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?     

You have $\int_{1}^{\infty}{\frac{1}{1+2n}\:dn}=\lim_{a\to\infty}\left(\int_{1}^{a}{\frac{1}{1+2n}\:dn}\right)$. Using the substitution you mentioned $u=1+2n$, we have our lower bound as $u(1)=1+2(1)=3$ and our upper bound as $u(a)=1+2(a)=1+2a$.
We also have $\frac{du}{dn}=2\implies dn=\frac{du}{2}$, therefore we have:
$$\int_{3}^{1+2a}{\frac{1}{u}\frac{du}{2}}=\frac{1}{2}\int_{3}^{1+2a}{\frac{1}{u}\:du}$$
Taking our previous limit, we have:
$$\lim_{a\to\infty}{\left(\frac{1}{2}\int_{3}^{1+2a}\frac{1}{u}\:du\right)}$$
Which is what you have in your answer book.
