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My Attempt

First, check the limit $$\lim_{n \to \infty} \frac{3^n + 4^n}{2^n + 5^n} = \lim_{n \to \infty} \frac{\left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n}{\left(\frac{2}{5}\right)^n + 1} = 0.$$ So, we cannot conclude anything. I used Comparison Test and Ratio Test.

Consider that $$\frac{3^n + 4^n}{2^n + 5^n} < 2\frac{5^n}{2^n + 5^n}.$$ If I can proof the convergence of the series on the right-side, then it's done by comparison test. I used ratio test, in order to proof the convergence of $$\sum_{n= 1}^{\infty} \frac{5^n}{2^n + 5^n}.$$ $$\lim_{n \to \infty} \frac{55^n}{2(2^n) + 5(5^n)} \frac{2^n + 5^n}{5^n} = 5 \lim_{n \to \infty} \frac{2^n + 5^n}{2(2^n) + 5(5^n)} = 1.$$ I didn't find the way to proof.

Any suggestion? Thanks in advanced.

Solution (@abiessu & @Thomas Andrew)

Consider that $$\frac{3^n + 4^n}{2^n + 5^n} < 2\frac{4^n}{2^n + 5^n}.$$ Proof this series converge. $$\sum_{n= 1}^{\infty} \frac{4^n}{2^n + 5^n}.$$

Proof (Ratio Test) $$\lim_{n \to \infty} \frac{44^n}{2(2^n) + 5(5^n)} \frac{2^n + 5^n}{4^n} = 4 \lim_{n \to \infty} \frac{2^n + 5^n}{2(2^n) + 5(5^n)} = \frac{4}{5}.$$ The series converge.

Hence $$\sum_{n = 1}^{\infty} \frac{3^n + 4^n}{2^n + 5^n}$$ converge.

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    $\begingroup$ Your sum has index $i$ but there is no $i$ in the summed term. Also your sum is finite, so convergence has no meaning. $\endgroup$
    – coffeemath
    Dec 22, 2022 at 17:48
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    $\begingroup$ You should compare with $2\frac {4^n}{2^n+5^n}$... $\endgroup$
    – abiessu
    Dec 22, 2022 at 17:50
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    $\begingroup$ Your terms are less than $\frac{4^n+4^n}{5^n}.$ $\endgroup$ Dec 22, 2022 at 17:52

3 Answers 3

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$$\left(\frac{2}{5}\right)^n + 1 > 1 $$ Hence

$$\frac{\left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n}{\left(\frac{2}{5}\right)^n + 1} < \left(\frac{3}{5}\right)^n + \left(\frac{4}{5}\right)^n $$

And both $\left(\frac{3}{5}\right)^n$ and $\left(\frac{4}{5}\right)^n$ are general terms of a convergent geometric series.

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My preferred method would be to note that $$ \frac{3^n+4^n}{2^n+5^n} \leq \frac{4^n+4^n}{0+5^n} = 2 \cdot \left(\frac{4}{5}\right)^n. $$ So, the series converges by the comparison test and the geometric series criterion.

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  • $\begingroup$ Also my preferred solution. My students know that I prefer the direct comparison test over limit comparison. I would have not done a comparison in the numerator though. $\endgroup$ Dec 22, 2022 at 18:25
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I prefer the Ratio Test. In fact, $$ \lim_{n\to\infty}\frac{\frac{3^n+4^n}{2^n+5^n}}{(\frac{4}{5})^n}=\lim_{n\to\infty}\frac{3^n+4^n}{4^n}\cdot\frac{5^n}{2^n+5^n}=1>0 $$ which implies that $\sum_{n=1}^\infty\frac{3^n+4^n}{2^n+5^n}$ converges since $\sum_{n=1}^\infty(\frac{4}{5})^n$ converges.

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  • $\begingroup$ Isn't it called the limit comparison test? $\endgroup$
    – Gary
    Dec 22, 2022 at 23:33

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