Answer:
The convergent series are;
[tex]\sum \limits _{n = 1}^\infty \left( \dfrac{1}{5} \right) ^n[/tex]
(And)
[tex]\sum \limits _{n = 1}^\infty \left( \dfrac{1}{10} \right) ^n[/tex]
Step-by-step explanation:
A series in mathematics is the sum of a sequence of numbers to infinity
A convergent series is a series that sums to a limit
From the given options, we have;
First option
[tex]\sum \limits _{n = 1}^\infty \dfrac{2\cdot n}{n + 1}[/tex]
As 'n' increases, 2·n becomes more larger than n + 1, and the series diverges
Second option
[tex]\sum \limits _{n = 1}^\infty \dfrac{n^2 - 1}{n - 2}[/tex]
As 'n' increases, n² - 1, becomes more larger than n - 2, and the series diverges
Third option
[tex]\sum \limits _{n = 1}^\infty \left( \dfrac{1}{5} \right) ^n[/tex]
As 'n' increases, [tex]\left( \dfrac{1}{5} \right) ^n[/tex], becomes more smaller and tend to '0', therefore, the series converges
Fourth option
[tex]\sum \limits _{n = 1}^\infty \left( \dfrac{1}{10} \right) ^n[/tex]
As 'n' increases, [tex]3 \times\left( \dfrac{1}{10} \right) ^n[/tex], becomes more smaller and tend to '0', therefore, the series converges
Fifth option
[tex]\sum \limits _{n = 1}^\infty \dfrac{1}{10} \cdot (3) ^n[/tex]
As 'n' increases, [tex]\dfrac{1}{10} \cdot (3) ^n[/tex], becomes more larger and tend to infinity, therefore, the series diverges.
