Calculate this sum $\sum_{x=1}^{{n}/{4}} \frac{1}{x}$ How can I calculate this sum
$$\sum_{x=1}^{{n}/{4}} \frac{1}{x}$$
for $n$ an odd integer?
Added later: For example, WolframAlpha outputs 4 - Pi/2 - Log[8] for HarmonicNumber[1/4], what are the steps that produce that? Can they be generalized for any odd number?
 A: The usual definition of the Harmonic numbers is:
$$H_n=\sum_{k=1}^n\frac1k$$
For example, $H_1=1$, $H_2=1+\frac12$,$H_3=1+\frac12+\frac13$, etc. However, this definition is only valid when $n$ is an integer.
When $n$ is not an integer, we use the alternate definition:
$$\sum_{k=1}^\infty\left(\frac1k-\frac1{k+n}\right)$$
For example, $H_2=\left(\frac11-\frac13\right)+\left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\left(\frac14-\frac16\right)+\dotsb=1+\frac12$, because everything other than $1$ and $\frac12$ cancels out. The great thing about this definition is that it works even when $n$ is not an integer!
You want to find:
$$H_{1/4}=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1/4}\right)\approx0.34976\dots$$
in closed form. This sum is usually done using techniques learned in calculus! However, there is a more elementary way of summing this.
(First, we need to know a few things:
$\displaystyle\sum_{k=1}^\infty\frac{(-1)^{k+1}}{2k-1}=1-\frac13+\frac15-\dotsb=\frac\pi4$
$\displaystyle\sum_{k=1}^\infty\frac{(-1)^{k+1}}k=1-\frac12+\frac13-\dotsb=\ln2$
If you've never seen these sums before, just trust me.)
Now:
\begin{align}
H_{1/4}=&\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1/4}\right)\\
=&\sum_{k=1}^\infty\left(\frac4{4k}+\frac{-4}{4k+1}\right)\\
=&\sum_{k=1}^\infty\left(\frac{0}{4k}+\frac{-2}{4k+1}+\frac{0}{4k+2}+\frac{2}{4k+3}\right)+\\
&\sum_{k=1}^\infty\left(\frac{2}{4k}+\frac{-2}{4k+1}+\frac{2}{4k+2}+\frac{-2}{4k+3}\right)+\\
&\sum_{k=1}^\infty\left(\frac{2}{4k}+\frac{0}{4k+1}+\frac{-2}{4k+2}+\frac{0}{4k+3}\right)
\end{align}
(You'll see why I split it up like that in a moment. To verify that they are equal, just add up the columns.)
The first sum is $-\frac25+\frac27-\frac29+\frac2{11}+\dotsb=-2(1-\frac13+\frac15-\dotsb)+2-\frac23=-\frac\pi2+\frac43.$
The second sum is $\frac24-\frac25+\frac26-\frac27+\dotsb=-2(1-\frac12+\frac13-\dotsb)+2-\frac22+\frac23=-2\ln2+\frac53.$
The third sum is $\frac24-\frac26+\frac28-\frac2{10}+\dotsb=\frac12-\frac13+\frac14-\frac15+\dotsb=\\-(1-\frac12+\frac13-\dotsb)+1=-\ln2+1.$
Add them all up and you get $H_{1/4}=4-\frac\pi2-3\ln2$.
A: The series could be defined by
\begin{align}
S_{m} = \sum_{n=1}^{\lfloor m/4 \rfloor} \frac{1}{n}
\end{align}
where $\lfloor x \rfloor$ represents the largest integer of $x$. Now, consider the series
\begin{align}
H_{n} = \sum_{k=1}^{n} \frac{1}{k}
\end{align}
which are known as the Harmonic numbers. From this it is seen that the series in question is given by
\begin{align}
S_{m} = H_{\lfloor m/4 \rfloor}.
\end{align}
Alternatively, since $\psi(x+1) + \gamma = H_{x}$ then $S_{m} = \gamma + \psi(\frac{m}{4} + 1)$, where $\gamma$ is the Euler-Mascheroni constant and $\psi(x+1)$ is the digamma function. 
Examples of values:
\begin{align}
S_{4} &= \sum_{n=1}^{\lfloor 4/4 \rfloor} \frac{1}{n} = 1 \\
S_{5} &= \sum_{n=1}^{\lfloor 5/4 \rfloor} \frac{1}{n} = \sum_{n=1}^{1} \frac{1}{n} = 1.
\end{align}
Now, making use of 
$$J_{m} = \sum_{k=1}^{m/4} \frac{1}{k} = \gamma + \psi\left( \frac{m+4}{4} \right)$$
then
\begin{align}
J_{1} = \gamma + \psi\left( \frac{5}{4} \right) = 4 + \gamma + \psi\left( \frac{1}{4} \right)
\end{align}
It is known that $\psi\left(\frac{1}{4} \right) = - \gamma - \frac{\pi}{2} - 3\ln 2$ which leads to the result
\begin{align}
J_{1} = 4 - \frac{\pi}{2} - 3 \ln 2. 
\end{align}
For the case of $S_{3}$ it is seen that
\begin{align}
J_{3} &= \gamma + \psi\left( \frac{7}{4} \right) = 4 + \gamma + \psi\left( \frac{3}{4} \right) = 4 + \frac{\pi}{2} - 3 \ln 2. 
\end{align}
A: As @Leucippus describes in his answer, your series is really a harmonic number. There is no way to compute Harmonic numbers exactly but there is a nice approximation.
$$ H_n \approx \ln n + \gamma + \frac{1}{2n} - \frac{1}{12n^2}$$ 
where $\gamma = 0.577215664901532860606512…$ is the Euler-Mascheroni constant.
A: Look here: http://en.wikipedia.org/wiki/Indefinite_sum
It turns out your sum is 
$$\psi(n/4+1)$$
where $\psi(x)$ is the digamma function
