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Sum of infinite series.

The first part of the question asks to derive the first five terms of the series expansion for ex. Next, find an expression for the nth term of the infinite series: (1/1!) + (4/2!) + (7/3!) + (10/4!) + … Which is (3n - 2)(n!). It then asks to sum the infinite series, which is what I am stuck on. I know you have to relate It to the series expansion for e^x.
(edited 4 years ago)
Reply 1
Avatar for mqb2766
mqb2766
21
Original post by Jrucks
The first part of the question asks to derive the first five terms of the series expansion for ex. Next, find an expression for the nth term of the infinite series: (1/1!) + (4/2!) + (7/3!) + (10/4!) + ...


So what are you stuck with?
a) Are you doing a power series (no differentiation) or a taylor series (differentiation)?
b) The nth term isn't hard. The denominator should be straightforward and if you difference the numerators, you should be able to spot what type of sequence the individual terms are.
Post your attempt and describe where you're stuck?
Reply 2
Avatar for Jrucks
Jrucks
OP
6
Original post by mqb2766
So what are you stuck with?
a) Are you doing a power series (no differentiation) or a taylor series (differentiation)?
b) The nth term isn't hard. The denominator should be straightforward and if you difference the numerators, you should be able to spot what type of sequence the individual terms are.
Post your attempt and describe where you're stuck?

I accidentally pressed the enter button midway through typing this; edited now. It is a Maclaurin expansion.
Original post by Jrucks
The first part of the question asks to derive the first five terms of the series expansion for ex. Next, find an expression for the nth term of the infinite series: (1/1!) + (4/2!) + (7/3!) + (10/4!) + … Which is (3n - 2)(n!). It then asks to sum the infinite series, which is what I am stuck on. I know you have to relate It to the series expansion for e^x.


ex=∑n=0∞xnn!\displaystyle e^x = \displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{n!}

∑n=0∞3n−2n!=−2+∑n=1∞3n−2n!=−2+3∑n=1∞1(n−1)!−2∑n=1∞1n!\displaystyle \sum_{n=0}^{\infty} \dfrac{3n - 2}{n!} = -2 + \sum_{n=1}^{\infty} \dfrac{3n - 2}{n!} = -2 + 3\sum_{n=1}^{\infty} \dfrac{1}{(n-1)!} - 2 \sum_{n=1}^{\infty} \dfrac{1}{n!}


Do they look similar to the series for exe^x? Try relating them to it.
(edited 4 years ago)
Reply 4
Avatar for mqb2766
mqb2766
21
Original post by Jrucks
I accidentally pressed the enter button midway through typing this; edited now. It is a Maclaurin expansion.



@RDKGames got there before me :-)
(edited 4 years ago)
Reply 5
Avatar for Jrucks
Jrucks
OP
6
Original post by RDKGames
ex=∑n=0∞xnn!\displaystyle e^x = \displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{n!}

∑n=0∞3n−2n!=−2+∑n=1∞3n−2n!=−2+3∑n=1∞1(n−1)!−2∑n=1∞1n!\displaystyle \sum_{n=0}^{\infty} \dfrac{3n - 2}{n!} = -2 + \sum_{n=1}^{\infty} \dfrac{3n - 2}{n!} = -2 + 3\sum_{n=1}^{\infty} \dfrac{1}{(n-1)!} - 2 \sum_{n=1}^{\infty} \dfrac{1}{n!}


Do they look similar to the series for exe^x? Try relating them to it.

Got it now; it was not spotting that (n)/(n!) can be written as (1)/(n-1)! that hindered my progress. Thanks!

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