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?     
 A: $$u=1+2x\Longrightarrow du=2dx\Longrightarrow dx=\frac{1}{2}du$$
Remember, not only you substitute the variable and nothing more: you also have to change the $\,dx\,$ and the integral's limits: 
$$u=1+2x\,\,,\,\text{so}\,\, x=1\Longrightarrow u=1+2\cdot 1 =3$$
A: 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.
A: Let $1 + 2x = u$ then $du = 2 dx \implies du = {1 \over 2} dx$
$x $ goes from $1$ to $\infty$, since we are using new variable $u$ here, it would have different bound. When $x \rightarrow \infty$, $ u \rightarrow \infty$ and when $x = 1$, $ u = 1 + 2 \cdot ( 1) = 3 $, so $u$ goes from $3$ to $\infty$
