Is there a closed form to the sum $\sum_{k=1}^n \frac{k}{2k - 1}$ ? If there is, what would that be? I know that the sum of the first n natural numbers is $\frac{n(n + 1)}{2}$ and that the sum of the first n odd numbers is $n^2$, so I wondered what would happen if you had the ratio of the $n^{th}$ natural number to the $n^{th}$ odd number and took the summation of that.
FYI, I haven't hit college yet so my knowledge of sums is basic, I pretty much know nothing that's more advanced than the sums I've mentioned. I'm hoping for an explanation of how to find the closed form of $\sum_{k=1}^n \frac{k}{2k - 1}$ or a way to be certain that there is none.
 A: Starting with
$$\frac{k}{2k-1} = \frac{1}{2} \left( \frac{2k-1}{2k-1} + \frac{1}{2k-1} \right)$$
the sum is
$$S_n = \frac{n}{2} + \frac{1}{2} \sum_{k=1}^n \frac{1}{2k-1}$$
As noted in the comments, the harmonic numbers are defined to be
$$H_p= \sum_{k=1}^p \frac{1}{k}.$$
Here $p=2n-1$,
so
$$\begin{aligned}
H_{2n-1} &= 1 + \frac{1}{2}+ \frac{1}{3} + \cdots + \frac{1}{2n-1} \\&=\left(1+\frac{1}{3}+\cdots+\frac{1}{2n-1} \right)\,+\, \left( \frac{1}{2}+\frac{1}{4} + \cdots+\frac{1}{2n-2} \right)\\
&= \left( 1+\frac{1}{3}+\cdots+\frac{1}{2n-1}\right) \,+\,\frac{1}{2} H_{n-1}
\end{aligned} $$
and $$S_n = \frac{n}{2} + \frac{1}{2} \left(H_{2n-1}-\frac{1}{2}H_{n-1}\right).$$
A: If you make it more general
$$S_n=\sum_{k=1}^n \frac{k}{ak +b}=\frac n a +\frac{b }{a^2}\left(H_{\frac{b}{a}}-H_{n+\frac{b}{a}}\right)\qquad \text{with} \quad a+b >0$$ where appear harmonic numbers.
If $n$ is large
$$S_n=\frac n a +\frac{b }{a^2} \left(H_{\frac{b}{a}}-\log (n)-\gamma
   -\frac{a+2 b}{2 a n}+O\left(\frac{1}{n^2}\right
   )\right)$$
