Use Limit Comparison Test to conclude if the series converges or diverges

Question

The series is:

$$\sum_{n=1}^\infty \frac{5^{n+2}-2}{2^{n+1}+3}.$$

I don't know which series I've to compare to start.

I'll aprecciate any help.

Answer 1

One could use the LCT as follows:

For all $n\ge 1,$ we find that \begin{eqnarray}5^{n+2}-2 &=& 5\cdot 5^{n+1}-2\\ &=& 4\cdot 5^{n+1}+5^{n+1}-2\\ &\ge& 4\cdot 5^{n+1}+23\\ &>& 4\cdot\left(5^{n+1}-5\right)\\ &=& 20\cdot\left(5^n-1\right)\\ &\ge& 20\cdot4^n,\end{eqnarray} and we also find that \begin{eqnarray}2^{n+1}+3 &<& 2^{n+1}+4\\ &=& 2\cdot\left(2^n+2\right)\\ &\le& 2\cdot\left(2^n+2^n\right)\\ &=& 4\cdot 2^n.\end{eqnarray}

Thus, for all $n\ge 1,$ we have $$\frac{5^{n+2}-2}{2^{n+1}+3}>\frac{20\cdot 4^n}{2^{n+1}+3}>\frac{20\cdot 4^n}{4\cdot 2^n}=5\cdot\frac{4^n}{2^n}=5\cdot\left(\frac42\right)^n=5\cdot 2^n.$$ Since $5\cdot 2^n>0$ for all $n$ and $5\cdot 2^n\not\to 0$ as $n\to\infty,$ then $\sum_{n=1}^\infty 5\cdot 2^n$ diverges. Thus, by LCT, $$\sum_{n=1}^\infty\frac{5^{n+2}-2}{2^{n+1}+3}$$ diverges, too.

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All that said, there are actually infinitely-many divergent series with positive terms to which we could compare the series in question, and find that the terms of the series in question are all (eventually) larger than those of the divergent series, thus concluding by the LCT that the series in question is divergent.

However, this is not necessarily the simplest approach. Instead, we could note that for all $n\ge 1,$ we have $$\frac{5^{n+2}-2}{2^{n+1}+3}>\frac{5^{n+2}-2}{2^{n+1}+2^{n+1}}=\frac{5^{n+2}-2}{2^{n+2}}=\left(\frac52\right)^{n+2}-\frac1{2^{n+1}}>\left(\frac52\right)^{n+2}-\left(\frac12\right)^{n+2}>0,$$ and since $$\left(\frac52\right)^{n+2}-\left(\frac12\right)^{n+2}\not\to 0$$ as $n\to\infty,$ then $$\frac{5^{n+2}-2}{2^{n+1}+3}\not\to 0$$ as $n\to\infty,$ and so $\sum_{n=1}^\infty\frac{5^{n+2}-2}{2^{n+1}+3}$ is divergent.

Answer 2

$$\sum_{n=1}^\infty \frac{5^{n+2}-2}{2^{n+1}+3}.>\sum_{n=1}^\infty \frac{4^{n}}{2^{n}}.$$ $$\sum_{n=1}^\infty \frac{4^{n}}{2^{n}}=\sum_{n=1}^\infty 2^n$$ so the series diverges