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Math Help - Midterm practice problem questions

  1. #1
    Junior Member
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    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.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by 3deltat View Post
    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

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

    Or note that If we use the integral test we get

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