Question

Evaluate:$\underset{n\to \infty }{lim}\frac{1+3+5+....nterms}{2+4+6+...nterms}$

• A. 2
• B. 1
• C. 3
• D. $\infty$

Answer 1

The correct option is B 1

We know that Sum of series of odd terms $=n^2$
And sum of series of even terms $=n(n+1)=n^2+n$
So the given limit is , $\underset{n\to \infty }{lim}\frac{n^2}{n^2+n}$
Dividing the numerator and denominator of the fraction by $n^2$, we get
$\underset{n\to \infty }{lim}\frac{1}{1+\frac{1}{n}}$
Substituting $n=\infty$ or $\frac{1}{n}=0$, we get
$=\frac{1}{1+1\left(0\right)}=\frac{1}{1}=1$