I'm trying to find the function this series converges to. $\sum \limits_{i=0}^{\infty}{{{n+i}\choose{n}}{(\frac{a}{x})}^{i}}$
I guess it converges to $(\frac{x}{x-a})^{n+1}$. Can someone help me?
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Arctic Char
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Are you sure that this is the correct question? I can move $(a/x)^n$ in front of the series, after which we clearly see that the series diverges. – Stockfish Jul 21 '20 at 07:15
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@Stockfish You're right. Thank you. The power of $a/x$ must be $ i $. I edited it. – MohammadJavad Vaez Jul 21 '20 at 07:18
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Start replacing $\frac a x$ by $y$ – Claude Leibovici Jul 21 '20 at 07:27
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@ClaudeLeibovici Thank you, and then? – MohammadJavad Vaez Jul 21 '20 at 07:42
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Another one: https://math.stackexchange.com/q/2431627. – Martin R Jul 21 '20 at 07:48
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@MartinR Yeah. Thanks. I don't know how to thank you – MohammadJavad Vaez Jul 21 '20 at 07:53
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@MartinR Thanks a lot – MohammadJavad Vaez Jul 21 '20 at 07:56
1 Answers
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By binomial theorem for any non-naturalindex, we have $$(1+z)^\nu=\sum_{k=0}^{\infty} {\nu \choose k} z^k.$$ Let $z=-x$ and $\nu=-n$, then $$(1-x)^{-n}=\sum_{k=0}^{\infty} {-n \choose k} (-x)^k$$ Now use the property that $$={-\nu \choose k}=(-1)^k {\nu+k-1 \choose k}$$ We get $$\sum_{k=0} {n+k-1 \choose k} x^k=(1-x)^{-n}$$ Replace $n$ by $n+1$ both sides, to get $$\sum_{k=0}^{\infty} {n+k \choose k} x^k=(1-x)^{-(n+1)}$$ So in your case $$\sum_{i=0}^{\infty} {n+i \choose i} (a/x)^i=(1-a/x)^{-(n+1)}=\left(\frac{x}{x-a}\right)^{n+1}$$
Z Ahmed
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