Let $S_n = \sum_{k=1}^n \frac{1}{k}$ and $I_n = \int_1^n \frac{x - [x]}{x^2} dx$. Then, $S_{10} + I_{10}$ is equal to
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:
The antiderivative is $\ln|x| + \frac{k}{x}$. Evaluating this from $k$ to $k+1$ yields:
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:
The second sum simplifies to:
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
Important Questions from Series
- In the sequence 6, 9, 14, $x$, 30, 41, a possible value of $x$ is
Let $a_0 = 0$ and define $a_n = \frac{1}{2}(1 + a_{n-1})$ for all positive integers $n \ge 1$.
The least value of $n$ for which $|1 - a_n| < \frac{1}{2^{10}}$ is __________.
(Answer in integer)
- Calculate the reciprocal of the coefficient of $z^3$ in the Taylor series expansion of the function $f(z) = \sin(z)$ around $z = 0$. (Provide the answer as an integer.)
- Let $a_1 = 1$ and $a_n = a_{n-1} + 4$, $n \ge 2$. Then,
$\lim_{n\to\infty} \left[\frac{1}{a_1a_2} + \frac{1}{a_2a_3} + \dots + \frac{1}{a_{n-1}a_n}\right]$
is equal to ________ - Let $S(x) = a_0 + \sum_{n=1}^\infty(a_n \cos (n x) + b_n \sin (n x))$ be the Fourier series of the$2 \pi$ periodic function defined by $f(x) = x^2 + 4 \sin (x) \cos(x)$, $-\pi \le x \le \pi$. Then
$|\sum_{n=0}^\infty a_n - \sum_{n=1}^\infty b_n|$
is equal to ________
