Convergence of Improper Integral and Series with Logarithmic Functions

[cal.queen92]

Homework Statement

The problem is divided into two sections:

a) does the improper integral: 2ln(x)/x^7 (from 1 to infinity) Converge or diverge? If it converges, to what value?

b) Determine whether the series: sigma n=1 to infinity (2ln(n)/n^7) converges or diverges.

Homework Equations

Integration by parts?

The Attempt at a Solution

For the first part, I made the limit as c--> infinity, and took out the 2, then I simply integrated by parts where:

u = ln(x) du= 1/x dx dv= x^7 dx v = (x^8)/8

and ended up with: I = 1/4 lim c--> inf (x^8*ln(x) + (x^8)/8) from 1 to c

When I work it out, I get it Diverges, as the end result is infinity... But it's supposed to converge...


As for the second part, I assumed that they were similar where I could use the integral series test (ending up with the same result as the first part) to get the answer, but again, the answer lead to convergence...

Am I using the correct techniques?

Thanks!

 

[SammyS]

... The problem is divided into two sections:

a) does the improper integral: 2ln(x)/x^7 (from 1 to infinity) converge or diverge? If it converges, to what value?
...

For the first part, I made the limit as c--> infinity, and took out the 2, then I simply integrated by parts where:

u = ln(x) du= 1/x dx dv= x^7 dx v = (x^8)/8

and ended up with: I = 1/4 lim c--> inf (x^8*ln(x) + (x^8)/8) from 1 to c

When I work it out, I get it Diverges, as the end result is infinity... But it's supposed to converge...As for the second part, I assumed that they were similar where I could use the integral series test (ending up with the same result as the first part) to get the answer, but again, the answer lead to convergence...

Am I using the correct techniques?

Thanks!

How did you integrate 2ln(x)/x⁷ ? That's the same as 2(x^-7)ln(x)

I would expect the anti-derivative to have x^-6 in it, not x⁸ !

 

[cal.queen92]

Thank you! That was it, didn't take the proper dv -- took x^7 as oppose to 1/x^7 giving x^-7.

Perfect!