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Question: How do you use the ratio test to test the convergence of the series \( \sum {\dfrac{{n + 1}}{{{3^n}}...

How do you use the ratio test to test the convergence of the series n+13n\sum {\dfrac{{n + 1}}{{{3^n}}}} from n=1 to infinity?

Explanation

Solution

Hint : To test the convergence of a series n=1an\sum\limits_{n = 1}^\infty {{a_n}} , we do a test called the ratio test, it is also known as D’Alembert’s ratio test or the Cauchy ratio test. Each term of the series is a real or a complex number, an{a_n} is not equal to zero and n is a large value. The ratio test is given as L=limxan+1anL = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| , if the value of L comes out to be smaller than 1 then the series converges, if it is equal to 1 then the test is inconclusive as the limit fails to exist and if the value comes out to be greater than 1 then the series is divergent. Doing this test on the given series we can find out whether the series is converging or not.

Complete step-by-step answer :
The series given is n+13n\sum {\dfrac{{n + 1}}{{{3^n}}}} , so the nth term of this series is given as - an=n+13n{a_n} = \dfrac{{n + 1}}{{{3^n}}}
Now to do the ratio test, we have L=limxan+1anL = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right|
Putting the value of an{a_n} in the above equation, we have –
L=limnn+13n+1n+13nL = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{\dfrac{{n + 1}}{{{3^{n + 1}}}}}}{{\dfrac{{n + 1}}{{{3^n}}}}}} \right|
As nn \to \infty , the limit clearly comes to be positive, so we can remove the modulus symbol.
L=limnn+13n+1×3nn L=limn3n3n.3×n+1n L=limn13(1+1n)  L = \mathop {\lim }\limits_{n \to \infty } \dfrac{{n + 1}}{{{3^{n + 1}}}} \times \dfrac{{{3^n}}}{n} \\\ \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{3^n}}}{{{3^n}.3}} \times \dfrac{{n + 1}}{n} \\\ \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{3}(1 + \dfrac{1}{n}) \\\
As n approaches infinity, 1n\dfrac{1}{n} approaches zero.
L=13(10) L=13 as13<1 L<1   \Rightarrow L = \dfrac{1}{3}(1 - 0) \\\ \Rightarrow L = \dfrac{1}{3} \\\ as\,\dfrac{1}{3} < 1 \\\ \Rightarrow L < 1 \;
Hence, the series n+13n\sum {\dfrac{{n + 1}}{{{3^n}}}} converges by the ratio test.

So, the correct answer is “The series n+13n\sum {\dfrac{{n + 1}}{{{3^n}}}} converges by the ratio test”.

Note : When infinitely many terms are added one after the other to a given starting quantity, the expression is called a series. It is represented as n=1an\sum\limits_{n = 1}^\infty {{a_n}} where \sum {} sign denotes the summation sign which indicates the addition of all the terms. When we get further and further in a sequence, the terms get closer and closer to a specific limit; this signifies the convergence of the series.