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Given the series

$$ \sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!} \quad \quad k \geq 1 $$ Find the interval of convergence.

I started by applying the Ratio test

$$ \lim_{n\to \infty}\left|\frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)(k+n)x^{n+1}}{(n+1)!}\cdot \frac{n!}{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}\right|$$

$$\lim_{n\to \infty}\left|\frac{(k+n)x}{(n+1)}\right|$$

to show that the series converges when $|x| \lt 1$.

However, when I test the end points of $(-1,1)$ for convergence, I end up with two series whose convergence I am unable to show. Namely, $$ \sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)}{n!} $$

and $$ \sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)(-1)^n}{n!} $$

How can I show that these two series converge or diverge?

rgarci0959
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2 Answers2

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Note that for $k\ge1$, we have $$ \frac{k(k+1)\cdots(k+n-1)}{n!}=\frac k1\frac{k+1}2\cdots\frac{k+n-1}{n}\ge1 $$ Thus, for $|x|=1$, the terms do not go to $0$.

robjohn
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Your series is $$ \sum_{n=0}^\infty\binom{-k}{n}(-x)^n=(1-x)^{-k} $$ This alone should show that there is no convergence at $x=1$ for positive $k$.

For the series at $x=-1$ consider that $$ \binom{-k}{n}(-1)^n=\binom{n+k-1}{n}=\binom{n+k-1}{k-1}=\frac{(n+1)(n+2)···(n+k-1)}{(k-1)!} $$ is a polynomial in $n$ of degree $k-1$.

Lutz Lehmann
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  • What is that binomial coefficient with the upper argument negative? And what is $;k;$ there related to $;n;$ or something? Thank you – DonAntonio Feb 29 '16 at 09:30
  • For $x=1$ the $n$th term is $e^{A_n}n^{k-1}$ where $\lim_{n\to \infty} A_n$ exists so the series is summable for $k<0$ but not for $ k\geq 0$. For $x=-1$ and k<1 the terms are of alternating sign and go to 0 so the series sums. For $x=\pm 1$ and $k\geq 1$ the terms do not go to zero. – DanielWainfleet Feb 29 '16 at 09:32
  • $k$ is a constant, from the name I assume integer. Look up generalized binomial coefficients, the numerator can be any real number, defined by exactly the fractions present in question and answer. With beta and gamma functions further generalization is possible. $$\binom{-k}{n}=\frac{(-k)(-k-1)(-k-2)···(-k-n+1)}{n!}=(-1)^n\frac{k(k+1)(k+2)···(k+n-1)}{n!}.$$ – Lutz Lehmann Feb 29 '16 at 09:33
  • @Joanpemo: see this answer. – robjohn Feb 29 '16 at 09:37
  • @robjohn Thank you very much. I actually knew that: it is just the development of negative Binomial exponents, but I wasn't used to $;\binom{-k}n;$ . – DonAntonio Feb 29 '16 at 09:45