Convergence/Divergence of a series.

Question

I was tasked to find how the series shown below: $$\sum_{n=r}^{\infty} \frac{\left( n-r \right)!}{n!}$$

depends on the integer $r$.

I was given the hint that for $r\le 1$, each term would be greater than or equal to the corresponding term of $\sum \frac{1}{n}$, and that I should do a comparison test with this term, which tells me that the given series is divergent for $r\le 1$ (since $\sum \frac{1}{n}$ is known to be divergent). Another part of the problem gives me a hint for $r\ge 2$, but that is irrelevant to my concern which I will state below.

My problem is, I do not know how to prove that for $r\le 1$, each term would be greater than or equal to the corresponding term of $\sum \frac{1}{n}$ (please spare me your anger, I've just returned from a 4-year AWOL). Can I please get at least a hint (I'd be fine without spoonfeeding) on how to prove it?

Answer 1

If $r \leq 0$, then

\begin{align*} (n-r)! &\geq n!\\ \frac{(n-r)!}{n!} &\geq 1 \end{align*}

So clearly the series diverges.

If $r=1$, then

$$\sum_{n=1}^{\infty} \frac{(n-1)!}{n!} = \sum_{n=1}^{\infty} (n-1) \to +\infty $$

Answer 2

Let $n-r=p$ $$S_r=\sum_{n=r}^{\infty} \frac{(n-r)!}{n!}= \sum_{p=0}^{\infty} \frac{p!}{(p+r)!}=\sum_{p=0}^{\infty} \frac{1}{(p+r)(p+r-1)...(p+1)}\sim \sum_{p=1}^{\infty} \frac{1}{p^r}.$$ By comparision test this will converge when $r>1.$