Sum of Infinite Series: Finding n and Correct to Three Decimal Places

  • Thread starter Thread starter remaan
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves finding the sum of the series from n=1 to infinity of the function (1/n^5), with a requirement to express the result correct to three decimal places. The original poster expresses confusion regarding the interpretation of n and the concept of partial sums.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the nature of the summation and whether integration can be applied. There are questions about the meaning of "correct to three decimal places" and how to determine when the sum stabilizes at that precision.

Discussion Status

The discussion is ongoing with various interpretations being explored. Some participants suggest that the series converges and that the sum can be approximated by calculating terms until the desired precision is reached. Others question the clarity of the instructions regarding decimal accuracy.

Contextual Notes

There is uncertainty about the definition of n in the context of the series and how to approach the summation without explicit limits on n. Participants also express concern about the implications of rounding and the accuracy of their results.

remaan
Messages
132
Reaction score
0

Homework Statement



Find the sum of the series n= 1 to infinity of ( 1/n ^5 ) correct to three decimal places .

Homework Equations



We have to find Sn by sub. n number of terms


The Attempt at a Solution



But , we are not given n, so how to find the sum in such a case, is there a way to find n ?

( I mean by n the number of Sn terms )

any help would be nice
 
Physics news on Phys.org
Any help will be great ..
 
You are given n, n=1 to infinity. The summation is defined as the sum of in this case, an infinite number of terms of \frac{1}{{n^5 }}. If the problem stated n = 1 to 3, for example, it would simply be \frac{1}{{1^5 }} + \frac{1}{{2^5 }} + \frac{1}{{3^5 }}
 
Do you mean that I should make an integral n=1 to infinity ?

even without telling that n=1 ?
 
Ya, I tried that and I got .25
and what I did is that I itegrated the given function and then evaluated the integtal
with replacing the infinity with t and this what I got..

So " it seems that we are not using the Sn , partial sums at all" Right ?
 
No, n is an index, not a variable in the usual sense. n counts from 1, 2, 3, 4, to infinity in integer steps. It sounds like an integral but it's completely different.
 
Yes, sure I understand all what you said,
But still did not answer my question,

Is the way I did correct ,, integration without any Sn ??
 
There is no integration involved, you're simply adding \frac{1}{{1^5 }} + \frac{1}{{2^5 }} + \frac{1}{{3^5 }} all the way to n = infinity. Now the question wants you to do this until the first 3 digits no longer change when you continue to add.
 
I was thinking that the anwer should be rounded to 3 decimal places
 
  • #10
Because it saya " correct to three decimal places " ??

What do you think ??
 
  • #11
If you start calculating each term, you'll quickly realize that the first 3 decimal places no longer change in the sum after a certain n. That means you've found the sum accurate to 0.001.
 
  • #12
Uha, but this is not always
It happenes to be correct in this case

Right ?
 
  • #13
As long as the series isn't divergent, each additional sum will add smaller and smaller amounts to the sum. However, I don't believe that the digits are guaranteed to stay the same as your summation approaches infinity.
 
  • #14
Ok, and when it says accurate to three decimal places that means

0.000x

where x is any number, I mean to say that three decimal places means three zeros
 
  • #15
remaan said:
Ok, and when it says accurate to three decimal places that means

0.000x

where x is any number, I mean to say that three decimal places means three zeros

Why are you messing around with this? Why don't you use the remainder estimate Tom Mattson suggested in your other thread? Rn is less than or equal to an integral.
 
  • #16
Uha, am ok with that, but I was wondering about the meaning of correct to three decimal places mean ?
 
  • #17
I think they probably just mean make sure that Rn<0.001. Making sure that really means you have 3 correct decimal places could be a little tricky if you have, let's say Sn=0.1234 and Rn<0.001. That really doesn't tell you exactly how to round off. But I don't think they mean to be tricky.
 
  • #18
Still, I don't feel that my question is anwered..

What does "THREE DECIMAL PLACES MEAN"

how many zeros after the point ??
 
  • #19
This is exactly what I mean ...
I wish I could an "exact" answer .. before my exam tommoro
 
  • #20
Three places after the decimal point.
 
  • #21
I know that it's became boring, but still "how many zeros after the point"?
 
  • #22
My suggestion is to find Rn<0.001. That means Rn has at least three zeros after the decimal point, right?
 
  • #23
mmm,, what I can see is there are two zeros after the point ?? right?
 
  • #24
This is getting silly. If Rn is LESS THAN 0.001 then there are three. 0.001 has two. Anything less has three.
 
  • #25
Ok,, thaks a lot I appretiate your effort !
 

Similar threads

Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
1K