Consider the series $a_n=1+\frac{2}{3}+\frac{4}{9}+...+\frac{2^n}{3^n}$. I need to find an upper bound (a value that is bigger than the series for every natural number n). I found that $a_n<n+1$. Is that enough or there is an exact value that is bigger than the series?
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4Hint: Geometric series. – Simply Beautiful Art Aug 24 '17 at 20:33
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Try $$a_n<3{}$$ – Did Aug 24 '17 at 20:37
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1$a_n < \frac{2}{3} a_n + 1$. – Daniel Schepler Aug 24 '17 at 20:48
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1What is asked is a constant upper bound. – Bernard Aug 24 '17 at 20:51
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I claim that $a_n \leq 3$. Notice $a_{n + 1} = 1 + \frac{2}{3} a_n$. So we can prove inductively that if $a_n \leq 3$, then \begin{align*} a_{n + 1} & = 1 + \frac{2}{3} a_n \\ & \leq 1 + \frac{2}{3} (3) \\ & = 3 . \end{align*} And voila, you have an upper bound. But we can show it's the optimal upper bound by observing that the sequence is increasing and bounded, and thus has a limit $L$. You can also see $$ L = 1 + \frac{2}{3} L \Rightarrow L = 3 .$$ So $3$ is the least upper bound to the sequence.
AJY
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For $n\in \Bbb N,$
$$a_n=\sum_{i=0}^n(\frac{2}{3})^i=\frac {1-(\frac {2}{3})^{n+1}}{1-\frac {2}{3}} $$
$$< \frac {1}{\frac {1}{3}} $$
$$\implies a_n <3.$$
hamam_Abdallah
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