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Does This Sequence Converge?

May 22, 2022/0 Comments/in Algebra, Calculus, College Algebra, limits, Math Test Prep, sequences, series /by Jill Hacker

Question

Does the sequence ${\frac{3n}{2n-1}}$ converge?

Solution

A sequence converges if there is a limit for its terms as the number terms goes to infinity.

Is there a limit of ${\frac{3n}{2n-1}}$ as $n$ goes to infinity? Let’s see:

$\frac{3n}{2n-1}=\frac{3n\cdot1/n }{\left (2n-1 \right )\cdot1/n}=\frac{3}{2-1/n}$ . Since the limit as $n$ goes to infinity of $1/n$ equals 0, the limit as $n$ goes to infinity of $\frac{3n}{2n-1}=$ $\frac{3}{2-0}=\frac{3}{2}$ .

There is a limit: $\frac{3}{2}$ . Therefore the sequence converges.

Caution

Graph of the sequence 3n/(2n-1). Value approaches 3/2 as n gets large.

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 $n$ th term is limited as $n$ goes to infinity.

To Learn More . . .

More information about sequences:

Solving Radical Equations

April 23, 2021/0 Comments/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.

Question

Does the equation $\sqrt{4x+3}+10=5$ have a real solution? If so, what is it?

Solution

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:

${\color{Blue} \sqrt{4x+3}+10=5}$

Subtract 5 from each side:

${\color{Blue} \sqrt{4x+3}=-5}$

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 ${\color{Blue} \sqrt{4x+3}=-5}$

and see what happens. OK, let’s try that. Follow the usual steps for solving a radical equation:

Isolate the radical. We did that above when we added 10 to each side.
Square both sides:

${\color{Blue} 4x+3=25}$

Subtract 3 from each side:

${\color{Blue} 4x=22}$

Divide by 4:

${\color{Blue} x=11/2}$

There’s a solution: $x=11/2$ . Now let’s see if it works. Substitute $x=11/2$ back into the left-hand side of the original equation. You should get:

${\color{Blue} \sqrt{4 \cdot 11/2+3}+10}$

That simplifies to:

${\color{Blue} \sqrt{25}+10= 15\neq5 }$

So what looked like the wrong way was indeed the wrong way, the solution $x=11/2$ does not work, and the answer to the question is No, the equation $\sqrt{4x+3}+10=5$ 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.