I just finished calculus 1,2 last semester and I am learning calculus 3 now. I saw this question and I post a solution as follow:

Prove series convergent,consider the limit:

$$\lim_{n\rightarrow\infty}\frac{\tan{n}}{1.5^n} = \lim_{n\rightarrow\infty}\frac{\frac{1}{\cos^2{n}}}{1.5^n\cdot\ln1.5} =\lim_{n\rightarrow\infty}\frac{1}{\ln1.5\cdot\cos^2n\cdot1.5^n} = 0$$

Hence, by the definition of limit, $$\forall\varepsilon\gt0,\exists N\gt0,\text{such that} ~n \gt N,~|\frac{\tan{n}}{1.5^n}| \lt \varepsilon$$ Let $\varepsilon = 1$, we have

\begin{align*} \ &\Longrightarrow|\frac{\tan{n}}{1.5^n}|\lt 1 \Longrightarrow \tan{n} \lt 1.5^n \\ &\Longrightarrow \frac{\tan{n}}{2^n}\lt \frac{1.5^n}{2^n} =(\frac{3}{4})^n \\ &\Longrightarrow \sum_{n=1}^{\infty}\frac{\tan{n}}{2^n}\lt\sum_{n=1}^{\infty}(\frac{3}{4})^n ~~~~\text{for some}\space n \gt N \end{align*}

Since the geometric series $\sum_{n=1}^{\infty}(\frac{3}{4})^n$ is convergent,by the comparison test,$\sum_{n=1}^{\infty}\frac{\tan{n}}{2^n} \text{is convergent.}$

But @i707107 told me that I cannot apply the L'Hospital's theorem for $\lim_{n\rightarrow\infty}\frac{\tan{n}}{1.5^n}$, so my whole proof is wrong. I did go back to check the scope of application of L'Hospital's theorem and I even read the proof in the appendix of our textbook, which only makes me more confused. So I am expecting somebody are able to explain why I can not apply L'Hospital's theorem here. Also, Since this question is beyond my knowledge right now, It would be perfect if you could tell me what material I should read first in order to totally understand it. Thanks very much.
L'Hospital's theorem


While $$\displaystyle\lim_{x\rightarrow \infty} 1.5^n$$ goes to $+\infty$, $$\displaystyle\lim_{x\rightarrow \infty} \tan x$$ does not go to $+\infty$ or $-\infty$ (in fact, I'm pretty sure that limit isn't defined at all), so it doesn't hit either of the cases needed in L'Hospital's Rule.

As your little snippet pointed out, we need either both limits diverging or both going to 0. We don't have that here.

  • $\begingroup$ Thanks for the help, Is it beacuse according to the graph of $\tan x$, It look like the limit does't not exist because it oscilliate between $+\infty$ or $-\infty$ as x goes $+\infty$. Hence it is not an indeterminated form of type $\frac{\infty}{\infty}$here, Thus we can not use L'Hospital's Rule? $\endgroup$ – SundayCat Oct 29 '13 at 21:36
  • $\begingroup$ Pretty much. You'd need something that definitely goes to one or the other (like $x^2$) for it to be an indeterminate form. $\endgroup$ – Dennis Meng Oct 29 '13 at 21:41
  • $\begingroup$ Just want to make this more clear :). For the indeterminate form of type $\frac{\infty}{\infty}$, we have to make sure that both nominator and denominator definitely go to $+\infty$ as x goes to $+\infty$ or a finite number. In our case, the value of $\tan x$ do not always goes to $+\infty$ in the interval $[-\infty,+\infty]$. so we can conclude that it is not an indeterminate form. Am I right? $\endgroup$ – SundayCat Oct 29 '13 at 21:59
  • $\begingroup$ Correct. We do not have a $\frac{\infty}{\infty}$ indeterminate form here, since $\tan x$ does not go to $+\infty$, it oscillates, as you noted earlier. $\endgroup$ – Dennis Meng Oct 29 '13 at 22:01
  • $\begingroup$ Thanks again for the great answer here. By the way, the scope of application of L'Hospital's theorem sometimes is really vague in my textbook. It would be more helpful if you could summarize and explain them in your answer here. Like why does the right hand side limit should be exist and why do the $ \f'(a), f'(b)$ not have to be existed? $\endgroup$ – SundayCat Oct 29 '13 at 22:09

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