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I would like to use the ratio and root test on the following series:

s = 1/2 + 1/3 + (1/2)^2 + (1/3)^2 + .. = a1 + a2 + a3 + ...

where a2 is (1/2)^2 + (1/3)^2 for example

I know we have a sum of two geometric series so the sum will be convergent but I'd like to find the following results after applications of the root and ratio test:

$lim_{n \rightarrow \infty} Inf\left(\frac{a_{n+1}}{a_n}\right) = 0 $

$lim_{n \rightarrow \infty} Sup\left(\frac{a_{n+1}}{a_n}\right) = +\infty $

$lim_{n \rightarrow \infty} Inf \sqrt[n]{a_n} = \frac{1}{\sqrt{3}} $

$lim_{n \rightarrow \infty} Sup \sqrt[n]{a_n} = \frac{1}{\sqrt{2}} $

I know that $\frac{(a_{n+1})}{(a_n)} = \frac{(3^{(n+1)} + 2^{(n+1)})}{(6.(3^n+2^n))}$ and $(a_n) = \frac{3^n + 2^n}{3^n . 2^n}$. But what to do afterwards ?

How to get these calculations ?

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  • $\begingroup$ You really fooled me by using inf and Inf in the same formula, with two different meanings! $\liminf_{n\to\infty}(a_{n+1}/a_n)=0$. Looks much better with some formatting! $\endgroup$ – Gerry Myerson May 18 '14 at 12:13
  • $\begingroup$ Corrected thank you ! $\endgroup$ – user3473016 May 18 '14 at 12:24
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So I believe you have decided to describe this series by, $$ \sum_{n = 1}^{\infty} \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n $$

Root Test: Now by the expansion of the Binomial Theorem for each $n \in \Bbb N$, $$ 0 \lt \sqrt[n]{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n } \lt \sqrt[n]{\left({\frac{1}{2} + \frac{1}{3}}\right)^n } = \left({\frac{1}{2} + \frac{1}{3}}\right) = \frac 5 6 $$

$$ 0 \lt \lim \sqrt[n]{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n } \lt 1 $$

See this. From this it follows that our series converges.

Ratio Test:

First note that $ \left({\frac{1}{2}}\right)^{n + 1} + \left({\frac{1}{3}}\right)^{n + 1} = \frac{1}{2}\left({\frac{1}{2}}\right)^{n} + \frac{1}{3}\left({\frac{1}{3}}\right)^{n} \lt \frac{1}{2}\left({\frac{1}{2}}\right)^{n} + \frac{1}{2}\left({\frac{1}{3}}\right)^{n} $ for each $n \in \Bbb N$

Therefore,

$$ 0 \lt \frac{ \left({\frac{1}{2}}\right)^{n + 1} + \left({\frac{1}{3}}\right)^{n + 1} }{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n} \lt \frac{1}{2}\frac{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n }{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n } = \frac{1}{2} $$

whence it follows that

$$ 0 \lt \lim \frac{ \left({\frac{1}{2}}\right)^{n + 1} + \left({\frac{1}{3}}\right)^{n + 1} }{ \left({\frac{1}{2}}\right)^n + \left({\frac{1}{3}}\right)^n} \lt \frac 1 2 \lt 1 $$

Clearly, our series again converges.


It is also probably worth mentioning that you can easily prove that if $\sum a_n$ and $\sum b_n$ are convergent series then $\sum a_n + b_n$ will converge too. And the sum will be the sum of the two convergent series. Therefore the convergence of the geometric series $\sum (\frac{1}{2})^n$ and $\sum (\frac{1}{3})^n$ imply the convergence of yours.

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  • $\begingroup$ Thank you two for these answers. But my question was not to demonstrate that this series was convergent but to prove the equalities I wrote and whose origin is missing .. Could you give me a hand ? $\endgroup$ – user3473016 May 18 '14 at 16:57
  • $\begingroup$ @user3473016: Apologies. I'm afraid my knowledge on the limit superior and inferior is close to nil. Sorry. Might I suggest you reword your question a bit too. $\endgroup$ – Ishfaaq May 19 '14 at 0:51

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