<s>[INST] Value of $S_{10}+I_{10}$ 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?
 [/INST] 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 $S_n+I_n=S_N+(\ln(n)-S_n+1)=\ln(n)+1)$.
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