Question

Let a sequence whose $n^{th}$ term is $a_n$ be define as $a_1=\frac{1}{2}$ and $(n-1)a_{n-1}=(n+1)a_n$ for $n\ge 2$ ; then

• A. $a_n=\frac{1}{n(n+1)}$
• B. $S_n=1-\frac{1}{n+1}$
• C. $\underset{n\to \infty }{lim}{S}_n=1$
• D. $S_n=\frac{1}{n+1}$

Answer 1

Answer: (B)

$S_n=1-\frac{1}{n+1}$

$a_1=\frac{1}{2}$

$(n+1)a_n=(n-1)a_{n-1}$, $n\ge 2$

For,

$n=2$, $3a_2=a_1$

$n=3$, $4a_3=2a_2$

$n=4$, $5a_4=3a_3$

.................................

$n=n$, $(n+1)a_n=(n-1)a_{n-1}$

Multiplying the above, we get

$3a_2\times 4a_3\times 5a_4...........\times (n+1)a_n=a_1\times 2a_2\times 3a_3\times 4a_4\times ..............\times (n-1)a_{n-1}$

$\Rightarrow 3a_2\times 4a_3\times 5a_4...........\times (n+1)a_n=\frac{1}{2}\times 2a_2\times 3a_3\times 4a_4\times ..............\times (n-1)a_{n-1}$

$\Rightarrow 3a_2\times 4a_3\times 5a_4...........\times \left(n\right)a_{n-1}(n+1)a_n=\frac{1}{2}\times 2a_2\times 3a_3\times 4a_4\times ..............\times (n-1)a_{n-1}$

$\Rightarrow n(n+1)a_n=1$

$\Rightarrow a_n=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

$\therefore S_n=\sum a_n={\sum }_{n=1}^n[\frac{1}{n}-\frac{1}{n+1}]=[\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}.......\frac{1}{n}-\frac{1}{n+1}]$

$\Rightarrow S_n=1-\frac{1}{n+1}$

Now, ${lim}_{n\to \infty }{S}_n={lim}_{n\to \infty }1-\frac{1}{n+1}=1-{lim}_{n\to \infty }\frac{1}{n+1}=1$