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Does This Sequence Converge?/in Algebra, Calculus, College Algebra, L'Hopital's rule, limits, Math Test Prep, sequences, series/by Jill Hacker
Does the sequence converge?
A sequence converges if there is a limit for its terms as the number terms goes to infinity.
Is there a limit of as goes to infinity? Let’s see:
. Since the limit as goes to infinity of equals 0, the limit as goes to infinity of .
There is a limit:. Therefore the sequence converges.
Graph of the 3n/(2n-1). Dots represent sequence values. Value approaches 3/2 as n gets large.
There may be confusion between convergence of a sequence and conversion of a series. Note the distinction: A sequence is a list of numbers: one number, then another number, then another, etc.; while a series is a sum of numbers: one number plus another number, plus another, etc. More confusing, there is a whole topic about convergence of series, with several kinds of tests for convergence.
To determine whether a sequence converges, don’t use the convergence tests that you use for series; instead, look at the limit. While a series converges only if the sum of the terms is limited, a sequence converges if its th term is limited as goes to infinity.
Paul Headley, who teaches at Northern Virginia Community College, has found an easier way to find the limit: L’Hopital’s rule. L’Hopital’s rule says that if the limits of both the numerator and the denominator of a rational function go to either zero or infinity, then the limit of the rational function is equal to the quotient of the limits of the derivatives of the numerator and the denominator.
In our example: To find the limit as goes to infinity of , start by noting that as goes to infinity, 3n and 2n-1 both go to infinity, so L’Hopital’s rule applies.
Next, replace the numerator and the denominator with their first derivatives, and you get . There’s the limit. Same answer, less sweat.
To Learn More . . .
More information about sequences:
Solving Radical Equations/in Accuplacer, Algebra, Math Test Prep/by Jill Hacker
This post about solving radical equations is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.
Does the equation have a real solution? If so, what is it?
This kind of equation is called a radical equation, because it contains a radical — in this case, a square root.
Let’s try to solve this radical equation:
Subtract 5 from each side:
The square root of something equals something negative? Really? The definition of square root specifies that it means a positive number. But this square root is supposed to equal something negative.
No real solution.
You may be tempted to keep going from
and see what happens. OK, let’s try that. Follow the usual steps for solving a radical equation:
There’s a solution: . Now let’s see if it works. Substitute back into the left-hand side of the original equation. You should get:
That simplifies to:
So what looked like the wrong way was indeed the wrong way, the solution does not work, and the answer to the question is No, the equation does not have a real solution.
This question is similar to question number 14 in the sample questions for the Accuplacer Advanced Algebra and Functions test.