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.

• I don't know how to express the sum in terms of $n$, but I am a little confused. Don't we have $\lim_{n\to\infty}S_n=\sum_{k=0}^\infty\frac{4^k}{4^k+5^k}\leq\sum_{k=0}^\infty(\frac45)^k=\frac1{1-\frac45}=5$ and hence $\lim_{n\to\infty}\frac{S_n}{n+1}=0$? – SmileyCraft Dec 16 '18 at 12:49
• Also, Wolfram Alpha gives a closed form involving the $q$-digamma function, which I've never heared of, and I doubt you have. What makes you think there is a closed form of the sum? – SmileyCraft Dec 16 '18 at 12:54
• Thanks, the first part of the question in a paper (which I doubt is wrong) asks to express $S_n$ in terms of $n$ before computing the limit of the quotient $\dfrac{S_n}{n+1}$ – Oussama Sarih Dec 16 '18 at 13:00
• @OussamaSarih, can you link to the paper, or reproduce verbatim what it says about expressing $S_n$? Given that the first three $S_n$'s are $1/2$, $17/18$, and $985/738$, I'd be a little surprised to see any kind of simple closed formula. (Even well written papers make occasional mistakes.) – Barry Cipra Dec 16 '18 at 13:04
• It's a homework sheet in french for highschool, I'll post a screenshot as soon as I get from one from the students who asked me to solve it. – Oussama Sarih Dec 16 '18 at 15:39

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 $$n$$.