Calculus - Infinite Series

Use the comparison test.

For every $n$, let $$ a_n = \dfrac{2^n + 3^n}{3^n + 4^n}. $$ The given series is $\sum_{n=1}^{\infty} a_n$. We will show that it converges.

For every $n$, $$ a_n = \dfrac{2^n + 3^n}{3^n + 4^n} \leq \dfrac{3^n + 3^n}{4^n} = 2 \biggl( \dfrac{3}{4} \biggr)^n. $$ For every $n$, let $b_n = 2(\tfrac{3}{4})^n$. So $a_n \leq b_n$ for every $n$. Since $\sum_{n=1}^{\infty} (\tfrac{3}{4})^n$ is a geometric series with common ratio $\tfrac{3}{4}$, it converges. Thus, $2 \sum_{n=1}^{\infty} (\tfrac{3}{4})^n$ converges also, i.e., $\sum_{n=1}^{\infty} b_n$ converges. Therefore, $\sum_{n=1}^{\infty} a_n$ converges by the comparison test.

Let us try to split:

$$\frac{2^n+3^n}{3^n+4^n}=\frac{2^n}{3^n+4^n}+\frac{3^n}{3^n+4^n}.$$

Then we consider each the following series:

$$\sum_{n=1}^\infty \frac{2^n}{3^n+4^n}\quad\text{and}\quad \sum_{n=1}^\infty \frac{3^n}{3^n+4^n}\tag 1$$

As we can see, we have $$\frac{2^n}{3^n+4^n}<\frac{2^n}{3^n}=\bigg(\frac{2}{3}\bigg)^n\quad \forall n\in\Bbb N$$ and $$\frac{3^n}{3^n+4^n}<\frac{3^n}{4^n}=\bigg(\frac{3}{4}\bigg)^n\quad \forall n\in\Bbb N$$

Convergence of the given two series in $(1)$ follows from the Comparison Test because
$$\sum_{n=1}^\infty \bigg(\frac{2}{3}\bigg)^n\quad\text{and}\quad \sum_{n=1}^\infty \bigg(\frac{3}{4}\bigg)^n$$ are both convergent. Hence, the series $$\sum_{n=1}^\infty \frac{2^n+3^n}{3^n+4^n}$$ is convergent (being a sum of two convergent series).

It's enough to notice that all terms $a_n$ of your sequence satisfy $$ 0< a_n < q^n \tag{1} $$ for an appropriately chosen $q<1$. For example, you can choose $q=0.9$, and $(1)$ will be true for all $n\ge1$. Then you can use the fact that the geometric series $$ \sum_{n=1}^\infty q^n = {q\over 1-q} $$ converges; therefore your series $\sum_{n=1}^\infty a_n$ converges too, by comparison test.