# I Alternating partial sums of a series

1. Oct 13, 2016

### spaghetti3451

Consider the Taylor series expansion of $e^{-x}$ as follows:

$\displaystyle{e^{-x}=1-x+\frac{x^{2}}{2}-\frac{x^{3}}{6}+\dots}$

For $x>0$, the partial sums $1$, $1-x$, $\displaystyle{1-x+\frac{x^{2}}{2}}$ bound $e^{-x}$ from above and from below alternately.

How do I prove this?

2. Oct 13, 2016

### haruspex

Be clear what you are trying to prove: that
1 > e-x, and
1-x < e-x, etc.

3. Oct 13, 2016

### spaghetti3451

Yes.

I am looking for a proof that covers all the cases $1>e^{-x}, 1-x<e^{-x}, \dots$ at once.

4. Oct 13, 2016

### haruspex

Ok, so what does it say about the sum of the rest of the terms in the expansion in each case?

5. Oct 13, 2016

### spaghetti3451

Well, it say that $1-e^{-x}>0, 1-x-e^{-x}<0, \dots$.

6. Oct 13, 2016

### haruspex

Ok, that edit happened while I was replying.
That was not clear from the OP. It looked like you were only asking for those three specific partial expansions.
Anyway, my hint in post #4 still applies.

7. Oct 13, 2016

### spaghetti3451

But, it's difficult to start the general proof since there are two separate cases.

Are we supposed to prove two inequalities, one for even partial sums and one for odd partial sums?

8. Oct 13, 2016

### mathwonk

this is a typical alternating series. each term is smaller than the previous one and the signs alternate. It follows that any partial sum ending in a plus sign is larger than any partial sum ending in a minus sign, hence also larger than the full sum, etc...I.e. the partial sums are jumping back and forth across the full sum, or final limit. draw a picture, as this is essentially obvious.

i.e. start from home and go a block north. then go back a 1/2 block south. you are obviously somewhere between home and the first block. then go back 1/4 block north. you are obviously somewhere between the 1/2 block point and the one block point.........

in answer to your question, yes there do seem to be two cases.

the best discussion i know of for series including alternating ones is in courant's calculus.

here is a free copy:

https://archive.org/details/DifferentialIntegralCalculusVolI

9. Oct 13, 2016

### spaghetti3451

The qualititative argument works well for me.

But, just for further info, do you think Taylor's theorem can be used to write a proof?