Question 11 and 12, Exercise 4.8

Solutions of Question 11 and 12 of Exercise 4.8 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Evaluate the sum of the series: $\sum_{k=1}^{n} \frac{1}{k(k+2)}$

Solution.

Let $T_k$ represent the $k$th term of the series. Then $$\begin{align*} T_k &= \frac{1}{k(k+2)}. \end{align*}$$ Resolving it into partial fractions: $$\begin{align*} \frac{1}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2} \ldots (1) \end{align*}$$ Multiplying both sides by $k(k+2)$, we get $$\begin{align*} 1 = A(k+2) + Bk \ldots (2) \end{align*}$$ Put $k=0$ in (2), we have $$\begin{align*} &1=2A + 0 \\ \implies & A = \frac{1}{2}. \end{align*}$$ Put $k+2=0 \implies k=-2$ in (2), we have $$\begin{align*} &1=0-2B\\ \implies &B = -\frac{1}{2}. \end{align*}$$ Using the values of $A$ and $B$ in equation (1), we get $$\begin{align*} \frac{1}{k(k+2)} &= \frac{1}{2k} - \frac{1}{2(k+2)}. \end{align*}$$ Thus, $$\begin{align*} T_k &= \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right). \end{align*}$$ Taking the sum, we have $$\begin{align*} S_n &= \sum_{k=1}^n T_k = \frac{1}{2} \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+2} \right). \end{align*}$$ This is a telescoping series, so most terms cancel out, and we are left with: $$\begin{align*} S_n &= \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right) \\ & = \end{align*}$$ This will be solved later.

Evaluate the sum of the series: $\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\ldots \ldots$ to infinity.

Solution.

Let $T_k$ represents the kth term of the series. Then $$\begin{align*} T_k &=\frac{1}{(3k-2)(3k+1)}. \end{align*}$$ Resolving it into partial fraction: $$\begin{align*} \frac{1}{(3k-2)(3k+1)} = \frac{A}{3k-2}+\frac{B}{3k+1} \ldots (1) \end{align*}$$ Multiplying with $(3k-2)(3k+1)$ $$\begin{align*} 1 = (3k+1)A+(3k-2)B \ldots (2) \end{align*}$$ Put $3k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have $$\begin{align*} &1 = \left(3\times\frac{2}{3}+1 \right)A+0 \\ \implies &A = \frac{1}{3}. \end{align*}$$ Now put $3k+1=0$ $\implies k=-\dfrac{1}{3}$ in (2), we have $$\begin{align*} &1 = 0+\left(3\left(-\frac{1}{3}\right)-2 \right)B \\ \implies &B = -\frac{1}{3}. \end{align*}$$ Using value of $A$ and $B$ in (1), we get $$\begin{align*} \frac{1}{(3k-2)(3k+1)} = \frac{1}{3(3k-2)}-\frac{1}{3(3k+1)} \end{align*}$$ This gives $$\begin{align*} T_k &=\frac{1}{3}\left[\frac{1}{3k-2}-\frac{1}{3k+1}\right] \end{align*}$$ Taking sum $$\begin{align*} S_n&=\sum_{k=1}^n T_k =\frac{1}{3}\sum_{k=1}^n\left[\frac{1}{3k-2}-\frac{1}{3k+1}\right] \\ & = \frac{1}{3}\left[\left(\frac{1}{1}-\frac{1}{4}\right) + \left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{10}\right)+\ldots\right.\\ &\left.+\ldots+\left(\frac{1}{3n-5}-\frac{1}{3n-2}\right)+\left(\frac{1}{3n-2}-\frac{1}{3n+1}\right) \right] \\ & = \frac{1}{3}\left[1-\frac{1}{3n+1} \right] \\ & = \frac{1}{3}\left[\frac{3n+1-1}{3n+1} \right] \\ & = \frac{1}{3}\left[\frac{3n}{3n+1} \right] \\ & = \frac{n}{3n+1} \end{align*}$$ Now $$\begin{align*} S_\infty& = \lim_{n \to \infty} S_n \\ &=\lim_{n \to \infty} \frac{n}{3n+1} \\ &=\lim_{n \to \infty} \frac{1}{3+1/n} \\ &=\frac{1}{3+0} = =\frac{1}{3} \\ \end{align*}$$

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