Q. 10 - Calculus

Question

Show that $\sum _{k=3}^9\frac{k}{k+1} \mathrm{is} \mathrm{equal} \mathrm{to} \sum _{k=4}^{10}\frac{k-1}{k}$by writing out the terms in each sum.

Step-by-Step Solution

Answer

It is shown that $\sum _{k=3}^9\frac{k}{k+1}=\sum _{k=4}^{10}\frac{k-1}{k}$

Step 1. Given Information

The given summations are $\sum _{k=3}^9\frac{k}{k+1}and \sum _{k=4}^{10}\frac{k-1}{k}$

Step 2. Explanation

The first sum can be written as follows,

$\sum _{k=3}^9\frac{k}{k+1}=\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\frac{6}{7}+\frac{7}{8}+\frac{8}{9}+\frac{9}{10}$

and the second sum can be written as,

$\sum _{k=4}^{10}\frac{k-1}{k}=\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\frac{6}{7}+\frac{7}{8}+\frac{8}{9}+\frac{9}{10}$

Hence, we can say that both the summations are equal.