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Find $$\lim_{n\rightarrow\infty} \frac{\log(2 + 3^{n})}{2n}$$

Hint: $$\log(2+3^{n}) = log(3^{n}) + \log(\frac{(2 + 3^{n})}{3^{n}})$$

Attempt:

If I apply the hint to the expression and do a little simplification I arrive at:

$$\frac{n log(3)}{2n} + \frac{\log(\frac{log(2 + 3^{n})}{3^{n}})}{2n}$$

Now if I take the limit of this expression:

$$\lim_{n\rightarrow\infty} \frac{ \log(3)}{2} + \lim_{n\rightarrow\infty} \frac{\log(\frac{\log(2 + 3^{n})}{3^{n}})}{2n}$$

Here is where I am stuck. I feel I can argue that the limit of the second term goes to 0 because the denominator will go to infinity faster than the numerator. As such I am left with $\frac{log(3)}{2}$ as my answer. My only concern is that I could've used that same train of thought with the original expression.

Advice?

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    $\begingroup$ The denominator on the right of your hint should be inside the log. $\endgroup$ – Arnaud Mortier Jan 27 '18 at 15:47
  • $\begingroup$ Upvoted the comment. I guess there are now complete answers for the OP, but hopefully the OP could actually solve the problem themselves already, with this hint fixed... $\endgroup$ – user491874 Jan 27 '18 at 15:50
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$log(2+3^n)=log(3^n(2/3^n+1)=log(3^n)+log(2/3^n+1)$, $log(3^n)/2n=log(3)/2$, $lim_{n\rightarrow+\infty}log(2/3^n+1)/n=0$.

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$$\frac{\log(2+3^n)}{2n}\sim \frac{\log(3^n)}{2n}=\frac{n\log(3)}{2n}\longrightarrow\frac{\log(3)}{2}\qquad\text{as $n\to\infty$}$$

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The hint is not quite right. $log(2+3^n)=log(3^n)+log(1+\frac{2}{3^n})$. Using this and the fact that $log(3^n)=n~log(3)$ you get $\frac{log(2+3^n)}{2n}=\frac{log(3)}{2}+\frac{log(1+\frac{2}{3^n})}{2n}$ which tends to $\frac{log(3)}{2}$ as $\frac{log(1+\frac{2}{3^n})}{2n}$ tends to $0$ (the top goes to $log(1)=0$ and the bottom to $\infty$).

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\begin{eqnarray}\lim_{n\rightarrow\infty} \frac{\log(2 + 3^{n})}{2n} &=&\lim_{n\rightarrow\infty} \frac{\log(2/3^n + 1)+\log3^n}{2n}\\\\&=&\lim_{n\rightarrow\infty} \frac{{1\over n}\log(2/3^n + 1)+\log 3}{2}\\&=&{\log 3\over 2}\end{eqnarray}

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$$\lim_{n\rightarrow\infty} \frac{log(2 + 3^{n})}{2n}=\lim_{n\rightarrow\infty} \frac{log( 3^{n})+log( \frac{2+3^n}{3^n})}{2n}=\lim_{n\to\infty}\dfrac{n\log 3}{2n}+\lim_{n\to\infty}\dfrac{\log {(1+\frac{2}{3^n})}}{2n}=\frac{\log 3}{2}+\frac{\log 1}{\infty}=\frac{\log 3}{2}$$

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Note that

$$\frac{\log(2 + 3^{n})}{2n}=\frac{\log 3^n+\log(\frac2{3^n} + 1)}{2n}=\frac{n\log 3+\log(\frac2{3^n} + 1)}{2n}\to\frac{\log 3}{2}$$

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Rewriting the hint properly:

Hint $$log(2+3^{n}) = log(3^{n}) + log(\frac{2 + 3^{n}}{3^{n}})$$

Then your own work leads you to $$\frac{n log(3)}{2n} + \frac{log(\frac{2 + 3^{n}}{3^{n}})}{2n}$$

Now if I take the limit of this expression:

You get $ \frac{ log(3)}{2} $ as you expected, together with something of the form $\frac{0}{\infty}$, QED.

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$$ \frac{\log(2 + 3^{n})}{2n}= \frac{\log(3^n(\frac2{3^n} + 1))}{2n}= \frac{n\log(3)+\log((\frac2{3^n} + 1))}{2n} \to\frac{\log(3)}{2}$$

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  • $\begingroup$ Nitpick: it would be better not to put $\lim_n$ signs in the first $3$ terms, as you don't know yet if the limit exists or not. $\endgroup$ – Gabriel Romon Jan 27 '18 at 16:01
  • $\begingroup$ @GabrielRomon happy nopw? $\endgroup$ – Guy Fsone Jan 27 '18 at 16:03
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We can generalize a little bit

Let $a_1,$ $\cdots,$ $a_p$ be a increasing sequence of nonnegative numbers, $a_p>1$ and $c>0$.

Calculate $$\lim_{n \to \infty} \frac{\ln \left(a_1^n+ \cdots+a_p^n +c \right)}{n}. $$

We have $$\frac{1}{n}\ln \left(a_1^n+ \cdots+a_p^n +c \right)=\frac{1}{n}\ln \left(a_p^n \left[\left(\frac{a_1}{a_p}\right)^n+ \cdots+1 +\frac{c}{a_p^n}\right] \right) $$ $$= \frac{n\ln (a_p)}{n} +\frac{1}{n} \ln \left( \left(\frac{a_1}{a_p}\right)^n+ \cdots+1 +\frac{c}{a_p^n} \right) $$ every term $\left(\frac{a_1}{a_p}\right)^n \to 0$ and the same $\frac{c}{a_p^n} \to 0$, so the above limit by continuity of the $\ln$, is $$\ln (a_p)=\lim_{n \to \infty}\frac{1}{n} \ln \left(a_1^n+ \cdots+a_p^n +c \right). $$

In this case $a_p=3$ and we multiply by $1/2$ the limit

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