# Thread: Midterm practice problem questions

1. ## Midterm practice problem questions

Hello!
I am going over some work for my midterm and some questions came up, and i was wondering if someone could help!

1) The integral of 1/((x^2)(x^2+9)^(1/2)
Clearly this is trig substitution...
x=3tangent
dx=3sec^2
(X^2+9)^1/2 = 3sec

Integral then becomes (1/27)(1/cos)/((sin^3)/(cos^3))

Integral turns into (Cot^2)(csc^2)

(I do not have the theta sign because i cant find it on my keyboard. When assume it is there after the tangent, cot, etc).

What do I do from there?

2) Is (n+1)! / n! = (n+1)n! / n!

3) The volume of the region, rotated about y=1, bounded by y=x^2 y=1 using cylindrical shells.

=2pi x Integral of ( ? )(y-y^2)dy

4) Compute the sum of the series as n -> infinity, starting at one, of ln (n/n+1). I know it is convergent, since the limit of n -> infinity of an goes to zero.

It is going to zero...

How do i find the sum?

Thank you all for your help! If I was unclear in the way I asked a question please let me know. This will help a lot.

2. Originally Posted by 3deltat
Hello!

4) Compute the sum of the series as n -> infinity, starting at one, of ln (n/n+1). I know it is convergent, since the limit of n -> infinity of an goes to zero.

It is going to zero...

How do i find the sum?
I think this is divergent

\displaystyle \begin{aligned}\sum_{n=1}^{\infty}\ln\left(\frac{n }{n+1}\right)&=\sum_{n=1}^{\infty}\bigg[\ln(n)-\ln(n+1)\bigg]\\ &=\left[\ln(1)-\ln(2)\right]+\left[\ln(2)-\ln(3)\right]+\left[\ln(3)-\ln(4)\right]+\cdots\\ &=\lim_{n\to\infty}-\ln(n+1)\\ &=-\infty \end{aligned}

Or note that If we use the integral test we get

\displaystyle \begin{aligned}\int_1^{\infty}\ln\left(\frac{x}{x+ 1}\right)dx&=\int_1^{\infty}\bigg[\ln(x)-\ln(x+1)\bigg]dx\\ &=\bigg[x\ln(x)-x-(x+1)\ln(x+1)+x+1\bigg]\bigg|_{x=1}^{x=\infty}\\ &=-\infty \end{aligned}