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Consider a function $R(n)$ defined across the naturals as $$R(n) = n \cdot 10^5$$ Consider another function $F(n)$ defined recursively across the naturals as $$F(n) = 1 + F(n - 1)\cdot \frac{n}{n-1}, \ n \ge 2, \ F(1) = 1$$ I am initially trying to prove that eventually, $F(n)$ will be greater than $R(n)$, as well as determined a closed form expression for $F(n)$. Clearly, $$\frac{R(n)}{n} = 10^5 \\ \frac{F(n)}{n} - \frac{F(n-1)}{n-1} = \frac{1}{n}$$ Then, we can consider the harmonic series. We'll get a telescoping series, and after some simplifications, we get: $$\sum_{i=1}^n \frac{1}{i} = \frac{F(n)}{n}$$ Since we know the harmonic series diverges to infinity as $n \rightarrow \infty$, then $\frac{F(n)}{n}$ also diverges to infinity as $n \rightarrow \infty$.

$\frac{R(n)}{n}$ is constant as $n$ grows, but $\frac{F(n)}{n}$ grows without bound. $$\frac{F(n)}{n} \gt \frac{R(n)}{n} \iff F(n) \gt R(n)$$ This solves the first problem. I'm stuck on how to use any of this information to determine a closed form expression for $F(n)$. Can we also find an analytical continuation of $F(n)$ to the reals?

Bill Dubuque
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vestieee
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    Better not use the notation $H(n)$ here, since it usually denotes harmonic numbers, which are closely related to your problem. Imo, you don't need to introduce any notation for $10^5n$. – Anne Bauval Dec 08 '24 at 15:31
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    Noted, will change that in the question. Thank you! – vestieee Dec 08 '24 at 15:32

2 Answers2

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Actually you can show by sum-integral comparison that :

$$ F_n \sim n\ln(n) $$

using that $\sum_{i\in\mathbb{N}^*}1/i \sim \ln(n)$

So that $F_n$ will exceed $H_n$ for $n$ large enough as you state.

But $F_n$ cannot have closed form : see here https://math.stackexchange.com/a/52579/763728

EDX
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The sequence $G(n):=\frac{F(n)}n$ satisfies $$G(1)=1,\quad G(n)=\frac1n+G(n-1)$$ hence coincides with the sequence of harmonic numbers, so $$F(n)=n\sum_{k=1}^n\frac1k.$$

Anne Bauval
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