Q. 42 - Calculus

Question

Write each expression in Exercises 41–43 in one sigma notation (with some extra terms added to or subtracted from the sum, as necessary). 

$\sum _{k=1}^{40}\frac{1}{k}-\sum _{k=0}^{39}\frac{1}{k+1}$

Step-by-Step Solution

Answer

One sigma notation is :$\sum _{k=1}^{39}\left(\frac{1}{k}-\frac{1}{k+1}\right)-\frac{39}{40}$

Step 1. Given information

Expression in sigma notation:

$\sum _{k=1}^{40}\frac{1}{k}-\underset{k=0}{\overset{39}{\sum }}\frac{1}{k+1}$

Step 2. Write expression into one sigma notation:

$$\begin{gathered} \sum _{k=1}^{39}\frac{1}{k}-\underset{k=0}{\overset{39}{\sum }}\frac{1}{k+1}=\left(\sum _{k=1}^{39}\frac{1}{k}+\frac{1}{40}\right)-\left(\frac{1}{0+1}\underset{k=1}{\overset{39}{+\sum }}\frac{1}{k+1}\right) \\ =\sum _{k=1}^{39}\frac{1}{k}+\frac{1}{40}-1-\sum _{k=1}^{39}\frac{1}{k+1} \\ =\sum _{k=1}^{39}\left(\frac{1}{k}-\frac{1}{k+1}\right)+\frac{1-40}{40} \\ =\sum _{k=1}^{39}\left(\frac{1}{k}-\frac{1}{k+1}\right)-\frac{39}{40} \end{gathered}$$