91Ó°ÊÓ

Consider the given question,

$\underset{k=0}{\overset{\mathrm{∞}}{∑}} \left(\frac{(1/2{)}^k}{\ln (k+2)}−\frac{(1/2{)}^{k+1}}{\ln (k+3)}\right)$

The first term of the given series is obtained by substituting $k=0$,

$$\begin{gathered} =\frac{(1/2{)}^k}{\ln (k+2)}−\frac{(1/2{)}^{k+1}}{\ln (k+3)} \\ =\frac{1}{\ln \left(2\right)}−\frac{(1/2)}{\ln \left(3\right)} \end{gathered}$$

First term is $\frac{1}{\ln \left(2\right)}−\frac{(1/2)}{\ln \left(3\right)}$.

The second term of the given series is obtained by substituting $k=1$,

$$\begin{gathered} =\frac{(1/2{)}^k}{\ln (k+2)}−\frac{(1/2{)}^{k+1}}{\ln (k+3)} \\ =\frac{(1/2)}{\ln \left(3\right)}−\frac{(1/2{)}^2}{\ln \left(4\right)} \end{gathered}$$

Second term is$\frac{(1/2)}{\ln \left(3\right)}−\frac{(1/2{)}^2}{\ln \left(4\right)}$.

The third term of the given series is obtained by substituting $k=2$,

$$\begin{gathered} =\frac{(1/2{)}^k}{\ln (k+2)}−\frac{(1/2{)}^{k+1}}{\ln (k+3)} \\ =\frac{(1/2{)}^2}{\ln \left(4\right)}−\frac{(1/2{)}^3}{\ln \left(5\right)} \end{gathered}$$

Third term is $\frac{(1/2{)}^2}{\ln \left(4\right)}−\frac{(1/2{)}^3}{\ln \left(5\right)}$.

The fourth term of the given series is obtained by substituting $k=3$,

$$\begin{gathered} =\frac{(1/2{)}^k}{\ln (k+2)}−\frac{(1/2{)}^{k+1}}{\ln (k+3)} \\ =\frac{(1/2{)}^3}{\ln \left(5\right)}−\frac{(1/2{)}^4}{\ln \left(6\right)} \end{gathered}$$

Fourth term is$\frac{(1/2{)}^3}{\ln \left(5\right)}−\frac{(1/2{)}^4}{\ln \left(6\right)}$.

The fifth term of the given series is obtained by substituting $k=4$,

$$\begin{gathered} =\frac{(1/2{)}^k}{\ln (k+2)}−\frac{(1/2{)}^{k+1}}{\ln {\displaystyle (}k+3)} \\ =\frac{(1/2{)}^4}{\ln \left(6\right)}−\frac{(1/2{)}^5}{\ln \left(7\right)} \end{gathered}$$

Fifth term is $\frac{(1/2{)}^4}{\ln \left(6\right)}−\frac{(1/2{)}^5}{\ln \left(7\right)}$.

The first and second terms in the sequence of partial sum is given below,

$$\begin{gathered} S_1=\frac{1}{\ln \left(2\right)}−\frac{(1/2)}{\ln \left(3\right)} \\ {S}_2=S_1+a_2 \\ =\frac{1}{\ln \left(2\right)}−\frac{(1/2)}{\ln \left(3\right)}+\frac{(1/2)}{\ln \left(3\right)}−\frac{(1/2{)}^2}{\ln \left(4\right)} \\ =\frac{1}{\ln \left(2\right)}−\frac{(1/2{)}^2}{\ln \left(4\right)} \end{gathered}$$

The third, fourth and fifth terms in the sequence of partial sum is given below,

$\begin{array}{r}{S}_3=S_2+a_3\\ =\frac{1}{\ln \left(2\right)}−\frac{(1/2{)}^2}{\ln \left(4\right)}+\frac{(1/2{)}^2}{\ln \left(4\right)}−\frac{(1/2{)}^3}{\ln \left(5\right)}\\ =\frac{1}{\ln \left(2\right)}−\frac{(1/2{)}^3}{\ln \left(5\right)}\\ S_4=S_3+a_4\\ =\frac{1}{\ln \left(2\right)}−\frac{(1/2{)}^3}{\ln \left(5\right)}+\frac{(1/2{)}^3}{\ln \left(5\right)}−\frac{(1/2{)}^4}{\ln \left(6\right)}\\ =\frac{1}{\ln \left(2\right)}−\frac{(1/2{)}^4}{\ln \left(6\right)}\\ S_5=S_4+a_5\\ =\frac{1}{\ln \left(2\right)}−\frac{(1/2{)}^4}{\ln \left(6\right)}+\frac{(1/2{)}^4}{\ln \left(6\right)}−\frac{(1/2{)}^5}{\ln \left(7\right)}\\ =\frac{1}{\ln \left(2\right)}−\frac{(1/2{)}^5}{\ln \left(7\right)}\end{array}$

The kth term in the sequence of the partial sums is given below,

$S_k=\left(\frac{1}{\ln \left(2\right)}−\frac{(1/2)}{\ln \left(3\right)}\right)+\left(\frac{(1/2)}{\ln \left(3\right)}−\frac{(1/2{)}^2}{\ln \left(4\right)}\right)+…\left(\frac{(1/2{)}^k}{\ln (k+2)}−\frac{(1/2{)}^{k+1}}{\ln (k+3)}\right)$

In each two consecutive pairs, the second term of a pair cancels with the first term of the subsequent pair.

Thus, the series is telescopic.

The general term in its sequence of partial sums is$S_k=\frac{1}{\ln \left(2\right)}−\frac{(1/2{)}^{k+1}}{\ln (k+3)}$.

The $S_k$in its sequence of partial sums is $S_k=\frac{1}{\ln \left(2\right)}−\frac{(1/2{)}^{k+1}}{\ln (k+3)}$.

The value of $\underset{k→∞}{\lim }{S}_k$is given below,

$\begin{array}{r}\underset{k→\mathrm{∞}}{lim} S_k=\underset{k→\mathrm{∞}}{lim} \left(\frac{1}{\ln \left(2\right)}−\frac{(1/2{)}^{k+1}}{\ln (k+3)}\right)\\ =\frac{1}{\ln 2}−0\left(As\frac{b^k}{\ln k}→0,k→\mathrm{∞}\right)\\ =\frac{1}{\ln 2}\end{array}$