Mathematical Physics Dilemma

Hello everyone.

I came across a physics question as follows,

A set of positive point charges are kept at the points on the x-axis, such that

$x=x_{0}$,$x=3x_{0}$,$x=5x_{0}$........$\infty$

And a set of negative point charges are placed on the x-axis, such that,

$x=2x_{0}$ , $x=4x_{0}$ , $x=6x_{0}$ ........ $\infty$

If the magnitude of the positive point charge is $+q$ and the magnitude of the negative point charge is $-q$ , assuming $\varepsilon_0$ is the absolute permittivity of free space, then prove that the electric potential at the origin is

$V=\frac{q\ln2}{4\pi\varepsilon_0x_{0}}$

This is how I proved it.

We know by the definition of electric potential due to a point charge,

$V=\frac{q}{4\pi\varepsilon_0x_{0}}$

where $x_{0}$ is the separation between the origin and the point where the point charge is placed. Also it is known that the electric potential due to a system of point charges is the algebraic sum of the electric potentials due to the individual point charges.

Hence by applying this to every single point charge given, we get

$V=\frac{q}{4\pi\varepsilon_0(x_{0})} + \frac{-q}{4\pi\varepsilon_0(2x_{0})} + \frac{q}{4\pi\varepsilon_0(3x_{0})} + \frac{-q}{4\pi\varepsilon_0(4x_{0})}+.............\infty$

Factoring out $\frac{q}{4\pi\varepsilon_0x_{0}}$ , we get

$V=\frac{q}{4\pi\varepsilon_0(x_{0})}[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.............\infty]$

We know that the infinite summation in the square brackets is the Maclaurin's series expansion of $\ln(1+x)$ where $x=1$

Thus we get,

$V=\frac{q\ln2}{4\pi\varepsilon_0x_{0}}$

But instead of substituting for the logarithm expansion if we did as I have done in the attached picture, then we get answer as zero. Please tell me where I have gone wrong.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold


- bulleted
- list

bulleted
list


1. numbered
2. list

numbered
list

Note: you must add a full line of space before and after lists for them to show up correctly

paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

example link

> This is a quote

This is a quote

    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"

# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"

Math

Appears as

Remember to wrap math in $ ... $ or \[ ... \] to ensure proper formatting.

2 \times 3

2 \times 3

2^{34}

2^{34}

a_{i-1}

a_{i-1}

\frac{2}{3}

\frac{2}{3}

\sqrt{2}

\sqrt{2}

\sum_{i=1}^3

\sum_{i=1}^3

\sin \theta

\sin \theta

\boxed{123}

\boxed{123}

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

First of all, you cannot apply the Maclaurin's series of $\ln(1 + x)$ outside its radius of convergence, which is $-1 < x < 1$ .
There are, however, other methods to show that $\ln 2 = \sum\limits_{k = 1}^\infty \frac{(-1) ^{k + 1}}{k}$

Your mistake comes from the subtraction of two divergent series as, $\sum\limits_{k = 1}^\infty \frac{1}{k}$ does not converge!

Ameya Daigavane - 5 years, 2 months ago

Okay now I get it. Thank you so much

Anirudh Chandramouli - 5 years, 2 months ago

Actually, if you change the order of summation for non-absolutely converging series, the sum changes. In fact, by a suitable rearangement of terms, the summation given above can be made equal to any real number. Plus, the Maclaurin's series of $\ln(1+x)$ has the following domain: $(-1, 1]$

A Former Brilliant Member - 5 years, 1 month ago

And secondly when you rearrange the order of the terms of a series that goes to infinite you are changing the value of the sum

Prit Savani - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$