Question

Solve : $\underset{n\to \infty }{lim}\frac{1+3+5+\dots ...+.(2n-1)}{2+4+6+..+2n}$

• A. $0$
• B. $1$
• C. $-1$
• D. $4$

Answer 1

Answer: (B)

$1$

Given: $\underset{n\to \infty }{lim}\frac{1+3+5+\dots ...+.(2n-1)}{2+4+6+..+2n}$

$\frac{1+3+5+7+9+....+(2n-1)}{2+4+6+8+...2n}=\frac{(1+2+3+4+\mathrm{5...2}n)-(2+4+6---+2n)}{2(1+2+\mathrm{3...}+n)}$

$=\frac{1+2+3+4+5+.....2n}{2+4+6+8+....+2n}-1=\frac{\frac{2n(2n+1)}{2}}{\frac{2n(n+1)}{2}}-1$

Use summation formula

$\underset{n\to \infty }{lim}\frac{1+3+5+\dots ...+(2n-1)}{2+4+6+..+2n}$

$=\underset{n\to \infty }{lim}\frac{\frac{2n(2n+1)}{2}}{\frac{2n(n+1)}{2}}-1$

$=\underset{n\to \infty }{lim}\frac{\frac{2n^2(2+\frac{1}{n})}{2}}{\frac{2n^2(1+\frac{1}{n})}{2}}-1$

$=\underset{n\to \infty }{lim}\frac{(2+\frac{1}{n})}{(1+\frac{1}{n})}-1$

$=\frac{2+0}{1+0}-1$ $[\because n\to \infty \Rightarrow \frac{1}{n}\to 0]$

$=2-1$

$=1$

Hence, option ${}^{\prime }{B}^{\prime }$ is correct.