Chapter 9.4, Problem 44E

To determine

Find and prove the formula for the sum of the first $n$ terms of the sequence.

Answer to Problem 44E

  $S_n=\frac{n}{2\left(n+2\right)}$

Explanation of Solution

Calculation:

Consider the given sequence,

  $\frac{1}{2\cdot 3},\frac{1}{3\cdot 4},\frac{1}{4\cdot 5},\frac{1}{5\cdot 6},\mathrm{.....}\frac{1}{\left(n+1\right)\left(n+2\right)}\mathrm{...}$

Now find the sum of terms,

  $\begin{array}{l}{S}_1=\frac{1}{2\cdot 3}=\frac{1}{6}\\ S_2=\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}=\frac{1}{6}+\frac{1}{12}\\ =\frac{3}{12}\\ S_3=\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\\ =\frac{6}{20}\\ S_4=\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\end{array}$

  $=\frac{10}{30}$

Observe that the numerator and denominator of $S_k$ is $\frac{k\left(k+1\right)}{2}$ and $\left(k+1\right)\left(k+2\right)$

Now the formula for the $k^{th}$ sum is,

  $\begin{array}{l}{S}_k=\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6}+\mathrm{.....}+\frac{1}{\left(k+1\right)\left(k+2\right)}\\ =\frac{k\left(k+1\right)}{2\left(k+1\right)\left(k+2\right)}\end{array}$

Now apply mathematical induction to prove the validity of the formula,

Assuming that the formula is valid for $n=k$ show that it is valid for $n=k+1$

  $\begin{array}{l}{S}_{k+1}=\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}+\frac{1}{5\cdot 6}+\mathrm{.....}+\frac{1}{\left(k+1\right)\left(k+2\right)}\\ =\frac{k\left(k+1\right)}{2\left(k+1\right)\left(k+2\right)}+\frac{1}{\left(k+2\right)\left(k+3\right)}\\ =\frac{k\left(k+1\right)\left(k+3\right)+2\left(k+1\right)}{2\left(k+1\right)\left(k+2\right)\left(k+3\right)}\\ =\frac{\left(k+1\right)\left[k\left(k+3\right)+2\right]}{2\left(k+1\right)\left(k+2\right)\left(k+3\right)}\\ =\frac{\left[k\left(k+3\right)+2\right]}{2\left(k+2\right)\left(k+3\right)}\\ =\frac{\left(k+1\right)\left(k+2\right)}{2\left(k+2\right)\left(k+3\right)}\end{array}$

We conclude that hypotheses is valid,

  $\begin{array}{l}{S}_n=\frac{n\left(n+1\right)}{2\left(n+1\right)\left(n+2\right)}\\ S_n=\frac{n}{2\left(n+2\right)}\end{array}$

Hence, the formula for the sum of the first $n$ terms of the sequence is, $S_n=\frac{n}{2\left(n+2\right)}$