Question about the integral test

[vande060]

Homework Statement

(∞, n=1) ∑ ne^-n

Homework Equations

The Attempt at a Solution

(∞, n=1) ∑ ne^-n

lim_{t--> ∞} ∫ xe^-x dx from 1 to t

here i tried to do integration by parts

u = x
du=dx
dv= e^-x dx
v= -e^-x

not sure where to take limits at this point, is it like this

lim_{t--> ∞} -xe^-x (from 1 to t) + lim_{t--> ∞} ∫ e^-xdx from 1 to ti can of course go further with the integral, but how does this look so far, i can't really make sense out of it :/

 

[Dick]

You forgot to say what you are actually trying to do. Are you trying to sum the series or just show it converges?

 

[jhae2.718]

When evaluating a definite integral using integration by parts, evaluate it as if it were an indefinite integral and then substitute your limits over the entire integrand.

 

[vande060]

You forgot to say what you are actually trying to do. Are you trying to sum the series or just show it converges?

oops use the integral test to determine whether the series is convergent or divergent

 

[vande060]

When evaluating a definite integral using integration by parts, evaluate it as if it were an indefinite integral and then substitute your limits over the entire integrand.

so ill try just taking the integral by parts, then apply the limit with where i end up over the entire integrand in that process like this?

I = ∫ xe^-x dx from 1 to t

u = x
du=dx
dv= e^-x dx
v= -e^-xI = -xe^-x (from 1 to t) + ∫ e^-x dx (from 1 to t)

I = -xe^-x (from 1 to t) -e^-x + C (from 1 to t)

I = lim _{t --> ∞} -xe^-x (from 1 to t) - lim _{t --> ∞} e^-x + C (from 1 to t)

 

[Dick]

oops use the integral test to determine whether the series is convergent or divergent

Ok, then. You know the integral converges, right? Now you just have to show lim x->infinity of x*e^(-x) exists.

 

[jhae2.718]

Basically, yes. Substitute your upper and lower limits into the integrand, and then take \lim_{t \to \infty}.

 

[Dick]

Checking convergence by doing a much simpler ratio test wouldn't hurt you a bit either. I don't know why you are doing an integral test on this one. Was that your choice?

 

[vande060]

Checking convergence by doing a much simpler ratio test wouldn't hurt you a bit either. I don't know why you are doing an integral test on this one. Was that your choice?

no choice, have not learned ration test yet, that's next chapter

Basically, yes. Substitute your upper and lower limits into the integrand, and then take lim

okedokie, so to wrap this up:

I = lim _{t --> ∞} -xe^-x (from 1 to t) - lim _{t --> ∞} e<sup>-x[/SUBP] + C (from 1 to t)

Ill break it into 2 pieces:

lim _{t --> ∞} -xe-x (from 1 to t) = -(0-1/e) = 1/e

- lim _{t --> ∞} e-x[/SUBP] + C (from 1 to t) = ( 0 - 1/e) = -1/e

so the answer is that the series converges on 2/e</sup>

 

[jhae2.718]

no choice, have not learned ration test yet, that's next chapter

The ratio test states that, for a series \sum_{n=0}^\infty a_n, for L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|:
if L > 1, the series diverges.
if L = 1 or DNE, the test fails
if L < 1, the series converges absolutely.

okedokie, so to wrap this up:

I = lim _{t --> ∞} -xe^-x (from 1 to t) - lim _{t --> ∞} e_-x + C (from 1 to t)

Yes, so substitute the limits in, take the limit as t > infinity, and what do you get? Note that the integral is definite so there is no need for C.

 

[jhae2.718]

Ill break it into 2 peices, starting with the second piece

- lim _{t --> ∞} e^{-x[/SUBP] + C (from 1 to t) this should go to 0}

^{You'll want to substitute in the limits of integration first, so you have:
\lim_{t \to \infty} \left[\left(-te^{-t}-e^{-t}\right)-\left(-e^{-1}-e^{-1}\right)\right]}

 

[jhae2.718]

so the answer is that the series converges on 2/e

Yes.