Tag Archive for: convergence

Does This Sequence Converge?

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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 nth 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.

To Learn More . . . 

More information about sequences: