Calculus Examples

Evaluate the Summation sum from n=1 to infinity of (2/3)^n
n=1(23)nn=1(23)n
Step 1
The sum of an infinite geometric series can be found using the formula a1-ra1r where aa is the first term and rr is the ratio between successive terms.
Step 2
Find the ratio of successive terms by plugging into the formula r=an+1anr=an+1an and simplifying.
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Step 2.1
Substitute anan and an+1an+1 into the formula for rr.
r=(23)n+1(23)nr=(23)n+1(23)n
Step 2.2
Cancel the common factor of (23)n+1(23)n+1 and (23)n(23)n.
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Step 2.2.1
Factor (23)n(23)n out of (23)n+1(23)n+1.
r=(23)n23(23)nr=(23)n23(23)n
Step 2.2.2
Cancel the common factors.
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Step 2.2.2.1
Multiply by 11.
r=(23)n23(23)n1r=(23)n23(23)n1
Step 2.2.2.2
Cancel the common factor.
r=(23)n23(23)n1
Step 2.2.2.3
Rewrite the expression.
r=231
Step 2.2.2.4
Divide 23 by 1.
r=23
r=23
r=23
r=23
Step 3
Since |r|<1, the series converges.
Step 4
Find the first term in the series by substituting in the lower bound and simplifying.
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Step 4.1
Substitute 1 for n into (23)n.
a=(23)1
Step 4.2
Simplify.
a=23
a=23
Step 5
Substitute the values of the ratio and first term into the sum formula.
231-23
Step 6
Simplify.
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Step 6.1
Multiply the numerator by the reciprocal of the denominator.
2311-23
Step 6.2
Simplify the denominator.
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Step 6.2.1
Write 1 as a fraction with a common denominator.
23133-23
Step 6.2.2
Combine the numerators over the common denominator.
2313-23
Step 6.2.3
Subtract 2 from 3.
23113
23113
Step 6.3
Multiply the numerator by the reciprocal of the denominator.
23(13)
Step 6.4
Cancel the common factor of 3.
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Step 6.4.1
Factor 3 out of 13.
23(31)
Step 6.4.2
Cancel the common factor.
23(31)
Step 6.4.3
Rewrite the expression.
2
2
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