Waste less time on Facebook — follow Brilliant.
×

Subtracting Till Infinity

When we have to Calculate a Sum Like this : \(\ S=1+\dfrac12+\dfrac14+\dfrac18+\dfrac1{16}+\cdots\infty \)

we just multiple by \(-\dfrac12\) and Then add the two equations to get our answer , but the same process is not allowed for other series , for example ,

\( \text{Let S =}1+3+3^2+3^3+3^4+3^5+\cdots\infty \\ -3S=-3-3^2-3^3-3^4-3^5-\cdots\infty \\ \implies -2S=1 \implies S=-\dfrac12 \)

why are these type of results considered incorrect ?? Everything i did was correct .

on the other hand sum of series like \(\displaystyle\sum_{n=0}^{\infty}2^n=-1\) and \(\displaystyle\sum_{n=0}^{\infty}n=-\dfrac1{12}\) are absolutely correct .

Note by Sabhrant Sachan
1 year, 6 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. 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"
MathAppears 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

Sort by:

Top Newest

the 3 latter sums all diverge,so this value isn't the "true" value of them,meaning that we only assign the series that value,but in the actual sense of equity,this makes no sense.You can read about ways for assigning values to divergent series here

Hummus A - 1 year, 6 months ago

Log in to reply

The series are divergent ,but the sum of the Last two Series are absolutely correct . In this way why can't we say that sum of other infinite [ divergent ]series will also be negative ?

Sabhrant Sachan - 1 year, 6 months ago

Log in to reply

no,the last 2 divergent series' values are as valid as the value which you gave the 2nd (divergent) series.We can assign that series a negative value,just like the other ones,but that doesn't mean the series doesn't diverge like the others.

Hummus A - 1 year, 6 months ago

Log in to reply

@Hummus A Then what is the significance of these sums if they are all meaningless ?

Sabhrant Sachan - 1 year, 6 months ago

Log in to reply

@Sabhrant Sachan they're not meaningless,but you need to be very careful with substituting that value.These series often come up in physics,you can read about it here

Here you can see an example of the series \(1+2+3..\) in physics

Hummus A - 1 year, 6 months ago

Log in to reply

This may explain your question: Ramanujan: Making sense of 1+2+3+... = -1/12 and Co.

It talks about this kind of geometry sequences. You may want to skip the 1+2+3+... = -1/12 part.

Lemuel Liverosk - 1 year, 6 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...