09 Apr 08
Originally posted by JirakonI think it's 1.
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)!]
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)]
09 Apr 08
Originally posted by mtthwI 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...
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)]
09 Apr 08
Originally posted by strokem1What 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.
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...
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.
10 Apr 08
Originally posted by adam warlockOver here we have two seperate terms for convergence of series.
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.
In this case the series itself converges to 0, this is obvious.
The SUM-series however converges to something something else.
10 Apr 08
Originally posted by TheMaster37With the conventions I was taught, a series always refers to a sum of a sequence. The sequence converges to zero.
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.
Wikipedia agrees - it must be true!
10 Apr 08
1 edit
Originally posted by JirakonThe clue didn't help then?
I think it's 1.
I know it's 1, but I don't know how.
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.
10 Apr 08
Originally posted by TheMaster37I 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.
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.
10 Apr 08
Originally posted by deriver69Ah yes, that's simpler than my method.
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
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.