Calculus problem (integral test)

[mooshupork34]

If anyone could explain how the following problem is done, it would be greatly appreciated!

Use the integral test to determine whether the series converges or diverges:

sigma starting at j=1 and going to infinity of:
je^-j

So far, I integrated the expression xe^-x (I just replaced j with x here) and got -xe^-x - e^-x. I'm confused as to how to find the limit, though.

 

[tkhunny]

Hint: \(\displaystyle \L\;e^{-x}\) gets very small very quickly.

Hint: \(\displaystyle \L\;xe^{-x}\) gets very small only a little more slowly.

 

[soroban]

Hello, mooshupork34!

Use the integral test to determine whether the series converges or diverges:

. . \(\displaystyle \L\sum_{j=1}^{\infty} je^{-j}\)

So far, I integrated \(\displaystyle xe^{-x}\) and got:\(\displaystyle \,-xe^{-x}\,-\,e^{-x}\)

So we have: \(\displaystyle \L\:-xe^{-x}\,-\,e^{-x} \;=\;-e^x(x\,+\,1) \;=\;\left\,-\frac{x\,+\,1}{e^x}\right]^{\infty}_1\)

Then: \(\displaystyle \L\:\lim_{x\to\infty}\left(-\frac{x\,+\,1}{e^x}\right) \:-\:\left(-\frac{1\,+\,1}{e^1}\right)\)

. . and apply L'Hopital to that limit . . .