# Math Help - series with integral test

1. ## series with integral test

$\sum_{n=3}^{\infty}\frac{n^2}{e^n}$

i have to tell if this series is convergent or divergent, so i figured i could try the integral test here. the problem is, once i get to the stage of substituting with:

$\int_3^{\infty} \ \frac{x^2}{e^x}\, dx$

i get a little bit lost because integration really isn't my strong point >_<

i realize that after i have integrated i need to put it in limit form and evaluate it, i just can't get the integration part done. please help?

2. I do not understand your "integral test"

You know that $\lim_{n \rightarrow +\infty} \frac{n^4}{e^n} = 0$

Therefore there exists N such that for every n > N, $\:\frac{n^4}{e^n} < 1$ or $\frac{n^2}{e^n} < \frac{1}{n^2}$

And you know that the series $\sum_{k=1}^{n} \frac{1}{n^2}$ converges

Another method is to calculate the limit of $\frac{u_{n+1}}{u_n}$

3. i just learned the integral test yesterday, it (apparently) says that if you have a positive, decreasing, & continuous function that you can replace your sequence with, you can use the improper integral of that, turn it into a limit, and if that converges, the sequence converges. if it diverges, the sequence diverges. (or so i was told)

edit:
& i don't really see why this is:

$\lim_{n \rightarrow +\infty} \frac{n^4}{e^n} = 0$

Therefore there exists N such that for every n > N, $\:\frac{n^4}{e^n} < 1$ or $\frac{n^2}{e^n} < \frac{1}{n^2}$

4. Originally Posted by buttonbear
$\sum_{n=3}^{\infty}\frac{n^2}{e^n}$

i have to tell if this series is convergent or divergent, so i figured i could try the integral test here. the problem is, once i get to the stage of substituting with:

$\int_3^{\infty} \ \frac{x^2}{e^x}\, dx$
This becomes a whole lot easier if you write it as $\int_3^\infty x^2e^{-x} dx$. Then integrate by parts, twice.

i get a little bit lost because integration really isn't my strong point >_<

i realize that after i have integrated i need to put it in limit form and evaluate it, i just can't get the integration part done. please help?

5. so, trying to integrate this by parts:
$\int_3^\infty x^2e^{-x} dx$

i first said
$\ u= x^2, dv=e^-xdx$
$\ du= 2xdx, v=-e^-x$

then i got
-e^-x*x^2- integral of -e^-x*2xdx
so i did parts again, with

$\ u= 2x, dv=-e^-xdx$
$\ du= 2dx, v=e^-x$

=-e^-x*x^2+e^-x*2x+ integral of 2e^-xdx
=-e^-x*x^2+e^-x*2x-2xe^-x

i really feel like i'm doing something wrong,
can someone help?

p.s.- i apologize for the last few lines..LaTex was giving me some trouble >.<

6. Originally Posted by buttonbear
i just learned the integral test yesterday, it (apparently) says that if you have a positive, decreasing, & continuous function that you can replace your sequence with, you can use the improper integral of that, turn it into a limit, and if that converges, the sequence converges. if it diverges, the sequence diverges. (or so i was told)
You are right, I forgot this
I apologize

7. oh, that's okay
i'm actually not that sure of myself at all when it comes to these kinds of problems, anyway

but, do you have any idea where i went wrong in my integration above?

8. You made a mistake at the last step when integrating 2e^-x

$\int 2e^{-x}dx = -2\:e^{-x}$

Finally

$\int x^2e^{-x}dx = (-x^2-2x-2)\:e^{-x}$

9. oh, okay, i thought i had messed up a lot more than that >.<

although, i'm not really sure why there's a negative sign in that middle term

10. There is one

Deriving $(-x^2-2x-2)\:e^{-x}$ gives $(-2x-2)\:e^{-x}-(-x^2-2x-2)\:e^{-x}=x^2\:e^{-x}$

While deriving $(-x^2+2x-2)\:e^{-x}$ gives $(-2x+2)\:e^{-x}-(-x^2+2x-2)\:e^{-x}=(x^2-4x+4)\:e^{-x}$

11. oh, wow, that is a big difference then!
i see how it derives to a different thing, but, i don't know how to get it
thanks though!

also, getting back to the bigger question
i plugged the integrated version back into the limit from 3 to infinity

and i ended up with something like..
$(-infinity^2- 2(infinity)- 2)e^-infinity- (-3^2- 2(3)- 2)e^-2$

i figured the first part of that was going to go to zero because of the e^-infinity

does this mean that the integral converges, & thus, the series converges?

(sorry again, LaTex hates me)

12. Taking $(-x^2-2x-2)\:e^{-x}$ in $+\infty$ gives 0 because $\lim_{x \rightarrow +\infty}x^k e^{-x}=0$ for every k real

Therefore the integral converges to $(3^2+2\cdot 3+2)\:e^{-3}=17 \:e^{-3}$ and thus the series converge

13. okay, well i must have made a mistake in my integrating when i missed the negative sign..

thank you so much

14. $\int x^2\:e^{-x} dx$

$\ u= x^2, dv=e^{-x}dx$
$\ du= 2xdx, v=-e^{-x}$

$\int x^2e^{-x} dx = -x^2\:e^{-x} - \int -2x\:e^{-x} dx=-x^2\:e^{-x} + \int 2x\:e^{-x} dx$

$\ u= 2x, dv=e^{-x}dx$
$\ du= 2dx, v=-e^{-x}$

$\int 2x\:e^{-x} dx = -2x\:e^{-x} - \int -2\:e^{-x} dx = -2x\:e^{-x} + \int 2\:e^{-x} dx = -2x\:e^{-x} - 2\:e^{-x}$