Solveeit Logo
Login

Question

Question: How do you use the ratio test to test the convergence of the series \[\sum{\dfrac{{{\left( 4n+3 \rig...

How do you use the ratio test to test the convergence of the series ∑(4n+3)n(n+7)2n\sum{\dfrac{{{\left( 4n+3 \right)}^{n}}}{{{\left( n+7 \right)}^{2n}}}} from n=1n=1 to infinity?

Explanation

Solution

In order to find the solution of the given question that is to use the ratio test to test the convergence of the series ∑(4n+3)n(n+7)2n\sum{\dfrac{{{\left( 4n+3 \right)}^{n}}}{{{\left( n+7 \right)}^{2n}}}} from n=1n=1 to infinity, apply the ratio test which states that Let an{{a}_{n}} is a series and L=lim⁡n→∞∣an+1an∣L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right| then if L < 1 then the series converges absolutely; if L > 1 then the series is divergent; if L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.

Complete step by step solution:
According to the question, given series in the question is as follows:
∑n=1∞(4n+3)n(n+7)2n\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( 4n+3 \right)}^{n}}}{{{\left( n+7 \right)}^{2n}}}}
Let us suppose that an=(4n+3)n(n+7)2n{{a}_{n}}=\dfrac{{{\left( 4n+3 \right)}^{n}}}{{{\left( n+7 \right)}^{2n}}}.
Now apply the ratio test in the given series which states that Let an{{a}_{n}} is a series andL=lim⁡n→∞∣an+1an∣L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right| then

