# Difficulty showing convergence or divergence of a sequence using DCT or LCT.

Using either the Direct or Limit Comparison Tests, determine if $\sum_{n=1}^{\infty}\frac{1+\sin^{2}(n)}{3^n}$ is convergent or divergent.

I seem to be completely stuck here.

I've chosen my series to be $\sum\frac{1}{3^n}$, which is clearly a convergent geometric series. But when I do either of the two tests, I get inconclusive answers.

\begin{align} \lim_{n\to\infty}\frac{1+\sin^2(n)}{3^n}\times\frac{3^n}{1}&=\lim_{n\to\infty}1+\sin^2(n)\\ &=\infty \end{align}

and

\begin{align} \frac{1+\sin^2(n)}{3^n}\leq\frac{1}{3^n}\Longrightarrow 1+\sin^2(n)\leq 1 \end{align}

but this inequality is not true.

But I just had a thought now... Am I allowed to choose the series $\sum\frac{2^n}{3^n}$ to help solve this? Clearly $1+\sin^2(n)\leq 2^n,\forall n\geq 1$, and $\sum\frac{2^n}{3^n}$ is also a convergent geometric series.

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$\limsup\limits_{n\to\infty}1+\sin^2(n)=2\lt\infty$ – robjohn Jun 13 '13 at 20:56
I'm sorry, but I don't know what lim sup is. I haven't learned anything like that yet. – agent154 Jun 13 '13 at 21:07
@agent154: it is essentially the same as what is in Git Gud's answer. That is, $1+\sin^2(n)\le2$. – robjohn Jun 13 '13 at 21:11

Better yet: $0\leq 1+(\sin (n))^2\leq 2$.

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Since $1\le1+\sin^2(n)\le2$, we can compare to the geometric series $$\sum_{n=1}^\infty\frac1{3^n}=\frac12\quad\text{and}\quad\sum_{n=1}^\infty\frac2{3^n}=1$$ In fact, this is actually the sum of three geometric series: \begin{align} \sum_{n=1}^\infty\frac{1+\sin^2(n)}{3^n} &=\sum_{n=1}^\infty\frac{1-\frac14(e^{2in}-2+e^{-2in})}{3^n}\\ &=\frac32\sum_{n=1}^\infty\frac1{3^n}-\frac12\mathrm{Re}\left(\sum_{n=1}^\infty\frac{e^{2in}}{3^n}\right)\\ &=\frac32\cdot\frac{1/3}{1-1/3}-\frac12\cdot\mathrm{Re}\left(\frac{e^{2i}/3}{1-e^{2i}/3}\right)\\ &=\frac34-\frac12\cdot\mathrm{Re}\left(\frac{e^{2i}/3}{1-e^{2i}/3}\cdot\frac{1-e^{-2i}/3}{1-e^{-2i}/3}\right)\\ &=\frac34-\frac12\cdot\mathrm{Re}\left(\frac{e^{2i}/3-1/9}{10/9-2\cos(2)/3}\right)\\ &=\frac34-\frac14\frac{3\cos(2)-1}{5-3\cos(2)}\\ &=\frac{4-3\cos(2)}{5-3\cos(2)}\\[9pt] &\doteq0.83996006708282 \end{align} As estimated, the sum is between $\frac12$ and $1$.

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We should swap votes on this question. – Git Gud Jun 13 '13 at 22:38
Your answer was first. That gets the voters who look while the question is new. I am no stranger to the capriciousness of voting. Besides, I'm capped for today. :-) – robjohn Jun 13 '13 at 22:41

You are correct may compare your series to a convergent geometric series; the series you have is $$\sum_{n=1}^{\infty}\frac{1+\sin^{2}(n)}{3^n}\quad \leq \quad\sum_{n=1}^{\infty}\frac{2}{3^n}\quad\leq \quad\sum_{n = 1}^\infty \left(\frac{2}{3}\right)^n\tag{*}$$

Now clearly you can do the direct comparison test with $t_n = \left(\dfrac{2}{3}\right)^n$. So your added idea will work just fine.

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