Have a problem:

Find the sum of the following convergent series, justify your answer:

Series from n=0 to infinity of n/e^(n)

Obviously, the terms go 0 + 1/e + 2/e^2 + 3/e^3 +....etc, but how exactly am I supposed to find the sum of this and justify the answer. I know it's not a geometric series so i can't use the a/(1-r) formula.

Thanks

If I remember my inf. seqs and series, you have to find the limit of that series as n -> inf.

So you want lim[sub]n->inf[/sub] n/e[sup]n[/sup], if the limit exists that's your convergent value. If the limit is inf or doesn't exist, series diverges.

Edit: This make sense to you or anyone else? n/e[sup]n[/sup] = 1/e[sup]2n[/sup]. If that is true, then I think you can use your formula.

Problem is I find my limit to be 0. So unless I'm a retard and been doing my math wrong this whole time, it's messed up.

The terms are obviously going towards 0, but added together that's defintily not the case. I plug it into my calculator and I get the sum to = 1, but I won't have the help of the calculator on the midterm

The limit *is* 0, but you aren't looking for the limit.

Arg it's been like 3 yrs since I've done this but I think if the limit is 0 then the test is inconclusive and you have to use other means.

I do seem to remember that under certain circumstances there is a way to use integration to get the sum you want but I'm blanking right now.

Let me look it up and I'll get back to you if no one else has helped you yet.

Just by looking at it the equation would have to tend to 0 for the question to make sense, unless infinity is counted as an answer. Because otherwise you'd have an infinite amount of non-zero, positive terms.

That being said, a quick ratio test gives an answer of one, and I can't be arsed to test it any other way, because it's half one in the morning and I've got a lecture at nine.

Couldn't tell you how to work it out, though.

EDIT: Whatever, this whole point is moot because people are faster than me.

Oh wait. JG is right.

The series is equal to n(Sum[1/e^n]). Sum the geometric series 1/e^n and then multiply by n.

I feel sorta stupid now =_= I was overthinking it x 10000000000

select seq.nextval from dual;

wrong. 1) you can't take an index variable out of a summation like that. 2) "multiply by n"? pray tell what 'n' equals in this case, el o el.

However, if you use the ratio test, and get 1, that indicates no info :/. I suppose the next would be to test with a p-series, but unsure of how to do that.

Whoops. I just utterly failed there. I was thinking of not just one something else but multiple something elses and put them together. My bad.

Yeah I postulated whether ∑(n/e[sup]n[/sup]) = ∑(1/e[sup]2n[/sup]). BTW, a blank ∑ is commonly understood as an inf. series @ n=1.

Pumping this into SciLab gives me =.9206736. Maybe it eventually hits one thanks to infinite series, but this is one area of math I prefer to let the computers do for me

(also ∑(1/e[sup]2n[/sup]) gives me 0.2909884)

Er wait...I fail at exponents. I was combining different bases. You can't do that. However...can you do this?

∑(ln(n/e[sup]n[/sup])) = ∑(ln(n) - ln(e[sup]n[/sup])) = ∑(ln(n) - n)

Of course lim[sub]n->inf[/sub] ln(n) - n goes to infinity so not the answer you're looking..

You're on your own!

That looks sound to me.

Well, i got an answer. I went to ask the "math tutor" that's on call in the commons area and even HE had no clue how to do it. Finally he e-mailed me back this morning (after I was promptly raped by my midterm). For those interested...


In this case, our “x” is actually just e (i. e., the thing to the powers in ((1)). We can now plug in the left hand sides of ((2)) and ((3)) into the right hand side of ((1)) to find the final answer: The following are some binomial series that we’re going to use. (these may be listed in your book along with e^x, sin x, cos x, ln x, etc..). The point is that these are some known series.

A) 1/(1+1/x)2=1-2/x+3/x2-4/x3+5/x4-...
B) 1/(1-1/x)2=1+2/x+3/x2+4/x3+5/x4+...

First, I’m going to rewrite what we have for this problem,

Sum of (n/e^n) = 1/e + 2/e^2 +3/e^3 +4/e^4 + ….
Factor out 1/ e,
Sum of (n/e^n) = (1/e) * (1 + 2/e + 3/ e^2 +4/e^3 + …) ((1))

Adding up the two series A and B above and dividing the result by 2 gives (notice that all terms with odd power cancel out),
(1/2) *(1/(1+1/x) 2 + 1/(1-1/x) 2 ) = 1+ 3/x2 + 5/x4+ …. ((2))

We begin to see the pattern here, these are some of the terms we want ( i.e., the terms in ((1)) above). We still need the terms in ((1)) that have odd exponents. These can be obtained similarly by now subtracting B from A and dividing the whole thing by 2, that’s,

(1/2) *(1/(1+1/x) 2 - 1/(1-1/x) 2 ) = -2/x - 4/x3 - 6/x5 -…. ((3))

(Again, note that the terms with even exponents cancel out, and the terms of interest appear but actually are negative).

Sum of(n/e^n)=(1/e)*((1/2) *(1/(1+1/e) 2+ 1/(1-1/e) 2 )-(1/2) *(1/(1+1/e) 2 - 1/(1-1/e) 2 ))

We subtracted the green part because we need +2/e + 4/e3 + 6/ e5 +…. in ((1)).

So, simplifying We get,

Sum of(n/e^n)=(1/e)*( 1/(1-1/e) 2)

And mscbuck demonstrates why I loathe sequences and series.

But the series definition of e^z is what gives you the most beautiful proof in all mathematics...