Question

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

Answer 1

Hint: To test the convergence of a series $ \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, $ {a_n} $ is not equal to zero and n is a large value. The ratio test is given as $ L = \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 $ \sum {\dfrac{{n + 1}}{{{3^n}}}} $ , so the nth term of this series is given as - $ {a_n} = \dfrac{{n + 1}}{{{3^n}}} $
Now to do the ratio test, we have $ L = \mathop {\lim }\limits_{x \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| $
Putting the value of $ {a_n} $ in the above equation, we have –
 $ L = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{\dfrac{{n + 1}}{{{3^{n + 1}}}}}}{{\dfrac{{n + 1}}{{{3^n}}}}}} \right| $
As $ n \to \infty $ , the limit clearly comes to be positive, so we can remove the modulus symbol.
 $
  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, $ \dfrac{1}{n} $ approaches zero.
 $
   \Rightarrow L = \dfrac{1}{3}(1 - 0) \\
   \Rightarrow L = \dfrac{1}{3} \\
  as\,\dfrac{1}{3} < 1 \\
   \Rightarrow L < 1 \;
  $
Hence, the series $ \sum {\dfrac{{n + 1}}{{{3^n}}}} $ converges by the ratio test.

So, the correct answer is “The series $ \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 $ \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.