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The infinite sum $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges. However, it is possible to find bounds from some $n$ to another integer $n$. Wolfram alpha is able to give a decimal approximation of the sum, so my question is, how can I find an approximate range, or bound for this harmonic series:

$$\sum_{n=10}^{100}\frac{1}{n}$$

What is an intuitive way to look at this?

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    $\begingroup$ Trapezoidal/midpoint approximations yield a rather precise $$\frac 1 2(\frac 1 i+\frac 1 j)+\log(\frac j i)\le\sum_{n=i}^j\frac 1 n\le \frac 1 i+\frac 1 j+\log(\frac{j-1/2}{i+1/2})$$ for $j>i$. $\endgroup$
    – A.S.
    Dec 23 '15 at 1:55
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In a graphical sense, your sum will be bounded by the two functions $\frac{1}{x}$ and $\frac{1}{x-1}$. These curves are your bounds that you want. Now to find approximations, all you are left to do is integrate: $$\int_{10}^{100}\frac{1}{x}<\sum_{n=10}^{100}\frac{1}{n} <\int_{10}^{100}\frac{1}{x-1}$$ $$\implies\log{\frac{100}{10}}<\sum_{n=10}^{100}\frac{1}{n}< \log{\frac{99}{9}}$$ $$\therefore 2.30<\sum_{n=10}^{100}\frac{1}{n}<2.40$$

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Another approximation can be found using the well known estimation of the $n$-th harmonic number $$H_{n}=\sum_{k=1}^{n}\frac{1}{k}=\log\left(n\right)+\gamma+r_{k}$$ where $r_{k}\sim\frac{1}{2k}$ and $\gamma$ is the Euler-Mascheroni constant. It is possible to show the bounds $$\log\left(n\right)+\gamma+\frac{1}{2\left(n+1\right)}<H_{n}<\log\left(n\right)+\gamma+\frac{1}{2n}$$ hence $$\sum_{k=10}^{100}\frac{1}{k}=\sum_{k=1}^{100}\frac{1}{k}-\sum_{k=1}^{9}\frac{1}{k}>\log\left(100\right)-\log\left(9\right)-\frac{46}{909}\approx2.35734$$ and $$\sum_{k=10}^{100}\frac{1}{k}=\sum_{k=1}^{100}\frac{1}{k}-\sum_{k=1}^{9}\frac{1}{k}<\log\left(100\right)-\log\left(9\right)-\frac{9}{200}\approx2.36294$$

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