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what is the value of $n$ so that the partial sum $S_n$ of the harmonic series ie $\sum _1 ^n \frac{1}{x} \geq y$ where $y$ is a natural number. I started with $2,3$ and got answers for $n$ as $4,11$. But for $5$ its $83$. Seeing this I have a strong belief that there's a formula which can be used to calculate $n$ quickly. Also if there is formula can it be used for irrational ,rational numbers. Note i tried integration but it obviously doesn't work for small $y$.

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From this answer, to compute the smallest $n$ such that $H_n$ exceeds an integer $N$,

$$log\left(n+\frac{1}{2}\right)+\gamma>N$$ $$log\left(n+\frac{1}{2}\right)>N-\gamma$$ $$n+\frac{1}{2}>e^{N-\gamma}$$ $$n>e^{N-\gamma}-\frac{1}{2}$$

so

$$n=\lceil e^{N-\gamma}-\frac{1}{2}\rceil$$

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The $n$-th harmonic number $H_n$ behaves like $\log(n)+\gamma$, where $\gamma$ is the Euler-Mascheroni constant. It follows that $H_n=N$ has a solution close to $n=\exp\left(N-\gamma\right)$.

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It can be shown by induction that:

$$\sum_{k=1}^{2^n} \frac1{k} > \frac{n}{2}$$

Hence

$$\sum_{k=1}^{4^n}\frac1{k} > n$$

This isn't optimal but nice.

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