I'm teaching infinite sequences and series in a precalc class for the first time, and among other things, we've been thinking about:
$$\sum_{n=1}^{\infty} n^{-k}$$ If we knew calculus, we could do an integral and show that this diverges for $k\le 1$, and converges for $k>1$. But we don't know calculus yet in my class! Rats.
So instead, I found a nice algebraic proof that this diverges for $k=1$ (and by comparison, for $k<1$). I also found a nice algebraic proof that this converges for $k=2$ (and by comparison, for $k>2$).
But this still leaves us with the question of what happens between $k=1$ and $k=2$. My vague poking at the question, as well as some internet searching, hasn't turned up much.
So: how can we show, without using calculus or other fancy techniques, that $\sum\limits_{n=1}^{\infty} n^{-k}$ converges for $k>1$? Is there a nice, simple proof?
We can prove that for $k>1$ the series converges proving by induction that
$$S_N=\sum_{n=1}^{N} \frac{1}{n^k} \le1+\frac1{k-1}\left(1-\frac1{N^{k-1}}\right)$$
or, as an alternative, by Cauchy condensation test, that is
$$\sum 2^n a_{2^n}=\sum\frac{2^n}{(2^n)^k}=\sum \frac1{2^{n(k-1)}}$$
which converges by ratio test.