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I am trying to figure out what the following sum converges to:

$$\sum_{n=0}^\infty {6+n\choose n}x^n(6+n),\qquad\qquad0<x<1$$

An answer would be great, but if you have an explanation, that'd be better! Thanks in advance for any help.

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Have you meant $$\sum_{n=0}^\infty\binom{6+n}n x^n(6+n)$$ or $$\sum_{n=0}^\infty\binom{6+n}n x^{n(6+n)}$$ ? – lab bhattacharjee Feb 22 '14 at 4:23
$$\sum_{n=0}^\infty\binom{6+n}n x^n(6+n)$$ Thanks. – ndm Feb 22 '14 at 4:40
First of all, $$ {6+n\choose n}x^n(6+n)= {6+n\choose n}x^n(7+n-1)=(7+n)\frac{(6+n)!}{n! 6!}x^n-{6+n\choose n}x^n$$ $$=\frac17{7+n\choose n}x^n-{6+n\choose n}x^n$$ Now I was trying to find the relation between $$\sum_{n=1}^\infty{6+n\choose n}x^n$$ and $$(1+x)^6+\frac{(1+x)^7}x+\frac{(1+x)^8}{x^2}+\cdots$$ which is an infinite Geometric Series – lab bhattacharjee Feb 22 '14 at 4:47
A not furthest relative : – lab bhattacharjee Feb 22 '14 at 5:07

2 Answers 2

up vote 0 down vote accepted

Related techniques: (I), (II). Follow the steps:

1) simplify $(n+6){ n+6\choose n} $ as

$$ (n+6){ n+6\choose n} = \frac{1}{6!}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2. $$

2) use the series identity

$$ \sum_{k=0}^{\infty} x^{n+1}=\frac{x}{1-x} \longrightarrow (*) $$

3) Applying the operators $D(xD)(x^2D)^5 $ to both sides of $(*)$ , where $D=\frac{d}{dx}$, gives

$$ \sum_{k=0}^{\infty}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2 x^{n+5}=D(xD)(x^2D)^5 \frac{x}{1-x} $$

$$ \implies \frac{1}{6!}\sum_{k=0}^{\infty}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2 x^{n}\\ =\frac{1}{6!\,x^5}D(xD)(x^2D)^5 \frac{x}{1-x} . $$

Note: The operator $(x^2 D)^5$ means

$$ (x^2 D)^5 = (x^2D)(x^2D)(x^2D)(x^2D)(x^2D). $$

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Great, thank you! – ndm Feb 22 '14 at 14:32
@Nmacgre: you are very welcome. – Mhenni Benghorbal Feb 22 '14 at 21:16

Starting fom the previous answers to this post, it seems to me (assuming no mistakes o my side) that $$\sum_{n=0}^\infty {6+n\choose n}x^n(6+n)=\frac{x+6}{(1-x)^8}$$

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Thanks, this is what I arrived at also (using Mhenni's solution). – ndm Feb 22 '14 at 14:33

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