Divergence of $\sum \left(\frac{k-2}{k}\right)^k$ via $\frac{1}{e^2}$

Question

I want to prove that: $$ \sum \left(\frac{k-2}{k}\right)^k$$ is divergent.

My approach would be to look at the limit of general term and notice that this goes to $e^{-2}$. How would I prove this? I have been trying to rearrange the term like:

$$\left(\frac{k-2 +2}{k-2}\right)^{-k}=\left(1+\frac{2}{k-2}\right)^{-k}$$ But I do not see how to get the "square" in the exponent too.

I want to use the fact that $(1+ \frac{1}{n})^n \rightarrow e$

Answer 1

$$ \left(\frac k {k-2}\right)^k = \left(1 + \frac 2{k-2}\right)^{k-2} \left(1 + \frac 2{k-2}\right)^2 = \left(\left(1 + \frac 1{(k-2)/2}\right)^{(k-2)/2}\right)^{\color{red}2} \left(1 + \frac 2{k-2}\right)^2 $$ then taking the reciprocal.

UPDATE

If you want to use $\lim_n (1+1/n)^n = \mathrm e$ only, we actually need to justify that $$ \lim_n \left(1 + \frac 2n\right)^n = \mathrm e^2. $$ To do this, note that $$ b_n = \left(1 +\frac 1 n\right)^{2n} = \left(1 + \frac 2n + \frac 1{n^2}\right)^n = \left(\frac {1+2n +n^2}{n^2}\right)^n \xrightarrow{n \to +\infty} \mathrm e^2, $$ while $$ a_n =\left(1 + \frac 2n\right)^n = \left(\frac {2+n}{n}\right)^n = \left(\frac {n^2+2n}{n^2}\right), $$ and $$ 1 \leqslant \frac {b_n}{a_n} = \left(\frac {1+2n+n^2}{n^2+2n}\right)^n = \left(1+\frac 1{n^2+2n}\right)^n = \left( \left(1 + \frac 1{n^2+2n}\right)^{n(n+2)}\right)^{1/(n+2)} \leqslant \mathrm e^{1/(n+2)} \xrightarrow{n\to +\infty} 1, $$ thus $\lim_n b_n/a_n = 1$ by squeezing theorem. Since $\lim_n b_n = \mathrm e^2$, using the arithmetic operation we have $$ \lim_n a_n =\left. \lim_n b_n \middle/ \lim_n \frac {b_n}{a_n} \right. = \mathrm e^2, $$ as we desire.

Answer 2

$\left(\frac {k-2} k\right)^{k}=\left(1-\frac 2 k\right)^{k} \to e^{-2}$. [ Write $(1-\frac 1 n)^{n}$ as $a_n^{-n/{n-1}}$ where $a_n=(1+\frac 1 {n-1})^{n-1}$ and conclude that $(1-\frac 1n )^{n} \to \frac 1 e$. For $(1-\frac 2n )^{n}$ split the sequence int0 even and odd terms. The argument is slightly lengthy but not difficult].

Answer 3

We can also use that

$$\left(1+\frac{2}{k-2}\right)^{-k}=e^{-k \log\left(1+\frac{2}{k-2}\right)}\to e^{-2}$$

indeed by standard limits

$$-k \log\left(1+\frac{2}{k-2}\right)=\frac{\log\left(1+\frac{2}{k-2}\right)}{\frac2{k-2}}\frac{-2k}{k-2}\to -2$$

Answer 4

With $m:=n-2$,

$$\left(\frac{n-2}n\right)^n=\frac1{\left(\left(1+\dfrac2m\right)^{m/2}\right)^2\left(1+\dfrac2m\right)^2}$$

and as this expression has the limit $e^{-2}$, the general term does not tend to zero.