# 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 $\sum_{n=1}^{\infty} \frac{\exp(-(\ln n)i)}n$.

• en.wikipedia.org/wiki/… – lab bhattacharjee Oct 19 '13 at 18:39
• But it converges if $i > 0$, by an integral test. – user43208 Oct 19 '13 at 18:45
• @user43208, what's your confusion? – lab bhattacharjee Oct 19 '13 at 18:53
• @labbhattacharjee Why don't you tell me? Do I misunderstand what $i$ is supposed to be? But if $i$ is a real positive number, then $\sum_{n \geq 2} \frac1{n^{i+1}} \leq \int_1^\infty \frac{dx}{x^{i+1}} = \frac1{i}$. What have I misunderstood? – user43208 Oct 19 '13 at 19:03
• The only "mistake" I could have made, it seems, is to miss the complex analysis tag and take $i$ for something other than a square root of $-1$. But it seems harsh to call that "my confusion". – user43208 Oct 19 '13 at 19:06

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.
• I already can show that my sum is $\sum_{n=1}^{\infty}\frac{1}{n}(\cos\log n-i\sin\log n)$, but how to divide it into pieces? With series $1/n$ it is easy, because I know its values for every $n$, but here.. I only know I can estimate cosine with $1$ and $-1$... Can you show me how to break it into pieces? – Anne Oct 19 '13 at 19:07
• One piece is $5\pi/3<\log n<7\pi/3$, another piece is $11\pi/3<\log n<13\pi/3$. Since $n$ increases by a factor of over two in both cases, $\sum (1/n)\cos\log n>\sum(1/n)/2>1/4$ – Empy2 Oct 19 '13 at 19:54
• I'm sorry, I don't get how did you obtain numbers like $5 \pi /3$ and so on. I don't understand. And my another question is, what about this part with sinus? – Anne Oct 19 '13 at 22:25
• $1/n^i=n^{-i}=\exp(-i\log n)=\cos(\log n)-i\sin(\log n)$ because $\exp(i\theta)=\cos\theta+i\sin\theta$. For all angles between $5\pi/3$ and $7\pi/3$, that is within $\pi/3$ of $2\pi$, $\cos\theta>1/2$. For all angles within $\pi/3$ of $3\pi$, $\cos\theta<-1/2$. – Empy2 Oct 20 '13 at 4:43
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.