Proving Divergence of Series \sum_{n=1}^{\infty} \frac{1}{n^{1+i}}

[Stumped1]

The question asks to prove that $$\sum_{n=1}^{\infty} \frac{1}{n^{1+i}}$$ diverges.

I am having trouble with this.

using the ratio test

$$\lim_{n\to\infty}\left|\frac{1}{(n+1)^{1+i}}\cdot\frac{n^{1+i}}{1}\right|$$

How can I simplify this further to find the limit?

Or is there another approach I should be taking?

Thanks for any help!

 

[Prove It]

Are you using $\displaystyle \begin{align*} i = \sqrt{-1} \end{align*}$ here?

 

[Stumped1]

That is correct.

Thanks for looking at this!

 

[alyafey22]

Hint

$ n^i = e^{i \log n } $

 

[chisigma]

The question asks to prove that $$\sum_{n=1}^{\infty} \frac{1}{n^{1+i}}$$ diverges.

I am having trouble w/ this.

using the ratio test

$$lim_{n-->\infty}|\frac{1}{(n+1)^{1+i}} \frac{n^{1+i}}{1}|$$

How can I simplify this further to find the limit?

Or is there another approach I should be taking?

Thanks for any help!

A simple and suggesting way to demonstrate that is to approximate...$\displaystyle \sum_{n=1}^{N} \frac{1}{n^{1 + i}} \sim \int_{1}^{N} \frac{d x}{x^{1 + i}} = i\ |x^{-i}|_{1}^{N} = i\ (N^{-i} - 1)\ (1)$

... and the limit of (1) for N tending to infinity doesn't exist...

Kind regards$\chi$ $\sigma$

 

[alyafey22]

$\displaystyle \sum_{n=1}^{N} \frac{1}{n^{1 + i}} \sim \int_{1}^{N} \frac{d x}{x^{1 + i}} = i\ |x^{-i}|_{1}^{N} = i\ (N^{-i} - 1)\ (1)$

Are you using the integral test for convergence ?

 

[chisigma]

Are you using the integral test for convergence ?

Yes, I do... o more precisely that is a sort of íntegral test for divergence'...

For the harmonic series is...$\displaystyle \sum_{n=1}^{N} \frac{1}{n} \sim \int_{1}^{N} \frac{d x}{x} = \ln N\ (1)$

... and the limit for N tending to infinity is infinity. In the problem proposed by Stumped the partial N-th sum remains bounded with N increasing but the limit for N tending to ininity doesn't exist...

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

Yes, I do... o more precisely that is a sort of íntegral test for divergence'...

For the harmonic series is...$\displaystyle \sum_{n=1}^{N} \frac{1}{n} \sim \int_{1}^{N} \frac{d x}{x} = \ln N\ (1)$

... and the limit for N tending to infinity is infinity. In the problem proposed by Stumped the partial N-th sum remains bounded with N increasing but the limit for N tending to ininity doesn't exist...

Kind regards

$\chi$ $\sigma$

The function you are integrating should be positive and decreasing in order to apply the integral test.

 

[chisigma]

The function you are integrating should be positive and decreasing in order to apply the integral test.

The problem is that we don't have to prove that a series of complex terms converges and we have to prove that it diverges. Now the so called 'integral test' is a test for convergence and it is a nonsense to use it to prove the divergence of a series...

Kind regards

$\chi$ $\sigma$

 

[Opalg]

Chisigma's comparison with an integral is an extremely strong heuristic indication that $$\sum_{n=1}^{N} \frac{1}{n^{1+i}}$$ oscillates as $N\to\infty$ (not converging, but also not tending to infinity, just gently oscillating between two limits). Strictly speaking, as ZaidAlyafey has pointed out, the integral test can only be used for series of positive terms, so it cannot be applied here. I think it is quite hard to give a rigorous proof that the series diverges. Here is my attempt.

The $n$th term of the series is $\frac1ne^{-i\log n} = \frac1n(\cos\log n - i\sin\log n)$. To prove that the series diverges, it will be enough to show that its real part $$\sum_{n=1}^\infty \tfrac1n\cos\log n$$ diverges.

The function $\cos x$ is positive if $x$ is within $\frac\pi2$ of an even multiple of $\pi$, and negative if $x$ is within $\frac\pi2$ of an odd multiple of $\pi$. More precisely, $\cos x > \frac12$ if $x$ lies in the interval from $2k\pi - \frac\pi3$ to $2k\pi + \frac\pi3$, for some integer $k$. The function $\log n$ increases extremely slowly, so there will be long runs of $n$ during which $\log n$ stays within one of those intervals, ensuring that $\cos \log n > \frac12$.

If we exponentiate the condition $2k\pi - \frac\pi3 < \log n < 2k\pi + \frac\pi3$, it says that $\exp\bigl(2k\pi - \frac\pi3\bigr) < n < \exp\bigl(2k\pi + \frac\pi3\bigr)$. For each $n$ in that interval, $\cos\log n$ will be greater than $\frac12$. Also, $\frac1n$ will be greater than $$\frac1{\exp\bigl(2k\pi + \frac\pi3\bigr)} = \frac1{e^{2k\pi}e^{\pi/3}}.$$ The number of terms in that interval is $\exp\bigl(2k\pi + \frac\pi3\bigr) - \exp\bigl(2k\pi - \frac\pi3\bigr) = e^{2k\pi}(e^{\pi/3} - e^{-\pi/3})$ (actually, the number of terms has to be an integer, but it will be the integer part of that last expression, or something close enough to make no difference). Therefore the sum of the terms in that interval will be $$\sum_{\exp(2k\pi - \frac\pi3) < n < \exp(2k\pi + \frac\pi3)}\frac1n\cos\log n > \frac{ e^{2k\pi}(e^{\pi/3} - e^{-\pi/3})}{2e^{2k\pi}e^{\pi/3}} = \tfrac12\bigl(1 - e^{-2\pi/3}\bigr) \approx 0.438.$$

In conclusion, there are infinitely many intervals during which the sum $$S_N = \sum_{n=1}^N \tfrac1n \cos \log n$$ increases by at least $0.438$ (and similarly infinitely many intervals during which it decreases by at least that amount). Therefore $$\lim_{N\to\infty}S_N$$ does not exist: the series diverges.

 

[alyafey22]

$$\sum_{n=1}^\infty \tfrac1n\cos\log n$$ diverges.

I was thinking of using Cauchy condensation law but unfortunately the series is not positive.

 

[Stumped1]

Thanks for all the help on this.

Now that I have a proof, I am still curious about the way first mentioned.

Using $$\frac{1}{n^{1+i}}=\frac{1}{n^i n}$$
and $$n^i=e^{ilogn}$$

we have $$\sum \frac{1}{e^{ilogn}n}$$

Since $$n $$is ever increasing, and $$e^z=e^{ilogn}$$ is periodic, is this enough on its own to prove this diverges?

Or should I write this different, or break this down more?

Or even on the right track?

Thanks so much for all the help!