# Infinite Series Limiting Value

EDIT:

Find the converging value of $\sum\limits_{n=2}^{\infty} \cfrac{(n-1) \lambda^n}{n!}$.

I just know that

$\sum\limits_{n=0}^{\infty} \cfrac{x^n}{n!}=e^x$ for all x.

So I can split up the summation into 2 terms:

$\sum\limits_{n=2}^{\infty} \cfrac{\lambda^n}{(n-1)!} -\sum\limits_{n=2}^{\infty} \cfrac{\lambda^n}{n!}$

Term 1:

Perform variable substitution setting $(n-1)=k \implies n=k+1$

$\sum\limits_{n=2}^{\infty} \cfrac{\lambda^n}{(n-1)!} = \sum\limits_{k=1}^{\infty} \cfrac{\lambda^{k+1}}{k!} = \lambda \sum\limits_{k=1}^{\infty} \cfrac{\lambda^{k}}{k!} = \lambda (e^{\lambda}+1)$

Term 2

$\sum\limits_{n=2}^{\infty} \cfrac{\lambda^n}{n!}=e^{\lambda}-1- \lambda$

Now we combine both terms getting:

$\lambda (e^{\lambda}+1) -e^{\lambda}+1+ \lambda$

I'm getting a different answer than below, but I still don't think this is right. Can someone please verify?

Thank you.

• Split it, $\sum\limits_{n=2}^\infty \frac{n\lambda^n}{n!} - \sum\limits_{n=2}^\infty \frac{\lambda^n}{n!}$. That the sum doesn't start at $0$ just means you have to subtract a few terms from the exponential. – Daniel Fischer Jul 29 '13 at 23:23
• Do you want to sum the series? – Mhenni Benghorbal Jul 29 '13 at 23:24
• I'm looking for the limit. What is the sum yes... @MhenniBenghorbal – user1527227 Jul 29 '13 at 23:25
• $\lambda^n=\lambda\cdot\lambda^{n-1}$. – David Mitra Jul 29 '13 at 23:27
• @DanielFischer Just make that an answer! Else we end up with unawnsered questions! Also, it has happened already that I answer and you have given the exact same hint/answer in the comments, and I feel bad =) – Pedro Tamaroff Jul 29 '13 at 23:30

Note that $$\sum\limits_{n = 2}^\infty {\frac{{(n - 1){\lambda ^n}}}{{n!}}} = \sum\limits_{n = 2}^\infty {\frac{{{\lambda ^n}}}{{\left( {n - 1} \right)!}}} - \sum\limits_{n = 2}^\infty {\frac{{{\lambda ^n}}}{{n!}}}$$

And further $$\sum\limits_{n = 2}^\infty {\frac{{{\lambda ^n}}}{{\left( {n - 1} \right)!}}} = \lambda \sum\limits_{n = 1}^\infty {\frac{{{\lambda ^n}}}{{n!}}} = \lambda \left( {{e^\lambda } - 1} \right)$$ while $$\sum\limits_{n = 2}^\infty {\frac{{{\lambda ^n}}}{{n!}}} = {e^\lambda } - 1 - \lambda$$

• @user1527227 Any time. – Pedro Tamaroff Jul 30 '13 at 0:10

Here is how you advance

$$\sum\limits_{n=2}^{\infty} \cfrac{(n-1) \lambda^n}{n!} =\sum\limits_{n=2}^{\infty} \cfrac{n \lambda^n}{n!} - \sum\limits_{n=2}^{\infty} \cfrac{ \lambda^n}{n!} = \sum\limits_{n=2}^{\infty} \cfrac{ \lambda^n}{(n-1)!}-(e^x-1-x)$$

$$= \sum\limits_{n=1}^{\infty} \cfrac{ \lambda^{n+1}}{n!}-(e^{\lambda}-1-\lambda)$$

$$=\lambda\left( e^{\lambda}-1 \right)-( e^{\lambda}-1-\lambda ).$$

• Doesn't ratio test just tell you whether it converges or diverges? I am looking for the sum. – user1527227 Jul 29 '13 at 23:25
• @user1527227: See the edit. – Mhenni Benghorbal Jul 29 '13 at 23:34
• Thanks @MhenniBenghorba! Did you do variable substitution on your summation for line 2? If so let's say $k=(n-1)$, then if n starts at 2, k=1. Should your starting limit on the second line be 1? – user1527227 Jul 29 '13 at 23:37
• What is $x$?.... – Ron Gordon Jul 29 '13 at 23:39
• I don't think this is the right solution. – user1527227 Jul 29 '13 at 23:53