 Posers and Puzzles

1. 09 Apr '08 20:49
How do I figure out what this series converges to?:

Sum: 1/[n! (n+2)], n goes from 0 to infinity.

In case there's any confusion, this is (n+1)/[(n+2)!]
2. 09 Apr '08 21:56
Originally posted by Jirakon
How do I figure out what this series converges to?:

Sum: 1/[n! (n+2)], n goes from 0 to infinity.

In case there's any confusion, this is (n+1)/[(n+2)!]
I think it's 1.

Too late to type up the answer now, so here's a clue. Consider:

f(x) = SUM{n = 0, inf} x^(n+2)/[n!(n+2)]
3. 09 Apr '08 22:16
Originally posted by mtthw
I think it's 1.

Too late to type up the answer now, so here's a clue. Consider:

f(x) = SUM{n = 0, inf} x^(n+2)/[n!(n+2)]
I thought it would be converging on zero... just due to the fact that as n gets infinitely large the bottom of the fraction is going to be exponentially larger than the top thus bringing you closer and closer to zero...
4. 09 Apr '08 23:07
Originally posted by strokem1
I thought it would be converging on zero... just due to the fact that as n gets infinitely large the bottom of the fraction is going to be exponentially larger than the top thus bringing you closer and closer to zero...
What converges to 0 is the expression of the general term of the series while n goes to infinity. But as you take more and more terms the series approaches a fixed value since it is obvious that that series converges.

For instance 1+1/2+1/4+1/8+1/16+1/32+...+1/2^n+...=2. What goes to 0 is 1/2^n that is the general term.

Actually the fact that the general term goes to 0 as n goes to infinity is a necessary but not sufficient condition for a series to converge.
5. 10 Apr '08 00:45
I think it's 1.

I know it's 1, but I don't know how.
6. 10 Apr '08 06:21
What converges to 0 is the expression of the general term of the series while n goes to infinity. But as you take more and more terms the series approaches a fixed value since it is obvious that that series converges.

For instance 1+1/2+1/4+1/8+1/16+1/32+...+1/2^n+...=2. What goes to 0 is 1/2^n that is the general term.

Actually the fact that the ...[text shortened]... to 0 as n goes to infinity is a necessary but not sufficient condition for a series to converge.
Over here we have two seperate terms for convergence of series.

In this case the series itself converges to 0, this is obvious.

The SUM-series however converges to something something else.
7. 10 Apr '08 09:14
Originally posted by TheMaster37
Over here we have two seperate terms for convergence of series.

In this case the series itself converges to 0, this is obvious.

The SUM-series however converges to something something else.
With the conventions I was taught, a series always refers to a sum of a sequence. The sequence converges to zero.

Wikipedia agrees - it must be true!
8. 10 Apr '08 09:181 edit
Originally posted by Jirakon
I think it's 1.

I know it's 1, but I don't know how.
The clue didn't help then?

If we write f(x) = SUM x^(n+2)/[n!(n+2)], note that f(1) gives the series you want.

df/dx = SUM x^(n+1)/n! = x(SUM x^n/n!)

You might recognise last sum: SUM x^n/n! = e^x

so df/dx = xe^x
=> f = (x - 1)e^x + c

To calculate c, set x = 0. We know f(0) = 0, so c = 1

=> f(x) = (x - 1)e^x + 1
=> f(1) = 1

One of those cases where the easiest way to solve a problem is actually to solve a more general problem, which has a special case that matches your original problem.
9. 10 Apr '08 10:18
Originally posted by TheMaster37
Over here we have two seperate terms for convergence of series.

In this case the series itself converges to 0, this is obvious.

The SUM-series however converges to something something else.
I was thought that the word series means to the actual value of the limiting sum. And that what is being summed is called "termo geral da série". I translated that to english as general term but maybe that isn't the correct technical term in english. Anyway this general term must tend to 0 for the series to converge. Now if the series converge to 0 either all terms are 0 or some kind of cancelling must occur between succesive terms.
10. 10 Apr '08 10:18
I think I would use
(n+1)/(n+2)! =(n+2-1)/(n+2)!
=(n+2)/(n+2)!-1(n+2)!
=1(n+1)!-1/(n+2)!

Use the method of differences and you are left with 1/1!-a successively small term which tends to 0 as the series has an infinite number of terms
11. 10 Apr '08 10:52
Originally posted by deriver69
I think I would use
(n+1)/(n+2)! =(n+2-1)/(n+2)!
=(n+2)/(n+2)!-1(n+2)!
=1(n+1)!-1/(n+2)!

Use the method of differences and you are left with 1/1!-a successively small term which tends to 0 as the series has an infinite number of terms
Ah yes, that's simpler than my method.

Having another look at the question, I think the way it's stated makes a difference. With the first formula, the idea of differentiating to get rid of the (n + 2) term leapt out at me. If I'd used the second (equivalent) formula I think I'd have been more likely to take your approach.