## 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.

# Maximum Value on an Interval

## Question

If a toy car moves according to the equation $v(t)=2/3t^3-4t^2+6t+2$, where t represents time and v represents velocity, what is the toy car’s maximum acceleration on the interval [0, 5]?

## Solution

Acceleration is the first derivative of velocity. So if $v(t)=2/3t^3-4t^2+6t+2$ represents the toy car’s velocity, then $a(t)=2t^2-8t+6$ represents its acceleration.

### How to find the maximum?

The maximum on the interval could be a local maximum somewhere within the interval, or it could be a value at one end or the other. You can calculate all three and compare them. To get started a sketch may help you get a sense of what the curve looks like.

To make a quick sketch of $a(t)=2t^2-8t+6$, factor the trinomial to $2(t-3)(t-1)$. That tells you the x-intercepts: $(3,&space;0)$ and $(5,&space;0)$. And the 6 at the end of $2t^2-8t+6$ tells you that the y-intercept is $(0,&space;6)$. Those three points are enough for a crude sketch. I’ve extended the domain to 5 on the right, because that’s the interval that’s given. I think the y-value is higher at $x=5$ than it is at $x=0$, because 5 is farther from 2, which appears to be the vertex.

As you can see from either the sketch or the equation (the leading coefficient is positive), this parabola opens upward. Setting the acceleration function’s first derivative equal to zero will tell you where the vertex is – but as you can see that vertex is a minimum, not a maximum, so that is not the route to pursue to find the maximum.

### Where the first derivative is zero or at  the interval’s end

Since the vertex is the only turning point on the interval, the maximum value on the interval must be at one of the interval’s ends. If the sketch is accurate, the acceleration function, $a(t)=2t^2-8t+6$, should be at its maximum in the interval at $t=5$.

To confirm, you may want to calculate acceleration at several key points. To find acceleration at the left-hand end of the interval, $t=0$, is easy: Substitute 0 for t and you get $a(0)=6$.

Next, try the local minimum. To find the y-value there, you first have to find the t-value. To do that, differentiate $a(t)=2t^2-8t+6$ and get $a’(t)=4t-8$. Set that to 0 to get

$4t-8=0$

$t=2$

(Alternatively, since the acceleration function is second degree, if you represent its equation as $at^2+bt+c$, you can find its vertex by setting t equal to $-b/(2a)$. For the equation $a(t)=2t^2-8t+6$$a=2$ , $b=-8$, and $c=6$, so $-b/(2a)=2$.)

$a(2)=2(2^2)-8(2)+6=-2$

Finally, to find $a(5)$ you plug 5 into $a(t)=2t^2-8t+6$, which is a bit more work than plugging in 0, but still not a big deal. Then

$a(5)=2(5^2)-8(5)+6=16$

The results:

 t A(t) 0 6 2 -2 5 16

As the crude sketch indicated, the highest value is at $t=5$. There $a(t)=16$, and that is the answer. If you can work with a calculator and are not limited to what you can sketch by hand, you may get a better picture, like this: