sequences and series - Closed form for sum

I wonder whether there is a closed form for this sum
$$S_n:=\displaystyle\sum_{k=0}^n \dfrac{4^k}{4^k+5^k}$$
The question asks to express the sum in terms of $n$ then to deduce the limit of $\dfrac{S_n}{n+1}$.
I tried to use the following sum as an auxiliary sum
$$T_n:=\displaystyle\sum_{k=0}^n\dfrac{5^k}{4^k+5^k}$$

noticing that $S_n + T_n = n+1$.
Any thoughts about this ? thanks.

Answer

As hinted by SmileyCraft, I cannot think of any simple closed form to express $S_n$.

My best guess is to use Big $\mathcal{O}$ notation. As you thought, $S_n = n+1 - T_n$ then write $T_n = \sum_{k=0}^n u_k$ with $u_k = \frac{1}{1+\left(\frac{4}{5}\right)^k}\underset{k\to\infty}{=}1 + \mathcal{O}\left(\frac{4}{5}\right)^k$. Now $\left(\frac{4}{5}\right)^k$ is a positive real sequence thus sommable in Big $\mathcal{O}$ notation.

This yields:
$$S_n \underset{n\to \infty}{=} n+1 - \left[(n+1) + \mathcal{O}(1) \right] = \mathcal{O}(1)$$

since $\sum_{k=0}^n \left(\frac{4}{5}\right)^k \underset{n\to \infty}{=} \mathcal{O}(1)$.

Then $\frac{S_n}{n+1} = \mathcal{O}(\frac{1}{n+1})$.

Note that it is very similar and perfectly equivalent to what SmileyCraft does but it does give you an expression of $S_n$ (though trivial) depending on $n$.