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zavuch27 [327]
3 years ago
10

Which of the following series are convergent?

Mathematics
1 answer:
Westkost [7]3 years ago
3 0

Answer:

All of them

Step-by-step explanation:

According to the ratio test, for a series ∑aₙ:

If lim(n→∞) |aₙ₊₁ / aₙ| < 1, then ∑aₙ converges.

If lim(n→∞) |aₙ₊₁ / aₙ| > 1, then ∑aₙ diverges.

(I) aₙ = 10 / n!

lim(n→∞) |(10 / (n+1)!) / (10 / n!)|

lim(n→∞) |(10 / (n+1)!) × (n! / 10)|

lim(n→∞) |n! / (n+1)!|

lim(n→∞) |1 / (n+1)|

0 < 1

This series converges.

(II) aₙ = n / 2ⁿ

lim(n→∞) |((n+1) / 2ⁿ⁺¹) / (n / 2ⁿ)|

lim(n→∞) |((n+1) / 2ⁿ⁺¹) × (2ⁿ / n)|

lim(n→∞) |(n+1) / (2n)|

1/2 < 1

This series converges.

(III) aₙ = 1 / (2n)!

lim(n→∞) |(1 / (2(n+1))!) / (1 / (2n)!)|

lim(n→∞) |(1 / (2n+2)!) × (2n)! / 1|

lim(n→∞) |(2n)! / (2n+2)!|

lim(n→∞) |1 / ((2n+2)(2n+1))|

0 < 1

This series converges.

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In this problem:

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\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\\\approx \frac{\frac{1}{3}}{3}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{1}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(1+\frac{2}{2}\right)^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{3}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)

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\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx \approx \frac{1}{9}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(\frac{3}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(2\right)^2}} +4\sqrt{1+\frac{1}{\left(\frac{5}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)

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