  1. if L<1 L < 1 then the series converges absolutely;
  2. if L>1L > 1 then the series is divergent;
  3. if L=1L = 1 or the limit fails to exist, then the test is inconclusive.
    We will have:
    ⇒L=lim⁡n→∞∣(4(n+1)+3)n+1((n+1)+7)2n+2(4n+3)n(n+7)2n∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{{{\left( 4\left( n+1 \right)+3 \right)}^{n+1}}}{{{\left( \left( n+1 \right)+7 \right)}^{2n+2}}}}{\dfrac{{{\left( 4n+3 \right)}^{n}}}{{{\left( n+7 \right)}^{2n}}}} \right|
    Now simplify the above expression by opening the brackets, we will have:
    ⇒L=lim⁡n→∞∣(4n+4+3)n+1(n+1+7)2n+2(4n+3)n(n+7)2n∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{{{\left( 4n+4+3 \right)}^{n+1}}}{{{\left( n+1+7 \right)}^{2n+2}}}}{\dfrac{{{\left( 4n+3 \right)}^{n}}}{{{\left( n+7 \right)}^{2n}}}} \right|
    We can rewrite the above expression as follows:
    ⇒L=lim⁡n→∞∣(n+7)2n(4n+4+3)n+1(n+1+7)2n+2(4n+3)n∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{\left( n+7 \right)}^{2n}}{{\left( 4n+4+3 \right)}^{n+1}}}{{{\left( n+1+7 \right)}^{2n+2}}{{\left( 4n+3 \right)}^{n}}} \right|
    After simplifying the above expression, we will have:
    ⇒L=lim⁡n→∞∣(n+7)2n(4n+7)n+1(n+8)2n+2(4n+3)n∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{\left( n+7 \right)}^{2n}}{{\left( 4n+7 \right)}^{n+1}}}{{{\left( n+8 \right)}^{2n+2}}{{\left( 4n+3 \right)}^{n}}} \right|
    We can rewrite the above expression as follows:
    ⇒L=lim⁡n→∞∣(n+7)2n(4n+7)n(4n+7)(n+8)2n(n+8)2(4n+3)n∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{\left( n+7 \right)}^{2n}}{{\left( 4n+7 \right)}^{n}}\left( 4n+7 \right)}{{{\left( n+8 \right)}^{2n}}{{\left( n+8 \right)}^{2}}{{\left( 4n+3 \right)}^{n}}} \right|
    ⇒L=lim⁡n→∞∣(n+7n+8)2n×(4n+74n+3)n×(4n+7)(n+8)2∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| {{\left( \dfrac{n+7}{n+8} \right)}^{2n}}\times {{\left( \dfrac{4n+7}{4n+3} \right)}^{n}}\times \dfrac{\left( 4n+7 \right)}{{{\left( n+8 \right)}^{2}}} \right|
    Now take n2n{{n}^{2n}} common from the first term and (4n)n{{\left( 4n \right)}^{n}} common from the second term in the above equation, we will get:
    ⇒L=lim⁡n→∞∣n2nn2n(1+7n1+8n)2n×(4n)n(4n)n(1+74n1+34n)n×(4n+7)(n+8)2∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{n}^{2n}}}{{{n}^{2n}}}{{\left( \dfrac{1+\dfrac{7}{n}}{1+\dfrac{8}{n}} \right)}^{2n}}\times \dfrac{{{\left( 4n \right)}^{n}}}{{{\left( 4n \right)}^{n}}}{{\left( \dfrac{1+\dfrac{7}{4n}}{1+\dfrac{3}{4n}} \right)}^{n}}\times \dfrac{\left( 4n+7 \right)}{{{\left( n+8 \right)}^{2}}} \right|
    After simplifying the above expression, we will have:
    ⇒L=lim⁡n→∞∣(1+7n1+8n)2n×(1+74n1+34n)n×(4n+7)(n+8)2∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| {{\left( \dfrac{1+\dfrac{7}{n}}{1+\dfrac{8}{n}} \right)}^{2n}}\times {{\left( \dfrac{1+\dfrac{7}{4n}}{1+\dfrac{3}{4n}} \right)}^{n}}\times \dfrac{\left( 4n+7 \right)}{{{\left( n+8 \right)}^{2}}} \right|
    After applying the limit to first two terms we get the result that lim⁡n→∞(1+7n1+8n)2n=1\displaystyle \lim_{n \to \infty }{{\left( \dfrac{1+\dfrac{7}{n}}{1+\dfrac{8}{n}} \right)}^{2n}}=1 and lim⁡n→∞(1+74n1+34n)n=1\displaystyle \lim_{n \to \infty }{{\left( \dfrac{1+\dfrac{7}{4n}}{1+\dfrac{3}{4n}} \right)}^{n}}=1, using this result in the above equation, we will have:
    ⇒L=lim⁡n→∞∣1×1×(4n+7)(n+8)2∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| 1\times 1\times \dfrac{\left( 4n+7 \right)}{{{\left( n+8 \right)}^{2}}} \right|
    Simplifying the above expression, we will have:
    ⇒L=lim⁡n→∞∣(4n+7)(n+8)2∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\left( 4n+7 \right)}{{{\left( n+8 \right)}^{2}}} \right|
    After solving the above expression by opening the bracket, we will have:
    ⇒L=lim⁡n→∞∣(4n+7)n2+18n+64∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\left( 4n+7 \right)}{{{n}^{2}}+18n+64} \right|
    Now take n2n{{n}^{2n}} common from the numerator and denominator of the above equation, we will get:
    ⇒L=lim⁡n→∞∣n2(4n+7n2)n2(1+18n+64n2)∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{n}^{2}}\left( \dfrac{4}{n}+\dfrac{7}{{{n}^{2}}} \right)}{{{n}^{2}}\left( 1+\dfrac{18}{n}+\dfrac{64}{{{n}^{2}}} \right)} \right|
    Simplifying the above expression, we will have:
    ⇒L=lim⁡n→∞∣(4n+7n2)(1+18n+64n2)∣\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\left( \dfrac{4}{n}+\dfrac{7}{{{n}^{2}}} \right)}{\left( 1+\dfrac{18}{n}+\dfrac{64}{{{n}^{2}}} \right)} \right|
    After applying the limit, we will get:
    ⇒L=0+01+0+0\Rightarrow L=\dfrac{0+0}{1+0+0}
    ⇒L=0\Rightarrow L=0
    As lim⁡n→∞∣an+1an∣=0<1\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|=0<1, according to the ratio test, the given series ∑n=1∞(4n+3)n(n+7)2n\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( 4n+3 \right)}^{n}}}{{{\left( n+7 \right)}^{2n}}}} converges.

Note: Students make mistakes while applying the ratio test and get confused between the statement of ratio test and root test. They apply the statement of root test for ratio test which states that Let an{{a}_{n}} is a series and L=lim⁡n→∞∣(an)1n∣L=\displaystyle \lim_{n \to \infty }\left| {{\left( {{a}_{n}} \right)}^{\dfrac{1}{n}}} \right| then if L < 1 then the series converges absolutely; if L > 1 then the series is divergent; if L = 1 or the limit fails to exist, then the test is inconclusive. This is completely wrong and leads to building wrong concepts.