CK-12-Calculus

8.3. Series Without Negative Terms http://www.ck12.org

The Integral Test

Another useful test for convergence or divergence of an infinite series without negative terms is theIntegral Test. It
involves taking the integral of the function related to the formula in the series. It makes sense to use this kind of test
for certain series because the integral is the limit of a certain series.
Theorem (The Integral Test)
Let∑∞k= 1 uk=∑∞k= 1 f(k)be a series without negative terms. Iff(x)is a decreasing, continuous, non-negative function
forx≥1, then:

1.∑∞k= 1 ukconverges if and only if

∫∞

1

f(x)converges.

2.∑∞k= 1 ukdiverges if and only if

∫∞

1

f(x)diverges.

In the statement of the Integral Test, we assumed thatukis a functionfofk. We then changed that function fto
be a continuous function ofxin order to evaluate the integral off. If the integral is finite, then the infinite series
converges. If the integral is infinite, the infinite series diverges. The convergence or divergence of the infinite series
depends on the convergence or divergence of the corresponding integral.

Example 7Determine if∑∞k= (^1) ( 2 k+^11 ) 32 converges or diverges.
Solution
We can use the Integral Test to determine convergence. Write the integral form:
∫∞
1

1

( 2 x+ 1 )^32

dx.

Next, evaluate the integral.

∫∞

1

1

( 2 x+ 1 )^32

dx=tlim→∞

∫t

1

( 2 x+ 1 )−^32 dx

Use the followingu−substitution to evaluate the integral:

u= 2 x+ 1

du= 2 dx

Then^12 ∫u−^32 du=^12 (−u− 11 //^22 )=−u^112 =−√^1 u.

Thus, limt→∞
∫t
1

( 2 x+ 1 )−^32 dx=limt→∞−√ 2 x^1 + 1 ]t 1 =limt→∞

(

−√^13 +−√ 2 t^1 + 1

)

=−√^13.

Since the integral is finite, the series∑∞k= (^1) ( 2 k+^11 ) 32 converges by the Integral Test.