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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.

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    $\begingroup$ 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 $\lim_{n\to\infty}\frac{S_n}{n+1}=0$? $\endgroup$ – SmileyCraft Dec 16 '18 at 12:49
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    $\begingroup$ 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? $\endgroup$ – SmileyCraft Dec 16 '18 at 12:54
  • $\begingroup$ 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 $\dfrac{S_n}{n+1}$ $\endgroup$ – Oussama Sarih Dec 16 '18 at 13:00
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    $\begingroup$ @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.) $\endgroup$ – Barry Cipra Dec 16 '18 at 13:04
  • $\begingroup$ 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. $\endgroup$ – Oussama Sarih Dec 16 '18 at 15:39
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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 $n$.

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