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Prove using ratio test: $$\frac{n+1+\log n}{2n} = \Theta(\log n).$$ If I recall correctly, theta means they grow at the same rate, which I told was true for this statement. I am having trouble showing it. I can set the ratio up by putting $\log n$ under the left side, but what do I do from there? Any help is greatly appreciated!

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    Make the left side one fraction first. You can then take limits, apply properties of $\theta$, etc. How have you done these in the past? – 2017-02-06
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    I think you have a typo here btw. The left side grows at least as fast as $n$ which is much faster growth than $\log(n) $ on the RHS – 2017-02-06
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    Did you perhaps mean $(n+1)\color{red}{\cdot} \frac{\log(n)}{2n}$?? – 2017-02-06
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    @BrevanEllefsen no it's (n+1)+log(n) all over 2n – 2017-02-06
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    Then it's false: $\frac{n+1+\log n}{2n} \xrightarrow[n\to\infty]{} \frac{1}{2}$, so it is $\Theta(1)$ -- definitely not $\Theta(\log n)$. – 2017-02-06
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    @cparks10 OK, that not what you wrote at the moment. What you wrote is $n+1+\frac{\log n}{2n}$ and what you **meant** to write was $\frac{n+1+\log(n)}{2n}$ – 2017-02-06
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    @ClementC. I agree. I still think that this is supposed to be a multiplication like my comment above reads in order to get $\Theta(\log n) $ – 2017-02-06
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    Yes, that's much more likely. – 2017-02-06
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    Proper notation is not $logn$ but $\log n,$ coded as \log n. That results in proper spacing in expressions like $a\log b$ and $a\log (b).$ I edited the question accordingly. – 2017-02-06
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    @MichaelHardy although the way you edited the post it's now harder to tell if it should be $\log\left(\frac{n}{2n}\right)$ or $\frac{\log n}{2n}$ — to appease this issue, I recommend writing $\log(n) $ here instead of $\log n$ – 2017-02-06
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    @BrevanEllefsen : It said $logn/2n$ and now it says $\log n/2n.$ Any ambiguity was already there; I didn't create it. And by one convention, $A/2n$ means $(A/2)n,$ whereas I suspect $A/(2n)$ was intended, and in fact you claim that was what was intended. – 2017-02-06
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    @MichaelHardy haha, fair enough with that last point. The OP did already say that it was "all over $2n$" already, but I see your point. I wasn't meaning to imply that you created the ambiguity, but that you didn't fix the ambiguity, and my comment was more aimed at the OP. I apologize for the confusion :) – 2017-02-06
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    @BrevanEllefsen the edit made by Michael Hardy to my original post is correct. The next problem in the assignment is like what you describe, multiplication instead of addition. I apologize for my novice understanding of stack overflow formatting. I can't express enough the gratitude I have for everyone doing such a great job helping me. Thank you very much! – 2017-02-06

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