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 $\sum_{n=0}^{\infty} (\frac{4}{7})^n \frac{1}{1+(\frac{3}{7})^n}$
 A: 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:
$$\frac{4^{n+1}}{3^{n+1}+7^{n+1}}\cdot \frac{3^n + 7^n}{4^n} = 4\frac{(3/7)^n + 1}{3(3/7)^n + 7}\to \frac47 < 1$$
A: You idea was fine, just remember that you always should factor the dominant term not the smaller one.
Here it should be $7^n$ so that you get $3^n+7^n=7^n\overbrace{\Big(1+\underbrace{(\frac 37)^n}_{\to 0}\Big)}^{\to 1}$ as you did on your last line.
Conclude by saying that since $\frac 1{1+(\frac 37)^n}\to 1$ then for $n$ large enough, this quantity is $<2$ and then compare your series to the geometric series $2\sum(\frac 47)^n$.
Simpler is indeed to do as VIVID indicated and say that $3^n+7^n>7^n$ and bound your series directly, I only wanted to show that your initial idea was just fine.
