Convergence of $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$. Does $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ converge?
Dividing the top and bottom by $4^n$ gives
\begin{equation*}
\frac{2^n+5^n}{3^n+4^n} = \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1}.
\end{equation*}
Hence,
\begin{equation*}
\lim_{n\to\infty} \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1} = \frac{0+\infty}{0+1} = \infty.
\end{equation*}
Thus, $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ diverges.
Is this correct? Thanks.
 A: Your argument is correct. However, you could come to the same conclusion through a different method. Let
$$a_n=\frac{2^n+5^n}{3^n+4^n}$$
and consider what happens as $n\to\infty$. In the numerator, $5^n$ will grow faster than $2^n$ and in the denominator $4^n$ will grow faster than $3^n$. Therefore
$$\lim_{n\to\infty}a_n=\lim_{n\to\infty} \frac{5^n}{4^n}=\lim_{n\to\infty}\left(\frac{5}{4}\right)^n=\infty$$
which implies by the term test that the series diverges.
A: Yes.                                                                                                              :)
A: That does work. What would be easier is to notice that $5^n$ dominates the numerator while $4^n$ dominates the denominator. Therefore $$\lim_{n \to \infty} \frac{2^n+5^n}{3^n+4^n} = \lim_{n \to \infty} \frac{5^n}{4^n} $$
This clearly diverges.
By the $n$th term test, you can see that the series also diverges.
A: $n>1$;
$a_n=\dfrac{2^n+5^n}{3^n+4^n} \gt$
$\dfrac{5^n}{4^n+4^n}=(1/2)(5/4)^n >$
$(1/2)(1+1/4)^n> (1/2)(1/4)n.$
Used : Binomial expansion
