Solving improper integrals and u-substitution on infinite series convergent tests This is the question:
Use the integral test to determine the convergence of $\sum_{n=1}^{\infty}\frac{1}{1+2n}$.
I started by writing:
$$\int_1^\infty\frac{1}{1+2x}dx=\lim_{a \rightarrow \infty}\left(\int_1^a\frac{1}{1+2x}dx\right)$$
I then decided to use u-substitution with $u=1+2n$ to solve the improper integral. I got the answer wrong and resorted to my answer book and this is where they went after setting $u=1+2n$:
$$\lim_{a \rightarrow \infty}\left(\frac{1}{2}\int_{3}^{1+2a}\frac{1}{u}du\right)$$
And the answer goes on...
What I can't figure out is where the $\frac{1}{2}$ came from when the u-substitution began and also, why the lower bound of the integral was changed to 3.
Can someone tell me?     

$$u=1+2x\Longrightarrow du=2dx\Longrightarrow dx=\frac{1}{2}du$$
Remember, not only you substitute the variable and nothing more: you also have to change the $\,dx\,$ and the integral's limits: 
$$u=1+2x\,\,,\,\text{so}\,\, x=1\Longrightarrow u=1+2\cdot 1 =3$$
