of $n$ and the value of the denominator.

$n$ | denominator
1 | 1
2 | ${3}^{2}$
3 | ${5}^{2}$
4 | ${7}^{2}$
5 | ${9}^{2}$

We can see now that our denominator can be represented as ${\left(2 n - 1\right)}^{2}$. If we try building our answer in sigma notation now, you'll find we almost have the correct answer.

${\sum}_{n = 1}^{5} \frac{1}{2 n - 1} ^ 2 = 1 + \frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \frac{1}{81}$.

We simply need to build in the alternating addition and subtraction. We do this by adding the term ${\left(- 1\right)}^{n - 1}$. See that when $n = 1$, ${\left(- 1\right)}^{1 - 1} = {\left(- 1\right)}^{0} = 1$. The same holds for when $n = 3$ and $n = 5$. When $n$ is even, this term becomes a negative one.

${\sum}_{n = 1}^{5} \frac{{\left(- 1\right)}^{n - 1}}{2 n - 1} ^ 2$.