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Does the sum: $$\sum_{n=1}^\infty \Big{(}\frac{1+3+3^2+3^3+...+3^n}{3^n+3^{n-1}}\Big{)}^n$$

converge? I already tried Cauchy's method, but I'm not sure how to solve the limit.

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    Please edit your post as it is very confusing. If you are not sure how to type in mathjax, then upload the original question using pdf file. – 2017-01-16

2 Answers 2

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This sum does not converge. Note $$\sum_{n=1}^\infty \Big{(}\frac{1+3+3^2+...+3^n}{3^{n-1}+3^n}\Big{)}^n=\sum_{n=1}^\infty \Big{(}\frac{1+3+3^2+...+3^{n-2}}{3^{n-1}+3^n}+1\Big{)}^n>\sum_{n=1}^\infty 1$$ which itself diverges.

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Using the root test:

$$L=\lim_{n\to +\infty}\sqrt[n]{a_n}\underbrace{=}_{\text{geom. prog.}}\lim_{n\to +\infty}\dfrac{\dfrac{3^{n+1}-1}{3-1}}{3^{n-1}+3^n}=\frac{1}{2}\lim_{n\to +\infty}\frac{1-\dfrac{1}{3^{n+1}}}{\dfrac{1}{3^2}+\dfrac{1}{3}}=\frac{1}{2}\dfrac{1}{\dfrac{1}{9}+\dfrac{1}{3}}=\frac{9}{8}>1$$ so, the series is divergent.

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    Shouldn't it be 3^n instead of 3^n+1? – 2017-01-16
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    No, it is correct. There are $n+1$ terms in the sum. – 2017-01-16