evaluate the sum

[camilus]

http://img409.imageshack.us/img409/3350/digitproblemdi8.jpg

[camilus]

A bit hard at first, but I came up with a rather elegant solution. Anyone want to give it a try?

It is currently the problem of the week at Purdue University.

[Absane]

\frac{9}{10}*(e^{10}-1)

[BenTheMan]

Even easier than that :)

[Absane]

Even easier than that :)

How so?

[BenTheMan]

Oh. Nevermind. It might not be that easy :)

[BenTheMan]

Where do you get 9/10 from?

I get the same thing, up to that factor of ten. Show me where I screwed up...

\sum_{n=1}^{\infty} \frac{1}{d(n)!} = 9\cdot\frac{1}{1!} + 90 \cdot \frac{1}{2!} + \cdots

Then this is is just

\sum_{n=1}^{\infty} \frac{1}{d(n)!} = 9 \Bigg{\sum_{n=0}^{\infty} \frac{10^n}{n!} - 1\Bigg} = 9 \cdot \Big{ e^{10} -1 \Big}

[Absane]

e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + ...

When doing the summation for the above question, it's:

9 + \frac{90}{2!} + \frac{900}{3!} + ...

9*(1 + \frac{10}{2!} + \frac{{100}}{3!} + ...)

9*(1 + \frac{10^{1}}{2!} + \frac{{10^{2}}}{3!} + ...)

Multiply by \frac{10}{10} and you should see it (I'm tired of typing tex).

[BenTheMan]

10^0 = 1?

[Absane]

10^0 = 1?

yes? Why does this matter?

Ugh.. I don't feel like typing out the explanation. Basically by multiplying by ten, you get e^10 -1.

[BenTheMan]

IAh yes, I screwed up.

Where do you get 9/10 from?

I get the same thing, up to that factor of ten. Show me where I screwed up...

\sum_{n=1}^{\infty} \frac{1}{d(n)!} = 9\cdot\frac{1}{1!} + 90 \cdot \frac{1}{2!} + \cdots

Then this is is just

\sum_{n=1}^{\infty} \frac{1}{d(n)!} = 9 \Bigg{\sum_{n=0}^{\infty} \frac{10^n}{n!} - 1\Bigg} = 9 \cdot \Big{ e^{10} -1 \Big}

This should be:

\sum_{n=1}^{\infty} \frac{1}{d(n)!} = 9 \Bigg{\sum_{n=0}^{\infty} \frac{10^n}{n!}\Bigg}-1,

which gives the same answer as yours.

Apologies!

[temur]

Your original "this" is correct, only that "-1" is not under the sum.

Now we knoe why camilus was asking about d(n). :-)

[Absane]

Now we knoe why camilus was asking about d(n). :-)

Yea I noticed that, too. However, I don't know why he was worrying about formula for it. I suppose he was hoping for an easy "plug and chug" so that the answer would present itself.

When I wrote out the summation formula, I saw pretty quick the taylor expansion for e^x mixed in there... I just needed to do some simple manpulation.

Also, didn't the OP originally post as Arcane?



As a side note... I am so excited! I may be able to write some sort of paper with my graph theory professor. This means that my first paper will be with a guy whose Erdos number is 1. Mine may be 2!

And sorry for last night Ben.. for some reason, I was a bit testy. I really did not feel like explaining the "error." :)

[temur]

Wow, Erdos number 2! You don't see them everyday.

[BenTheMan]

And sorry for last night Ben.. for some reason, I was a bit testy. I really did not feel like explaining the "error."

No worries man. My arithmetic is pretty dreadful anyway.

How do you calculate Erdos number?

[Absane]

How do you calculate Erdos number?

Erdos himself has the number zero. Anyone that has coauthored a paper with Erdos has Erdos number 1. Anyone that coauthors/coauthored a paper with someone that coauthored a paper with Erdos has Erdos number 2. Etc...

http://upload.wikimedia.org/wikipedia/commons/thumb/f/fe/Erdosnumber.png

[D H]

Congrats! So, are you going to try for a finite Erdős-Bacon number?

[camilus]

wow absane I think you're the only one who got it.

it is indeed {9 \over 10}(e^{10}-1)

it comes directly from

\sum_{n = 0}^{\infty} {x^n \over (n+1)!} = {e^x \over x} - {1 \over x} = {e^x - 1 \over x}

[camilus]

http://i159.photobucket.com/albums/t121/camilus23/etothex.jpg

[BenTheMan]

Bah I was off by a factor of ten.

[temur]

wow absane I think you're the only one who got it.


I don't mean to be strange, but it would be fair to say Absane was the first to get it.

[Absane]

http://i159.photobucket.com/albums/t121/camilus23/etothex.jpg

I see you that you do mathematics in pen.