<s>[INST] Convergence of $\sum_{n=0}^{\infty} \frac{4^n}{3^n+7^n}$ I need help with this.
$\sum_{n=0}^{\infty}  \frac{4^n}{3^n+7^n}$
I know that it converges but i can not proove why.
I tried to rewrite it, it seems to be a geometric serie. I tried to do a common factor between $3^n+7^n  \rightarrow 3^n(1+\frac{7^n}{3^n})$
So I have $\sum_{n=0}^{\infty} (\frac{4}{3})^n \frac{1}{1+(\frac{7}{3})^n}$ And I do not know if that helps.
I can also make different the common factor and I would have $\sum_{n=0}^{\infty} (\frac{4}{7})^n \frac{1}{1+(\frac{3}{7})^n}$
 [/INST] Comparison test: Observe that
$$\frac{4^n}{3^n+7^n} \le \frac{4^n}{7^n} = \left(\frac47\right)^n$$
which is a convergent geometric series

Root test:
Observe that
$$\frac{4}{\sqrt[n]{7^n+7^n}}\le \sqrt[n]{\frac{4^n}{3^n + 7^n}} \le \frac{4}{\sqrt[n]{7^n}} \\
\frac{4}{7\sqrt[n]{2}}\le \sqrt[n]{\frac{4^n}{3^n + 7^n}} \le \frac{4}{7}$$
Now, by Squeeze theorem
$$\lim_{n\to\infty} \sqrt[n]{\frac{4^n}{3^n + 7^n}} = \frac47 < 1$$

Ratio test: Calculate the limit of the ratio of two consecutive terms:
$$\frac{4^{n+1}}{3^{n+1}+7^{n+1}}\cdot \frac{3^n + 7^n}{4^n} = 4\frac{(3/7)^n + 1}{3(3/7)^n + 7}\to \frac47 < 1$$
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