Question

For the following series, write formulas for the sequences ${\mathit{a}}_{\mathbf{n}}\mathbf{,}{\mathit{S}}_{\mathbf{n}}\mathbf{,}\mathbf{\text{and}}{\mathit{R}}_{\mathbf{n}}$, and find the limits of the sequences as $\mathit{n}\mathbf{\to }\mathbf{\infty }$ (if the limits exist).

$\frac{\mathbf{3}}{\mathbf{1}\mathbf{·}\mathbf{2}}\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}\mathbf{·}\mathbf{3}}\mathbf{+}\frac{\mathbf{7}}{\mathbf{3}\mathbf{·}\mathbf{4}}\mathbf{-}\frac{\mathbf{9}}{\mathbf{4}\mathbf{·}\mathbf{5}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

Short answer

Hence, the required formulae is:

$$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k=0 \\ \underset{k\to \infty }{\lim }{S}_k=1 \\ \underset{k\to \infty }{\lim }{R}_k=0 \end{gathered}$$

Step-by-step solution

Sum of the Series

The sum of the k terms of the series which is in geometric progression, with first term a and common ratio r, is given by:

${\text{S}}_k=\frac{a\left(1-r^k\right)}{1-r}$

Find the General Term

The given series is: $\frac{3}{1·2}-\frac{5}{2·3}+\frac{7}{3·4}-......$

The series can be represented as:

$\frac{3}{1·2}-\frac{5}{2·3}+\frac{7}{3·4}-......=\sum _{k=1}^{\infty }{\left(-1\right)}^{k+1}\frac{2k+1}{k^2+k}$

Clearly, the general term will be:

$\begin{array}{rcl}{a}_k& =& {\left(-1\right)}^{k+1}\frac{2k+1}{k^2+k}\\ & =& {\left(-1\right)}^{k+1}\frac{2k+1}{k\left(k+1\right)}\\ & & \end{array}$

Using partial fraction method, we have:

$\begin{array}{rcl}\frac{2k+1}{k\left(k+1\right)}& =& \frac{A}{k}+\frac{B}{k+1}\\ 2k+1& =& \left(A+B\right)k+A\\ A& =& 1\\ B& =& 1\\ & & \end{array}$

So, we get:

$\begin{array}{rcl}{a}_k& =& {\left(-1\right)}^{k+1}\frac{2k+1}{k\left(k+1\right)}\\ & =& {\left(-1\right)}^{k+1}\left\{\frac{1}{k}+\frac{1}{k+1}\right\}\\ \underset{k\to \infty }{\lim }{a}_k& =& 0\end{array}$

Find the Sum

Now, let us substitute $k=1,2,3,..........,\left(n-1\right),n$, we get:

$\begin{array}{rcl}{S}_k& =& \frac{1}{1}+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-.......+{\left(-1\right)}^k\left\{\frac{1}{k-1}+\frac{1}{k}\right\}+{\left(-1\right)}^k\left\{\frac{1}{k}+\frac{1}{k+1}\right\}\\ & =& 1+\frac{{\left(-1\right)}^{k+1}}{k+1}\\ & & \end{array}$

Find the Sequences

Now, we have:

$\begin{array}{rcl}S& =& \underset{k\to \infty }{\lim }{S}_k\\ & =& \underset{k\to \infty }{\lim }\left[1+\frac{{\left(-1\right)}^{k+1}}{k+1}\right]\\ & =& 1\end{array}$

The sequence can be calculated as follows:

$\begin{array}{rcl}{R}_k& =& S-S_k\\ & =& 1-1-\frac{{\left(-1\right)}^{k+1}}{k+1}\\ & =& \frac{-{\left(-1\right)}^{k+1}}{k+1}\\ \underset{k\to \infty }{\lim }{R}_k& =& 0\\ & & \end{array}$

Hence, this is the required answer.