Value of $S_{10}+I_{10}$

Question

Let $S_n=\sum_{i=1}^n \frac{1}{k}$ and$$I_n=\int_{1}^{n} \frac{x-[x]}{x^2} dx$$ then what is the value of $S_{10}+I_{10}$ ?
$S_{10}=1+1/2+1/3+.....+1/10$ and $$I_{10}=\int_{1}^{10} \frac{x-[x]}{x^2} dx$$
$$=ln 10-\int_{1}^{10} \frac{[x]}{x^2} dx$$
But I am stuck here.How to proceed further?

Answer 1

You just have to use $$ \begin{align*} \int_{1}^{10} \frac{[x]}{x^2} dx & =\int_1^2\frac{1}{x^2} dx+2\int_2^3\frac{1}{x^2} dx +3\int_3^4\frac{1}{x^2} dx+\cdots +9\int_9^{10}\frac{1}{x^2} dx \\ & =\sum_{k=1}^{9} k \int_k^{k+1}\frac{1}{x^2} dx \\ & = \sum_{k=1}^{9} k [-\tfrac{1}{u}]_k^{k+1} \\ &= \sum_{k=1}^{9} k\big( \tfrac{1}{k} -\tfrac{1}{k+1}\big) \\ &=\sum_{k=1}^{9} 1 -(1 -\tfrac{1}{k+1})\\ &= \sum_{k'=2}^{10}\tfrac{1}{k'} = S_{10}-1 \end{align*} $$

Since $[x]=k$ on each interval $[k,k+1)$ ; you see that it is easy to generalize for other indices values than 10... (and $S_n+I_n=S_N+(\ln(n)-S_n+1)=\ln(n)+1)$.

Answer 2

Another similar approach is to use the Abel summation formula $$\sum_{k=1}^{n}\frac{1}{k}=\sum_{k=1}^{n}1\cdot\frac{1}{k}=1+\int_{1}^{n}\frac{\left[t\right]}{t^{2}}dt.$$