# 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 $u=1+2n$:

$$\lim_{a \rightarrow \infty}\left(\frac{1}{2}\int_{3}^{1+2a}\frac{1}{u}du\right)$$

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?

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$$u=1+2x\Longrightarrow du=2dx\Longrightarrow dx=\frac{1}{2}du$$

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

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Oh yes, I forgot that you have to change the limits of integration when u-substituting to the u-substitution limits of integration. – ODP Aug 28 '12 at 16:07

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:

$$\lim_{a\to\infty}{\left(\frac{1}{2}\int_{3}^{1+2a}\frac{1}{u}\:du\right)}$$

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 $\infty$