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?

  • $\begingroup$ $\frac{k(x+1)(k+2)\cdots (k+n-1)}{n!}\neq \frac{(k+n-1)!}{n!}$ $\endgroup$
    – 5xum
    Feb 29, 2016 at 9:11
  • $\begingroup$ Thanks for pointing that out, I've edited the post in light of this. $\endgroup$
    – rgarci0959
    Feb 29, 2016 at 9:20

2 Answers 2


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 $0$.


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 $k-1$.

  • $\begingroup$ What is that binomial coefficient with the upper argument negative? And what is $\;k\;$ there related to $\;n\;$ or something? Thank you $\endgroup$
    – DonAntonio
    Feb 29, 2016 at 9:30
  • $\begingroup$ 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. $\endgroup$ Feb 29, 2016 at 9:32
  • $\begingroup$ $k$ 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. $$\binom{-k}{n}=\frac{(-k)(-k-1)(-k-2)···(-k-n+1)}{n!}=(-1)^n\frac{k(k+1)(k+2)···(k+n-1)}{n!}.$$ $\endgroup$ Feb 29, 2016 at 9:33
  • $\begingroup$ @Joanpemo: see this answer. $\endgroup$
    – robjohn
    Feb 29, 2016 at 9:37
  • $\begingroup$ @robjohn Thank you very much. I actually knew that: it is just the development of negative Binomial exponents, but I wasn't used to $\;\binom{-k}n\;$ . $\endgroup$
    – DonAntonio
    Feb 29, 2016 at 9:45

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