Sum of two series equals zero

**rman144** · June 30th 2008, 03:13 PM

I've been messing around with the plotter from the site:
Dirichlet Series Plotter

And it seems like any imaginary part put into the equation is a reflection about the x-axis of the negative imaginary part. I need someone to cofirm the equation in the attachment because I don't think this is adding up correctly.

**CaptainBlack** · July 2nd 2008, 10:55 PM

Originally Posted by rman144

I've been messing around with the plotter from the site:
Dirichlet Series Plotter

And it seems like any imaginary part put into the equation is a reflection about the x-axis of the negative imaginary part. I need someone to cofirm the equation in the attachment because I don't think this is adding up correctly.

Assuming $b \in \mathbb{R}$ :

Well the imaginary part of:

$\sum_{n=1}^{\infty}\frac{1}{n^{1/2-ib}} +\sum_{n=1}^{\infty}\frac{1}{n^{1/2+ib}}$

if it exists, is zero, since the partial sums of the first series are equal to the conjugates of the partial sums of the second. However since you cannot rearrange conditionally convergent series what I say below about the real part also applies here.

The real part is:

$2 \sum_{n=1}^{\infty}\frac{\cos(b\ln(n))}{n^{1/2}}$

If this is zero (or even converges, since it does not satisfy the conditions of the alternating serier test) I don't know.

(The integral does not converge, but that also means nothing as the conditions for the integral test for convergence are not satisfied by this series)

If I were a betting person I think I would put my money on non-convergence.

(Note $b=0$ is divergent)

It might be interesting to consider other definitions of convergence for this problem. (for some reason I cannot find any references on line for these at present),
but I can givew an example:

For the dubiously convergent series $\sum_{n=1}^{\infty} a_n$ , define the "sum" to be:

$S^*=\lim_{\alpha \to 0} \sum_{n=1}^{\infty} a_n e^{-\alpha n}$

RonL

**CaptainBlack** · July 3rd 2008, 12:51 AM

Originally Posted by CaptainBlack

It might be interesting to consider other definitions of convergence for this problem. (for some reason I cannot find any references on line for these at present),
but I can givew an example:

For the dubiously convergent series $\sum_{n=1}^{\infty} a_n$ , define the "sum" to be:

$S^*=\lim_{\alpha \to 0} \sum_{n=1}^{\infty} a_n e^{-\alpha n}$

RonL

Further online research after a bit of lateral thinking about search strings give this reference.

The above example appears to be equivalent to Abel summation

RonL

**rman144** · July 3rd 2008, 03:50 PM

I know it is unusual to answer my own question, but the resulting series diverges. Test a few values on the site:

On-Line Calculator

And you will see that the values progressively grow. I don't have a rigid proof, but the numbers would indicate a divergent series.

**CaptainBlack** · July 3rd 2008, 08:39 PM

Originally Posted by rman144

I know it is unusual to answer my own question, but the resulting series diverges. Test a few values on the site:

On-Line Calculator

And you will see that the values progressively grow. I don't have a rigid proof, but the numbers would indicate a divergent series.

Well I can't get that calculator to sum series properly for me, but the sum:

$<br /> S_N=\sum_{n=1}^{N}\frac{\cos(\ln(n))}{n^{1/2}}<br />$

does not grow without bound as $N \to \infty$ it ossilates with increasing amplitude (at least for the first 10,000,000 terms).

RonL

**rman144** · July 3rd 2008, 08:49 PM

I think you may have made a typo there. The equation you just posted is missing an imaginary component and would thus be zeta(1/2). My equation was zeta(1/2+bi). You are correct however in saying the equation oscillates. However, although it oscillates, it diverges to greater value with each oscillation (on average).

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