91Ó°ÊÓ

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:

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

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

The series can be represented as:

$\frac{3}{1·2}-\frac{5}{2·3}+\frac{7}{3·4}-......=\underset{k=1}{\overset{∞}{∑}}{\left(-1\right)}^{k+1}\frac{2k+1}{k^2+k}$

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:

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

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

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

Now, we have:

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

The sequence can be calculated as follows:

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

Hence, this is the required answer.