Chapter 12.4, Problem 4CFU

To determine

To find: the convergence or divergence and indicate any restrictions on method’s use.

Answer to Problem 4CFU

if an infinite series has sum or limit, the series is convergent.

If a series is not convergent it is divergent.

By ratio test, if the value of $r<1$ , then the series is convergent, if $r>1$ then the series is divergent.

Explanation of Solution

Given:

Convergent and divergent.

Concept used:

if an infinite series has sum or limit, the series is convergent.

If a series is not convergent it is divergent.

If the series is geometric series and having common ratio less than $1$ then it is convergent.

If the series is geometric series and having common ratio more than $1$ then it is divergent.

By using ratio test, the infinite series having the general expression $t_n$ can check whether it is convergent or divergent.

Let’s suppose that the value is:

  $\underset{n\to \infty }{\lim }\frac{a_n+1}{a_n}=r$ .

By ratio test, if the value of $r<1$ , then the series is convergent, if $r>1$ then the series is divergent.

Calculation:

Convergent and divergent series: if an infinite series has sum or limit, the series is convergent.

If a series is not convergent it is divergent.

If the series is geometric series and having common ratio less than $1$ then it is convergent.

If the series is geometric series and having common ratio more than $1$ then it is divergent.

By using ratio test, the infinite series having the general expression $t_n$ can check whether it is convergent or divergent.

Let’s suppose that the value is:

  $\underset{n\to \infty }{\lim }\frac{a_n+1}{a_n}=r$ .

By ratio test, if the value of $r<1$ , then the series is convergent, if $r>1$ then the series is divergent.

If $r=1$ then ratio test fails to discuss the convergence or divergence for the series.

By comparison test, a series of positive term is convergent for $n>1$ , each term of the series is equal to or less then the value of the corresponding terms of some convergent series of positive terms.

A series of positive terms is divergent, if for $n>1$ , each term of the series is equal to or greater than the value of the corresponding term of some divergent series of positive terms.

If the series is arithmetic series then it is divergent.

If the series is harmonic series.

  $1+\frac{1}{2}+\frac{1}{3}+\mathrm{....}$ is divergent.

If the series of the form is:

  $1+\frac{1}{2^p}+\frac{1}{3^p}+\mathrm{....}$

Is convergent if $p>1$ .