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 $$\sum_{n=0}^{\infty} (\frac{4}{7})^n \frac{1}{1+(\frac{3}{7})^n}$$

• Use comparison with the geometric series $\Sigma(\frac47)^n$. Jul 17, 2021 at 17:47
• You should know this result: if $0\le a_n\le b_n$ for all $n$ and $\Sigma b_n$ converges, then so does $\Sigma a_n$. Jul 17, 2021 at 17:47

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: $$\frac{4^{n+1}}{3^{n+1}+7^{n+1}}\cdot \frac{3^n + 7^n}{4^n} = 4\frac{(3/7)^n + 1}{3(3/7)^n + 7}\to \frac47 < 1$$

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.

• Hi, thanks. But why do I have to factor the dominant? Jul 17, 2021 at 23:09
• Because the minority term divided by the dominant one will converge to $0$, and we will be able to discard this negligible quantity $(\frac 37)^n$. If you do the opposite, you are left with a quantity $(\frac 73)^n$ which grows very large and which you cannot ignore.
– zwim
Jul 17, 2021 at 23:22