Why does $$\sum_{i=10^x}^{10^{(x+1)}-1} \frac{1}{i}$$ seem to tend towards $\ln(10)$ as $x $ increases?
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2Trusting that $r=i$, compare your sum to the related definite integral. – lulu May 01 '19 at 13:16
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3You could try and use the fact that $H_n = \ln n + \gamma + o(1)$ as $n\to\infty$, where $H_n := \sum\limits_{i=1}^{n}\frac{1}{i}$ and $\gamma$ is the Euler-Mascheroni constant. – Minus One-Twelfth May 01 '19 at 13:22
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$$\sum_{i=10^x}^{10^{(x+1)}-1} \frac{1}{i}=H_{10^{x+1}-1}-H_{10^x-1}=\log10^{x+1}-\log10^x+o(1)=\log10+o(1)$$ where $H_n$ denotes the n-th harmonic number
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