Alternative proof for $ \sum \left(\frac{k-2}{k}\right)^k$ is divergent.

Question

In an earlier question I asked help to prove that: $$ \sum \left(\frac{k-2}{k}\right)^k$$ is divergent, using the fact the the general term converges to $\frac{1}{e^2}$ and by the limit test the series diverges, is there a more straightforward method I am missing? One that does not use $e$ necessarily.

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See original post: Series diverges

Answer 1

if $n>2$ then $\Big(\dfrac{n-2}{n}\Big)^n>0$ and $$ \\\lim_{n\to+\infty}{\Big(\frac{n-2}{n}\Big)^n}=\lim_{n\to+\infty}{\Big(1-\frac{2}{n}\Big)^n}=e^{-2}>0 $$

Answer 2

Suppose we don't know anything about $e$. We can still show that $(1-2/k)^k\not\to0$ as $k\to\infty$, using the fact that the (alternating) binomial expansion can be truncated after a few terms to give an inequality, since $2^k/k!\lt1$ for $k\ge4$.

$$\begin{align} \left(k-2\over k \right)^k &=\left(1-{2\over k}\right)^k\\ &=1-{k\choose1}\left(2\over k\right)+{k\choose2}\left(2\over k\right)^2-{k\choose3}\left(2\over k\right)^3+\cdots\\ &=1-2+2-{8\over6}{(k-1)(k-2)\over k^2}+{16\over24}{(k-1)(k-2)(k-3)\over k^3}-{32\over120}{(k-1)(k-2)(k-3)(k-4)\over k^4}+\cdots\\ &\gt1-2+2-{4\over3}+{2\over3}\left(1-{6\over k}\right)-{4\over15}\\ &={1\over15}-{4\over k} \end{align}$$

A nice way to wrap things up is now to note that, for $k\ge105$, we have

$$\left(k-2\over k\right)^k\gt{1\over15}-{4\over105}={1\over35}$$

and therefore $(1-2/k)^k\not\to0$ as $k\to\infty$, which in turn implies $\sum(1-2/k)^k$ diverges.

Remarks: In addition to $2^k/k!\lt1$ for $k\ge4$, which permitted truncation after a subtraction, we used the crude inequalities $(k-1)(k-2)\lt k^2$ and $(k-1)(k-2)(k-3)(k-4)\lt k^4$ and the slightly less crude inequality $(k-1)(k-2)(k-3)=k^3-6k^2+11k-6\gt k^3-6k^2$.