# Limit Test for Divergence Tutoring

Introduction to limit test for divergence tutoring:

Limit test can also be called as nth term test. For divergence, the condition for limit test will be

if `lim_(n->oo)` an ? 0, then `sum` an will diverge

Thus, this condition satisfies only if the series terms within the limit don't go to zero.

Learning limit test for divergence through tutoring  is very interactive and fun. In tutoring, students get more opportunities to clear their doubts, and they can get help at any time through online tutoring. Here, we are going to see few example and practice problems to show how tutoring help you for learning limit test for divergence.
Learn Limit Test for Divergence with Example Problems from Tutoring:

Example problem 1:

Determine whether the series `sum_(n=1)^oo` `((-1)^n4^n)/(4^n + 4^(n+1)n^-2)` is convergent or divergent.

Solution:

Step 1: Given series

`sum_(n=1)^oo` `((-1)^n4^n)/(4^n + 4^(n+1)n^-2)`.

Step 2: Find an for the test.

`sum_(n=1)^oo` `((-1)^n4^n)/(4^n + 4^(n+1)n^-2)` = `sum_(n=1)^oo` `(-1)^n(4^n)/(4^n + 4^(n+1)n^-2)`.

`sum_(n=1)^oo` `((-1)^n4^n)/(4^n + 4^(n+1)n^-2)` = `sum_(n=1)^oo` `(-1)^n`an

Therefore,

an = `(4^n)/(4^n + 4^(n+1)n^-2)`.

We can write an as follows,

an = `(4^n)/(4^n + 4^n4^1n^-2)`.

an = `(1)/(1 + 4n^-2)`.

an = `(1)/(1 + 4/(n^2))`.

an = `(1)/((n^2 + 4)/(n^2))`.

= `(n^2)/((n^2 + 4))`.

Step 3: Check the condition for divergence

`lim_(n->oo)` an = `lim_(n->oo)` `(n^2)/(n^2 + 4)` = 1

Therefore,

`lim_(n->oo)` an ? 0

Since the given series is not equal to zero, it satisfies the condition of limit test for divergence.

Step 4: Solution

Hence, the given series `sum_(n=1)^oo` `((-1)^n4^n)/(4^n + 4^(n+1)n^-2)`is divergent series.

Example problem 2:

Solve and determine whether the series `sum_(n=1)^oo` `(4sin (npi + pi/2))/(sqrt(16n))` is convergent or divergent.

Solution:

Step 1: Given series

`sum_(n=1)^oo` `(4sin (npi + pi/2))/(sqrt(16n))`.

The given series can be written as follows

`sum_(n=1)^oo` `(4sin (npi + pi/2))/(sqrt(16n))`= `sum_(n=1)^oo` `(4sin (npi + pi/2))/(4sqrt(n))`

= `sum_(n=1)^oo` `(sin (npi + pi/2))/(sqrt(n))`

Step 2: Find an for the test.

We know that sin (n? + `pi/2` ) = (-1)n

`sum_(n=1)^oo` `(sin (npi + pi/2))/(sqrtn)` = `sum_(n=1)^oo` `(-1)^n(1)/(sqrtn)`.

`sum_(n=1)^oo` `(sin (npi + pi/2))/(sqrtn)` = `sum_(n=1)^oo` `(-1)^n` an

Therefore,

an = `(1)/(sqrtn)`.

Step 3: Check the conditions

`lim_(n->oo)` an = `lim_(n->oo)` `(1)/(sqrtn)` = 0

Therefore,

`lim_(n->oo)` an = 0

Since an of the given series is equal to zero, it satisfies the condition of limit test for divergence.

Step 4: Solution

Hence, the given series `sum_(n=1)^oo` `(4sin (npi + pi/2))/(sqrt(16n))`is convergent.
Learn Limit Test for Divergence with Practice Problems from Tutoring:

1) Determine whether the series `sum_(n=1)^oo` `((-1)^n(n^2))/(n^2 + 4)` is convergent or divergent.

2) Determine whether the series `sum_(n=1)^oo` `(2cos (npi))/(sqrt(4n))` is convergent or divergent.

Solutions:

1) The given series `sum_(n=1)^oo` `((-1)^n(n^2))/(n^2 + 4)` is divergent.

2) The given series `sum_(n=1)^oo` `(2cos (npi))/(sqrt(4n))` is convergent.