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Infinite series summation

Find the sum of the infinite series with
Unparseable latex formula:

n\6{th} \ term\ \frac{(n-1)^3}{n!}


I know the answer should be 1+e but cannot see how to do it. Can anyone point me in the right direction please?
Original post by brianeverit
Find the sum of the infinite series with
Unparseable latex formula:

n\6{th} \ term\ \frac{(n-1)^3}{n!}


I know the answer should be 1+e but cannot see how to do it. Can anyone point me in the right direction please?


Here's a rather longwinded method using series for xrexx^re^x for r=0,1,2,3.

Expand the numerator in your original series.


We note that:

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

And setting x=1 we have e=∑n=0∞1n!\displaystyle e=\sum_{n=0}^{\infty}\frac{1}{n!}

Similarly


xex=∑n=0∞xn+1n!\displaystyle xe^x=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!} and differentiating we get:

(x+1)ex=∑n=0∞(n+1)xnn!\displaystyle (x+1)e^x=\sum_{n=0}^{\infty} \frac{(n+1)x^n}{n!}

And setting x=1 we have

2e=∑n=0∞(n+1)n!\displaystyle 2e=\sum_{n=0}^{\infty}\frac{(n+1)}{n!}

Spoiler

(edited 9 years ago)
Reply 2
Avatar for BabyMaths
BabyMaths
As well the method suggested above by ghostwalker you can also get there by adjusting the index on the sum:

E.g. ∑n=1∞nn!=∑n=1∞1(n−1)!=∑n=0∞1n!=e\displaystyle \sum_{n=1}^{\infty} \frac{n}{n!}=\sum_{n=1}^{\infty} \frac{1}{(n-1)!}=\sum_{n=0}^{\infty} \frac{1}{n!}=e

You can then use the same method together with this result to obtain ∑n=1∞n2n!\displaystyle \sum_{n=1}^{\infty} \frac{n^2}{n!} and so on.

Eventually you get

∑n=1∞1n!=e−1\displaystyle \sum_{n=1}^{\infty} \frac{1}{n!}=e-1

∑n=1∞nn!=e\displaystyle \sum_{n=1}^{\infty} \frac{n}{n!}=e

∑n=1∞n2n!=2e\displaystyle \sum_{n=1}^{\infty} \frac{n^2}{n!}=2e

∑n=1∞n3n!=5e\displaystyle \sum_{n=1}^{\infty} \frac{n^3}{n!}=5e

and you can put all the pieces together.

Probably even more long winded than ghostwalker's suggestion.

This reminds me a STEP question...from last year?

Edit: It was 2012 and it was quite similar.

http://www.thestudentroom.co.uk/showthread.php?t=2044552&page=6&p=38368043
(edited 9 years ago)
Reply 3
Avatar for brianeverit
brianeverit
OP
9
Original post by ghostwalker
Here's a rather longwinded method using series for xrexx^re^x for r=0,1,2,3.

Expand the numerator in your original series.


We note that:

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

And setting x=1 we have e=∑n=0∞1n!\displaystyle e=\sum_{n=0}^{\infty}\frac{1}{n!}

Similarly


xex=∑n=0∞xn+1n!\displaystyle xe^x=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n!} and differentiating we get:

(x+1)ex=∑n=0∞(n+1)xnn!\displaystyle (x+1)e^x=\sum_{n=0}^{\infty} \frac{(n+1)x^n}{n!}

And setting x=1 we have

2e=∑n=0∞(n+1)n!\displaystyle 2e=\sum_{n=0}^{\infty}\frac{(n+1)}{n!}

Spoiler



Thank you. That's just what I needed to enable me to do it.

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