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?
On
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 $$\sum_{k=10}^{100}\frac{1}{k}=\sum_{k=1}^{100}\frac{1}{k}-\sum_{k=1}^{9}\frac{1}{k}<\log\left(100\right)-\log\left(9\right)-\frac{9}{200}\approx2.36294$$
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: $$\int_{10}^{100}\frac{1}{x}<\sum_{n=10}^{100}\frac{1}{n} <\int_{10}^{100}\frac{1}{x-1}$$ $$\implies\log{\frac{100}{10}}<\sum_{n=10}^{100}\frac{1}{n}< \log{\frac{99}{9}}$$ $$\therefore 2.30<\sum_{n=10}^{100}\frac{1}{n}<2.40$$