Check the solution to "Write the first 5 terms of the sequence..."

Riley Quinn

Answered question

2022-03-08

Write the first 5 terms of the sequence with general terms

a_{n}

and indicate whether they are convergent or divergent
1)

a_{n} = \frac{3 n - 5}{n + 8}

a_{n} = (2 n + 1) \times 3^{n}

a_{n} = \frac{1}{2^{n}} + n = n + 2^{- n}

Setartetaliffvz5

Beginner2022-03-09Added 1 answers

a_{n} = \frac{3 n - 5}{n + 8}

a_{1} = \frac{3 - 5}{1 + 8} = \frac{- 2}{9}

a_{2} = \frac{6 - 5}{2 + 8} = \frac{1}{10}

a_{3} = \frac{9 - 5}{3 + 8} = \frac{4}{11}

a_{4} = \frac{12 - 5}{4 + 8} = \frac{7}{12}

a_{5} = \frac{15 - 5}{5 + 8} = \frac{10}{13}

Test convergence:

lim_{n \to \infty} a_{n}

= lim_{n \to \infty} \frac{3 n - 5}{n + 8}

= lim_{n \to \infty} \frac{3 - \frac{5}{n}}{1 + \frac{8}{n}}

= 3 \neq 0

Therefore, the series diverges
2)

a_{n} = (2 n + 1) \times 3^{n}

a_{1} = 3 \times 3 = 9

a_{2} = 5 \times 3^{2} = 45

a_{3} = 7 \times 3^{3} = 189

a_{4} = 9 \times 3^{4} = 729

a_{5} = 11 \times 3^{5} = 2673

L = lim_{n \to \infty} \frac{a_{n + 1}}{a + n}

= lim_{n \to \infty} \frac{(2 n + 3) \times 3^{n + 1}}{(2 n + 1) \times 3^{n}}

= 3 lim_{n \to \infty} \frac{2 n + 3}{2 n + 1}

= 3 > 0

Thus, this series diverges

Theresa Bullock

Beginner2022-03-10Added 1 answers

(3) Given

a_{n} = \frac{1}{2^{n}} + n = n + 2^{- n}

a_{1} = \frac{3}{2}

a_{2} = 2 + \frac{1}{4} = \frac{9}{4}

a_{3} = 3 + \frac{1}{8} = \frac{25}{8}

a_{4} = 4 + \frac{1}{16} = \frac{65}{16}

a_{5} = 5 + \frac{1}{32} = \frac{161}{32}

lim_{n \to \infty} a_{n} = lim_{n \to \infty} \frac{1}{2^{n}} + n

= \infty \neq 0

Answer: the series diverges

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