Test absolute convergence of series $\sum_{k=1}^{\infty} (\frac{1}{k} - \frac{1}{k!})$

Question

So I need to test if the series $\sum_{k=1}^{\infty} (\frac{1}{k} - \frac{1}{k!})$ is absolutely convergent or not. So far, I've decided to go ahead with the ratio test, actually with a corollary of it that states that the series $\sum_{k=1}^{\infty} a_k$ is absolutely convergent if

$$\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n} \right| < 1$$

I tried to calculate it out but I'm stuck at:

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

I am uncertain if I can just expand the absolute value into the individual terms or leave it out completely so I can in turn calculate the limit of the individual terms too?

Any hints on how to simplify this and show the corollary above would be greatly appreciated!

Answer 1

I'd suggest using the limit comparison test with $\frac{1}{k}$ to show divergence.

If $a_k$ and $b_k$ are positive-termed sequences and $\lim_{k\rightarrow \infty} \frac{a_k}{b_k} = c$ where $c$ is positive and finite, then the series $\sum_{k=1}^\infty a_k$ and $\sum_{k=1}^\infty b_k$ either both converge or both diverge.

In this case choose $a_k = \frac{1}{k} -\frac{1}{k!}$ and $b_k = \frac{1}{k}$. Then we have \begin{align*} \lim_{k\rightarrow \infty} \frac{a_k}{b_k} & = \lim_{k\rightarrow \infty} \frac{\frac{1}{k} -\frac{1}{k!}}{\frac{1}{k}} \\ & = \lim_{k\rightarrow \infty} \left( 1-\frac{1}{(k-1)!} \right) \end{align*} It should be pretty clear that this limit is finite and positive, therefore both series diverge.

Answer 2

You can use the ratio test, just have to be careful. Notice that the summands can be written as $$a_n = \frac{n! - n}{n \cdot n!}.$$

Now we proceed with the ratio test. Observe:

\begin{align*} \frac{(n+1)! - (n+1)}{(n+1) \cdot (n+1)!} \cdot \frac{n \cdot n!}{n! - n} &= \frac{(n+1)! - (n+1)}{(n+1)^2} \cdot \frac{n}{n! - n} \\ &= \frac{(n+1) (n! - 1)}{(n+1)^2} \cdot \frac{n}{n((n-1)! - 1)} \\ &= \frac{n! -1}{(n-1)! - 1} \\ & \geq \frac{n! - 1}{(n-1)!} \\ &= \frac{n!}{(n-1)!} - \frac{1}{(n-1)!} \\ &= n - \frac{1}{(n-1)!} \\ &\to \infty \end{align*}

Thus the sum is divergent. Note that the fourth line uses the inequality $p < q \implies \frac{1}{p} > \frac{1}{q}$. The original expression is bounded below by something that goes to infinity, so the desired quantity also goes to infinity.

Answer 3

By Stirling's formula, $$\frac{1}{k} - \frac{1}{k!}\sim \frac{1}{k}-\frac{1}{\sqrt{2\pi k}}\left(\frac{e}{k}\right)^k=\frac{1}{k}+O\left(\frac{1}{k^k}\right)$$ as $k\to+\infty$. The series diverges since the harmonic series diverges.