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 .

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TopNewestthe 3 latter sums all diverge,so this value isn't the "true" value of them,meaning that we only

assignthe 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 agoLog in to reply

– Sambhrant Sachan · 1 year ago

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 ?Log in to reply

– Hummus A · 1 year ago

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.Log in to reply

– Sambhrant Sachan · 1 year ago

Then what is the significance of these sums if they are all meaningless ?Log in to reply

here

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 itHere you can see an example of the series \(1+2+3..\) in physics – Hummus A · 1 year ago

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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 ago

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