Question

Express $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}..........+\frac{1}{13}$using summation notation.

Short answer

The summation notation is:$\sum _{n=1}^{13}{\left(-1\right)}^{n+1}\left(\frac{1}{n}\right)$, where $n$is positive integers.

Step-by-step solution

Step 1. Given information:

Series is :$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...........+\frac{1}{13}$

Step 2. To find the general term of the series.

$$\begin{gathered} 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{13} \\ ={\left(-1\right)}^2.\frac{1}{1}+{\left(-1\right)}^3.\frac{1}{2}+{\left(-1\right)}^4.\left(\frac{1}{3}\right)+{\left(-1\right)}^5.\left(\frac{1}{4}\right)+.......+{\left(-1\right)}^{14}.\left(\frac{1}{13}\right) \\ ={\left(-1\right)}^{(1+1)}.\frac{1}{1}+{\left(-1\right)}^{(2+1)}.\frac{1}{2}+{\left(-1\right)}^{(3+1)}.\left(\frac{1}{3}\right)+{\left(-1\right)}^{(4+1)}.\left(\frac{1}{4}\right)+.......+{\left(-1\right)}^{(13+1)}.\left(\frac{1}{13}\right) \end{gathered}$$

From above expansion it is clear that the general term of the series can be written as

$a_n={\left(-1\right)}^{n+1}.\left(\frac{1}{n}\right)$, where $n$is positive integers.

Step 3. To write the series as summation notation.

Apply the limits $n=1$to $n=13$.

So summation notation is$$\begin{gathered} \sum _{n=1}^{13}{a}_n \\ =\sum _{n=1}^{13}{(-1)}^{n+1}.\left(\frac{1}{n}\right) \end{gathered}$$