Divergence of $ \sum_{n=1}^{\infty} \frac{1}{n^{i+1}}$ Does anyone have any idea how to prove that the series $ \sum_{n=1}^{\infty} \frac{1}{n^{i+1}}$ diverges? Can somebody help with it? Maybe it would be somehow easier to write it as $ \sum_{n=1}^{\infty} \frac{1}{e^{\log(n)(i+1)}}$, but I don't know..
Here $i$ designates the imaginary unit. Another way to write the sum is $\sum_{n=1}^{\infty} \frac{\exp(-(\ln n)i)}n$.

Remember the proof that $\sum_{n=1}^{\infty}1/n$ diverges.  You break the series into pieces, each of which has a sum at least 1/2.
To prove this diverges, break it into similar pieces, but this time one piece has $\cos\log(n)>1/2$, another piece has $\cos\log(n)<-1/2$, and other pieces are in-between.
Show that the sum is $$\sum_{n=1}^{\infty}\frac{1}{n}(\cos\log n-i\sin\log n)$$, and the sum within each of many pieces is above 1/4, or below -1/4.  So it never settles down.
I think there is a value, this is of course $\zeta(i+1)$, but it doesn't equal the series.
