In finding the interval of convergence of a series, I needed to test the endpoints of the interval to see if it converged there. Doing so amounted to using the test for divergence and examining the limit of the general term of the series at the endpoints. However, while I was able to determine that the limit was definitely greater than zero, I could not explicitly find it.

The series was:

$$\sum_{n=1}^{\infty} \frac{n!x^n}{6\cdot 13 \cdot 20 \cdot \cdots \cdot (7n-1)}$$

I found the radius of convergence to be 7, and then proceeded to test the endpoints ($x=7$ and $x=-7$).

For $x=7$ the series becomes

$$\sum_{n=1}^{\infty} \frac{n!7^n}{6\cdot 13 \cdot 20 \cdot \cdots \cdot (7n-1)}.$$

Using the divergence test, I tried to find the limit of the general term of this series:

$$\lim_{n \to \infty}a_n=\lim_{n \to \infty}\frac{n!7^n}{6\cdot 13 \cdot 20 \cdot \cdots \cdot (7n-1)}$$

This is a limit in the form of $\frac{\infty}{\infty}$, but L'Hopital's Rule won't apply here. With some rewriting, however, I did manage to simplify the limit to

$$\lim_{n \to \infty}\frac{((1)(2)(3)(4)\cdots(n))(7^n)}{6\cdot 13 \cdot 20 \cdot \cdots \cdot (7n-1)} $$ $$=\lim_{n \to \infty}\frac{(1\cdot7)(2\cdot7)(3\cdot7)(4\cdot7)\cdots(n\cdot7)}{6\cdot 13 \cdot 20 \cdot \cdots \cdot (7n-1)} $$ $$=\lim_{n \to \infty}\frac{7\cdot 14 \cdot 21 \cdot 28 \cdots \cdot (7n)} {6\cdot 13 \cdot 20 \cdot \cdots \cdot (7n-1)} $$

From here, we can conclude that the limit is definitely greater than one (we know it's greater than $\frac{7}{6}$, and greater than $\left(\frac{7}{6}\right)\left(\frac{14}{3}\right)\cdots$). But what exactly is the limit? Is there something obvious that I'm missing here?

From intuition, I reasoned that after each increase in $n$, $a_n$ is $\frac{7n}{7n-1}$ times its previous value, where $n$ is the value after the increase. E.g. $a_2$ is $\frac{14}{13}$ times $a_1$. As $n$ increases without bound, the ratio of successive terms tends to one, but never actually equals one. So every successive term is always at least a little bit bigger than the previous one, and so my guess is that the product is infinite.

However, the above argument is purely intuition-based and has no rigorous proof. I had a few ideas as to how to prove this, including:

  • Gamma Function
  • Stirling's Approximation
  • Converting to a series by taking logarithms (not sure if this would work in this case)

How would one calculate the limit of this sequence?


2 Answers 2


First, does it even matter? As your $a_n$ are all $\ge 1$, they certainly don't converge to $0$, which makes the series $\sum a_n$ divergent, which is what you wanted to determine.

Then again, we can do the following: Onstead of $a_n$, consider its logarithm$$\ln a_n=\ln\frac 76+\ln\frac{14}{13}+\ldots +\ln\frac{7n}{7n-1}.$$ As $\ln (1+x)\approx x$ (to make this precise you may show something like: for $x>0$ small enough, $\frac 12x<\ln (1+x)<x$). Hence the last summand of $\ln a_n$ is $\approx \frac1{7n-1}\approx \frac17\cdot\frac 1n$ and this makes $\ln a_n$ grow like the harmonic series, i.e., $a_n\to +\infty$.


$$\sum_{n = 1}^{\infty} \frac {n! x^n} {6 \cdot 13 \cdots (7n-1)} = \sum_{n = 1}^{\infty} \prod_{k = 1}^{n} \frac {kx} {7k - 1} > \sum_{n = 1}^{\infty} \prod_{k = 1}^{n} \frac {x} {7} = \sum_{n = 1}^{\infty} \left (\frac {x} {7} \right)^n.$$ This means, wherever $|x| < 7$ the series will converge. It will diverge everywhere else.


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