Mathematics-I (3110014)

$\mathrm{Here}, {\mathrm{u}}_{\mathrm{n}}=\frac{\sqrt{\mathrm{n}}}{{\mathrm{n}}^2+1}$

$\mathrm{Here}, {\mathrm{u}}_{\mathrm{n}}=\frac{\sqrt{\mathrm{n}}}{{\mathrm{n}}^2\left(1+{\displaystyle \frac{1}{{\mathrm{n}}^2}}\right)}$

$\mathrm{Here}, {\mathrm{u}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^{{\displaystyle \frac{3}{2}}}\left(1+{\displaystyle \frac{1}{{\mathrm{n}}^2}}\right)}$

$\mathrm{Let}, {\mathrm{v}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^{{\displaystyle \frac{3}{2}}}}$

$\mathrm{Now}, \mathrm{L}=\underset{\mathrm{n}\to \infty }{\lim } \frac{{\mathrm{u}}_{\mathrm{n}}}{{\mathrm{v}}_{\mathrm{n}}}$

$\mathrm{Now}, \mathrm{L}=\underset{\mathrm{n}\to \infty }{\lim } \frac{\frac{1}{{\mathrm{n}}^{{\displaystyle \frac{3}{2}}}\left(1+\frac{1}{{\mathrm{n}}^2}\right)}}{{\displaystyle \frac{1}{{\mathrm{n}}^{{\displaystyle \frac{3}{2}}}}}}$

$\mathrm{Now}, \mathrm{L}=\underset{\mathrm{n}\to \infty }{\lim } \frac{{\mathrm{n}}^{\frac{3}{2}}}{{\mathrm{n}}^{\frac{3}{2}}\left(1+\frac{1}{{\mathrm{n}}^2}\right)}$

$\mathrm{Now}, \mathrm{L}=\underset{\mathrm{n}\to \infty }{\lim } \frac{1}{1+\frac{1}{{\mathrm{n}}^2}}$

$\mathrm{Now}, \mathrm{L}= \frac{1}{1+\frac{1}{\infty }}$

$\mathrm{Now}, \mathrm{L}= 1 ; \mathrm{Which} \mathrm{is} \mathrm{finite} \mathrm{and} \mathrm{non}-\mathrm{zero}.$

$\mathrm{Since}, \sum {\mathrm{v}}_{\mathrm{n}}=\sum {\mathrm{n}}^{\frac{3}{2}} \mathrm{is} \mathrm{convergent} \mathrm{by} \mathrm{p}-\mathrm{test}. [ \mathrm{p}=\frac{3}{2}>1 ]$

$\mathrm{Hence}, \mathrm{by} \mathrm{Limit} \mathrm{Comparison} \mathrm{test}, \sum {\mathrm{u}}_{\mathrm{n}}=\sum \frac{\sqrt{\mathrm{n}}}{{\mathrm{n}}^2+1} \mathrm{is} \mathrm{also} \mathrm{convergent}.$