Harmonic Series Calculation

The harmonic series is defined as $S_n = \sum_{k=1}^n \frac{1}{k}$.

For $n=10$, $S_{10} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{10}$.

Integral Calculation

The integral is defined as $I_n = \int_1^n \frac{x - [x]}{x^2} dx$. The term $x - [x]$ represents the fractional part of $x$, often denoted as $\{x\}$.

We can split the integral into intervals:

$I_n = \sum_{k=1}^{n-1} \int_k^{k+1} \frac{x - [x]}{x^2} dx$

Within the interval $[k, k+1)$, the floor function $[x]$ is equal to $k$. So, $x - [x] = x - k$.

The integral over one interval becomes:

$\int_k^{k+1} \frac{x-k}{x^2} dx = \int_k^{k+1} \left(\frac{1}{x} - \frac{k}{x^2}\right) dx$

The antiderivative is $\ln|x| + \frac{k}{x}$. Evaluating this from $k$ to $k+1$ yields:

$\left[ \ln x + \frac{k}{x} \right]_k^{k+1} = \left(\ln(k+1) + \frac{k}{k+1}\right) - \left(\ln k + \frac{k}{k}\right) = \ln\left(\frac{k+1}{k}\right) + \frac{k}{k+1} - 1$

Summing these results from $k=1$ to $n-1$:

$I_n = \sum_{k=1}^{n-1} \left( \ln\left(\frac{k+1}{k}\right) + \frac{k}{k+1} - 1 \right) = \sum_{k=1}^{n-1} \ln\left(\frac{k+1}{k}\right) + \sum_{k=1}^{n-1} \left(\frac{k}{k+1} - 1\right)

The first sum is a telescoping series:

$\sum_{k=1}^{n-1} \ln\left(\frac{k+1}{k}\right) = \ln n$

The second sum simplifies to:

$\sum_{k=1}^{n-1} \left(\frac{k - (k+1)}{k+1}\right) = \sum_{k=1}^{n-1} \frac{-1}{k+1} = -\left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)$

Recognizing that $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} = S_n - 1$, the second sum is $-(S_n - 1)$.

Therefore, the integral simplifies to:

$I_n = \ln n - (S_n - 1)$

Final Sum Calculation

We need to find $S_{10} + I_{10}$. Substitute $n=10$ into the expression for $I_n$:

$I_{10} = \ln 10 - (S_{10} - 1)$

Now, add $S_{10}$:

$S_{10} + I_{10} = S_{10} + (\ln 10 - S_{10} + 1) = S_{10} + \ln 10 - S_{10} + 1 = \ln 10 + 1