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 $\sum_{n=1}^{\infty} \frac{\exp(-(\ln n)i)}n$.
 A: 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.
A: Consider the extracted sequence of partial sums $p_k=\sum_{n=1}^{2^k}\frac{\exp(-(\ln n)\mathbf i)}n$. In any difference $p_{k+1}-p_k$ one has a sum of $2^k$ terms, all of which are${}\geq2^{-(k+1)}$ in absolute value, and whose argument lies in a sector of angles less than $\ln 2<0.7$ radians wide. The latter means the projection of the unit complex number $\exp(-(\ln n)\mathbf i)$ in the direction $\exp(-k(\ln2)(\ln \frac32)\mathbf i)$ halfway this sector of angles is always at least $\cos\frac{0.7}2>0.9$. This implies $|p_{k+1}-p_k|>0.9/2=0.45$, and the sequence is not a Cauchy sequence, nor is the sequence of all partial sums, which therefore does not converge.
