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

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

2 Answers 2

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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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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.

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  • $\begingroup$ Hi, thanks. But why do I have to factor the dominant? $\endgroup$
    – Sofi Garde
    Commented Jul 17, 2021 at 23:09
  • $\begingroup$ 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. $\endgroup$
    – zwim
    Commented Jul 17, 2021 at 23:22

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