Determine the convergence of the series \sum_{n=1}^\infty \frac{3e^{7n}}{e^{3n}+ 6} . Explain...


Determine the convergence of the series {eq}\displaystyle \sum_{n=1}^\infty \frac{3e^{7n}}{e^{3n}+ 6} {/eq}. Explain why it is convergent or divergent.

Ratio test for Checking the Convergence:

Consider the series {eq}\sum\limits_{n = 1}^\infty {{a_n}} \;,\;{a_n} \ne 0,\forall {\text{n}},\;{\text{and}}\;\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| \to L {/eq} for some {eq}L {/eq}.

1.) If L<1, then {eq}\sum _{n=1}^{\infty }a_{n} {/eq} converges.

2.) If L>1, then {eq}\sum _{n=1}^{\infty }a_{n} {/eq} diverges.

3.) If L=1, then we can't make any conclusion.

Answer and Explanation:

Here the given series is:

{eq}\displaystyle \sum_{n=1}^\infty \frac{3e^{7n}}{e^{3n}+ 6}=\sum_{k=1}^\infty \frac{3e^{7k}}{e^{3k}+ 6} {/eq}


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Learn more about this topic:

Testing for Convergence & Divergence by Comparing Series

from AP Calculus BC: Exam Prep

Chapter 21 / Lesson 5

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