Does the series converge or diverge? $\sum\limits_{n = 0}^\infty {\dfrac{{{5^n}}}{{{3^n} + {4^n}}}} $.
Answer
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Hint: We are given the infinite series. We have to determine whether the series converges or diverges. First, we will simplify the expression by multiplying and dividing the expression by some constant. Then, rewrite the expression by writing 1 in the numerator. Then, we will apply the limits to the expression. Then, compare the absolute value of the expression by 1 and write the result of the limit. Then, determine whether the limit of the partial sums is finite or infinite. If it is finite then the series converges. If the limit is infinite, then the series diverges.
Complete step by step solution:
We are given the series, $\sum\limits_{n = 0}^\infty {\dfrac{{{5^n}}}{{{3^n} + {4^n}}}} $
First, divide and multiply the expression by ${4^n}$.
$ \Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\dfrac{5}{4}} \right)}^n}}}{{{{\left( {\dfrac{3}{4}} \right)}^n} + 1}}} $
Now, rewrite the expression in the form of product of numerator and remaining expression.
$ \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{5}{4}} \right)}^n} \times } \sum\limits_{n = 0}^\infty {\dfrac{1}{{{{\left( {\dfrac{3}{4}} \right)}^n} + 1}}} $
Now, we will apply the limit to the expression.
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{5}{4}} \right)^n}\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{{\left( {\dfrac{3}{4}} \right)}^n} + 1}}$
Since $\left| {\dfrac{5}{4}} \right| > 1$, therefore ${\left( {\dfrac{5}{4}} \right)^\infty } = \infty $.
Since $\left| {\dfrac{3}{4}} \right| < 1$, therefore ${\left( {\dfrac{3}{4}} \right)^\infty } = 0$
Therefore, we get:
$ \Rightarrow \infty \cdot \dfrac{1}{{0 + 1}}$
$ \Rightarrow \infty \cdot 1 = \infty $
Since the expression has no finite limit, which means the series is divergent.
Final Answer: Thus, the series $\sum\limits_{n = 0}^\infty {\dfrac{{{5^n}}}{{{3^n} + {4^n}}}} $ is divergent.
Additional Information:
In the infinite series, if the limit is applied to the partial sums and the result is equal to some real number, then the series is convergent. On the other hand, if the limit is applied on the partial sums and the output is equal to infinity, then the series is divergent.
Note:
In such types of questions students mainly make mistakes while applying the limits to the expression because in such types of questions we generally do not apply the left-hand limit to n. Also, students can make mistakes while determining whether the expression is infinity or zero.
Complete step by step solution:
We are given the series, $\sum\limits_{n = 0}^\infty {\dfrac{{{5^n}}}{{{3^n} + {4^n}}}} $
First, divide and multiply the expression by ${4^n}$.
$ \Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{{\left( {\dfrac{5}{4}} \right)}^n}}}{{{{\left( {\dfrac{3}{4}} \right)}^n} + 1}}} $
Now, rewrite the expression in the form of product of numerator and remaining expression.
$ \Rightarrow \sum\limits_{n = 0}^\infty {{{\left( {\dfrac{5}{4}} \right)}^n} \times } \sum\limits_{n = 0}^\infty {\dfrac{1}{{{{\left( {\dfrac{3}{4}} \right)}^n} + 1}}} $
Now, we will apply the limit to the expression.
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{5}{4}} \right)^n}\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{{\left( {\dfrac{3}{4}} \right)}^n} + 1}}$
Since $\left| {\dfrac{5}{4}} \right| > 1$, therefore ${\left( {\dfrac{5}{4}} \right)^\infty } = \infty $.
Since $\left| {\dfrac{3}{4}} \right| < 1$, therefore ${\left( {\dfrac{3}{4}} \right)^\infty } = 0$
Therefore, we get:
$ \Rightarrow \infty \cdot \dfrac{1}{{0 + 1}}$
$ \Rightarrow \infty \cdot 1 = \infty $
Since the expression has no finite limit, which means the series is divergent.
Final Answer: Thus, the series $\sum\limits_{n = 0}^\infty {\dfrac{{{5^n}}}{{{3^n} + {4^n}}}} $ is divergent.
Additional Information:
In the infinite series, if the limit is applied to the partial sums and the result is equal to some real number, then the series is convergent. On the other hand, if the limit is applied on the partial sums and the output is equal to infinity, then the series is divergent.
Note:
In such types of questions students mainly make mistakes while applying the limits to the expression because in such types of questions we generally do not apply the left-hand limit to n. Also, students can make mistakes while determining whether the expression is infinity or zero.
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