Answer to Question #189967 in Calculus for Moel Tariburu

Question

Write out the first five-term of the sequence, determine whether the sequence converges, if so find its limit (i) {√(n^2+3n)-n}_(n=1)^(+∞) (ii) {((n+3)/(n+1))^n }_(n=1)^(+∞)

Answer 1

(i) We have

${a_n} = \sqrt {{n^2} + 3n} - n$

Then

${a_1} = \sqrt {1 + 3} - 1 = 1$

${a_2} = \sqrt {4 + 6} - 2 = \sqrt {10} - 2$

${a_3} = \sqrt {9 + 9} - 3 = \sqrt {18} - 3 = 3\sqrt 2 - 3$

${a_4} = \sqrt {16 + 12} - 4 = \sqrt {28} - 4 = 2\sqrt 7 - 4$

${a_5} = \sqrt {25 + 15} - 5 = \sqrt {40} - 5 = 2\sqrt {10} - 5$

Since

$\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} + 3n} - n} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {\sqrt {{n^2} + 3n} - n} \right)\left( {\sqrt {{n^2} + 3n} + n} \right)}}{{\sqrt {{n^2} + 3n} + n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2} + 3n - {n^2}}}{{\sqrt {{n^2} + 3n} + n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{3n}}{{\sqrt {{n^2} + 3n} + n}} = \mathop {\lim }\limits_{n \to \infty } \frac{3}{{\sqrt {\frac{{{n^2}}}{{{n^2}}} + \frac{3}{n}} + 1}} = \frac{3}{{\sqrt {1 + 0} + 1}} = \frac{3}{{1 + 1}} = \frac{3}{2}$

then the sequence converges and its limit is $\frac{3}{2}$ .

(ii) We have

${a_n} = {\left( {\frac{{n + 3}}{{n + 1}}} \right)^n}$

Then

${a_1} = {\left( {\frac{{1 + 3}}{{1 + 1}}} \right)^1} = 2$

${a_2} = {\left( {\frac{{2 + 3}}{{2 + 1}}} \right)^2} = \frac{{25}}{9}$

${a_3} = {\left( {\frac{{3 + 3}}{{3 + 1}}} \right)^3} = \frac{{27}}{8}$

${a_4} = {\left( {\frac{{4 + 3}}{{4 + 1}}} \right)^4} = \frac{{{\rm{2401}}}}{{625}}$

${a_5} = {\left( {\frac{{5 + 3}}{{5 + 1}}} \right)^5} = \frac{{{\rm{1024}}}}{{243}}$

Since

$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{n + 3}}{{n + 1}}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{n + 1 + 2}}{{n + 1}}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{2}{{n + 1}}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{1}{{\frac{{n + 1}}{2}}}} \right)^{\frac{{n + 1}}{2} \cdot \frac{{2n}}{{n + 1}}}} = \mathop {\lim }\limits_{n \to \infty } {e^{\frac{{2n}}{{n + 1}}}} = {e^2}$

then the sequence converges and its limit is ${e^2}$ .