# Thread: Finding bounds for a series

1. ## Finding bounds for a series

Use the integral test to find upper and lower bounds on the limit of the series:

$\displaystyle \sum_{j=1}^{\infty} je^{-j}$

So do I first turn this series into an integral? So this would then be...

$\displaystyle \int_{1}^{\infty} je^{-j}$

Is this right so far? If so, I'm stuck right here.

2. Originally Posted by larson
Use the integral test to find upper and lower bounds on the limit of the series:

$\displaystyle \sum_{j=1}^{\infty} je^{-j}$

So do I first turn this series into an integral? So this would then be...

$\displaystyle \int_{1}^{\infty} je^{-j}$

Is this right so far? If so, I'm stuck right here.
integral test is way too much trouble. but since they asked for it...

yes, that is ok. do integration by parts on this guy. let $\displaystyle u = j$ and $\displaystyle dv = e^{-j}$

3. Originally Posted by Jhevon
integral test is way too much trouble. but since they asked for it...

yes, that is ok. do integration by parts on this guy. let $\displaystyle u = j$ and $\displaystyle dv = e^{-j}$
So... $\displaystyle u = j ... \frac {du}{dx} = 1 ... du = dx$

Then... $\displaystyle dv = e^{-j}dx ... \frac {dv}{dx} = e^{-j} ... v = \frac {e^{-j}}{-1}$

Now it gets messy...

$\displaystyle \int{u}dv=uv-\int{v\cdot{du}}$

$\displaystyle j\frac {e^{-j}}{-1} - \int{\frac {e^{-j}}{-1}dx}$

Is this right as of now?

4. Originally Posted by larson
So... $\displaystyle u = j ... \frac {du}{dx} = 1 ... du = dx$

Then... $\displaystyle dv = e^{-j}dx ... \frac {dv}{dx} = e^{-j} ... v = \frac {e^{-j}}{-1}$

Now it gets messy...

$\displaystyle \int{u}dv=uv-\int{v\cdot{du}}$

$\displaystyle j\frac {e^{-j}}{-1} - \int{\frac {e^{-j}}{-1}dx}$

Is this right as of now?
let's not make our notation make the problem more complicated than it already is. $\displaystyle \frac {e^{-j}}{-1} = - e^{-j}$

so your integral is: $\displaystyle -je^{-j} + \int e^{-j}~dj$

and you can take it from there. start plugging in the limits. of course, we have to use limits for the infinite integration limit

5. Originally Posted by Jhevon
let's not make our notation make the problem more complicated than it already is. $\displaystyle \frac {e^{-j}}{-1} = - e^{-j}$

so your integral is: $\displaystyle -je^{-j} + \int e^{-j}~dj$

and you can take it from there. start plugging in the limits. of course, we have to use limits for the infinite integration limit
So this then equals $\displaystyle -je^{-j} + \frac{e^{-j}}{-1}$ which of course means...

$\displaystyle -je^{-j} - e^{-j}$ so now just plug in the limits, aka 1 and $\displaystyle \infty$

And after you do that I get... $\displaystyle 2e^{-1}$ which is the lower bound, then to find the upper one wouldn't it just be... $\displaystyle 3e^{-1}$