## 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:

$\frac{3n}{2n-1}=\frac{3n\cdot1/n&space;}{\left&space;(2n-1&space;\right&space;)\cdot1/n}=\frac{3}{2-1/n}$. 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 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.

### Update

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 $n$ goes to infinity of ${\frac{3n}{2n-1}}$, start by noting that as $n$ 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 $\frac{3}{2}$. There’s the limit. Same answer, less sweat.

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 $\inline \fn_jvn \sqrt{4x+3}+10=5$ 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:

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

Subtract 5 from each side:

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

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

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:

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

• Subtract 3 from each side:

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

• Divide by 4:

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

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

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

That simplifies to:

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

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

## Is This Relation a Function?

This post about whether a relation is a function is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

One or more of the graphs below represent 𝑦 as a function of 𝑥. Which one or ones?

 A B C D

## Solution

Function/not function?

A function is a relation – that is, a correspondence between x and  y – for which every x-value relates to only one y-value. No vertical line will pass through more than one point on a function’s graph, but a vertical line may pass through more than one point on the graph of a relation that is not a function. Thus you can use the vertical line test on a relation to see whether it is a function.

Look at relation A. Imagine sliding a vertical line across it. At no point will the line intersect the relation more than once. This relation is a function.

Now look at relation B. The y-axis, which is a vertical line, intersects the relation in three places. Thus relation B fails the vertical line test and relation B is not a function.

Relation C: Again, there are places where you could draw a vertical line and cross the relation more than once. For example, the y-axis crosses the relation twice. This relation is not a function.

Relation D: There is only one x-value where it looks like a vertical line could cross this relation twice: the x-value at the right-hand end of the lower piece, which is the x-value at left-hand end of the upper piece. Are there two y-values at this x-value? The lower piece, in which the circle is filled in, does take this x-value. But the upper piece, in which the circle is not filled in, does not take the value. Thus there is no x-value that both pieces take, and there is no x-value for which there is more than one value of  This relation is a function.

𝑨𝒏𝒔𝒘𝒆𝒓: A and D

## Multiplying a Binomial by a Trinomial

This post about multiplying a binomial by a trinomial is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.

## Question

Multiply:

$\fn_jvn {\color{DarkBlue}(x-2)(x^2+3x+3)}$

## Solution

The first expression — the contents of the first set of parentheses — is called a binomial because it has two terms. The second expression — the contents of the second set of parentheses — is called a trinomial because it has three terms. To multiply a binomial  by a trinomial, multiply each term in the binomial by each term in the trinomial.

You need a way to organize this. Try this: Multiply x, the first term in the binomial, by each term in the trinomial. Then multiply -2, the second term in the binomial, by each term in the trinomial. Then add all those results together.

• Multiply x, the first term in the binomial, by each term in the trinomial.

$\fn_jvn {\color{DarkBlue}x(x^2+3x+3)=x^3+3x^2+3x}$

• Multiply -2, the second term in the binomial, by each term in the trinomial.

$\fn_jvn {\color{DarkBlue}-2(x^2+3x+3)=-2x^2-6x-6}$

• Add the results, adding like terms to like terms. If you stack the expressions vertically, that will help you keep like terms lined up:

$\fn_jvn {\color{DarkBlue}x^3+3x^2+3x}\\ {\color{DarkBlue} \hspace{1.24cm}-2x^2-6x-6}$

$\fn_jvn {\color{DarkBlue}x^3+x^2-3x-6}$