Series sum questio

1. Jun 14, 2010

nhrock3

they take members from the marked series
how they make the power to be from n-1 to n+1
??

and why if we take the member of n=0 n=1
the index starts from n=1 and not n=2

??

2. Jun 14, 2010

estro

If I got right your question.

$$\sum_{n=0}^\infty \frac {1}{n!} (\frac {1}{u})^{n-1} =u+ \sum_{n=1}^\infty \frac {1}{n!} (\frac {1}{u})^{n-1}= u+1+ \sum_{n=2}^\infty \frac {1}{n!} (\frac {1}{u})^{n-1}$$

Remember: 0! = 1
(כתב יפה)

Last edited: Jun 14, 2010
3. Jun 17, 2010

LCKurtz

You are apparently expecting the sum you underlined to be:

$$\sum_{n=2}^\infty \frac 1 {n!}\left(\frac 1 u\right)^{n-1}$$

because the n = 0 and n = 1 terms were written separately. But notice what happens in this sum if you change the variables:

k = n -1 or n = k + 1. You get

$$\sum_{k=1}^\infty \frac 1 {(k+1)!}\left(\frac 1 u\right)^{k}$$

which, if you call the index n instead of k, is what is written.