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Sum converging to 0

by Bipolarity
Tags: converging
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Bipolarity
#1
Nov9-12, 02:40 PM
P: 783
Does there exist a sequence of real nonzero numbers whose sum converges to 0?
I would think there isn't, but I'm interested in people's opinions and arguments.

For any nonzero m, a series of nonzero numbers whose sum converges to m can easily be constructed using the formula: [itex] \sum ^{\infty}_{n=1}m(0.5)^{n} [/itex]

But that is for nonzero m, what if you wanted to construct a series whose sum converged to 0?

BiP
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Vargo
#2
Nov9-12, 02:47 PM
P: 350
1+-1+.5+-5+.25+-.25+.125+-.125+......
Bipolarity
#3
Nov9-12, 03:30 PM
P: 783
Quote Quote by Vargo View Post
1+-1+.5+-5+.25+-.25+.125+-.125+......
Can you find an explicit representation for that seqence (i.e. with sigma notation) ?

BiP

Bipolarity
#4
Nov9-12, 03:57 PM
P: 783
Sum converging to 0

Eureka!!! I believe I found it!

[tex] \sum^{\infty}_{n=1} (-1)^{n+1} (\frac{1}{2})^{ \frac{2n-3+(-1)^{n+1}}{4}} [/tex]

I believe it converges to 0, but can anyone verify this?

BiP
Edgardo
#5
Nov9-12, 04:10 PM
P: 686
If [itex]\sum ^{\infty}_{n=1}m(0.5)^{n}[/itex] converges to m, then shouldn't [itex]m-\sum ^{\infty}_{n=1}m(0.5)^{n}[/itex] converge to 0?
Bipolarity
#6
Nov9-12, 04:15 PM
P: 783
Quote Quote by Edgardo View Post
If [itex]\sum ^{\infty}_{n=1}m(0.5)^{n}[/itex] converges to m, then shouldn't [itex]m-\sum ^{\infty}_{n=1}m(0.5)^{n}[/itex] converge to 0?
Yes, but [itex]m-\sum ^{\infty}_{n=1}m(0.5)^{n}[/itex] is not a series... unless you can express it as one with nonzero terms.

BiP
lurflurf
#7
Nov9-12, 08:55 PM
HW Helper
P: 2,264
Who cares about expressing it as one with nonzero terms? A series is a series is a series.
This is simple
How about
[tex]\sum_{n=0}^\infty \frac{(\pi)^{2n+1}}{(2n+1)!} (-1)^n[/tex]
Bipolarity
#8
Nov9-12, 10:18 PM
P: 783
Quote Quote by lurflurf View Post
Who cares about expressing it as one with nonzero terms? A series is a series is a series.
The problem requires it.

BiP
haruspex
#9
Nov9-12, 11:07 PM
Homework
Sci Advisor
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Thanks
P: 9,922
Quote Quote by Bipolarity View Post
Yes, but [itex]m-\sum ^{\infty}_{n=1}m(0.5)^{n}[/itex] is not a series... unless you can express it as one with nonzero terms.
How about (1-λ)λn-(1-μ)μn for n >= 0, where λ is algebraic and μ is not?
Millennial
#10
Nov10-12, 05:02 AM
P: 295
How about taking a sequence [itex](a_x)_x[/itex] which satisfies [itex]\displaystyle \lim_{x\to\infty}a_x = 0[/itex] and then using the series [itex]\displaystyle \sum_{x=0}^{\infty} (-1)^x b_x[/itex], where the sequence [itex]b_x[/itex] is defined as [itex]b_{2x} = b_{2x+1} = a_x[/itex]?
lurflurf
#11
Nov10-12, 07:44 PM
HW Helper
P: 2,264
So you dislike the pi example and the usual example
[tex]\sum_{k=0}^\infty a_k b_x[/tex]
where a_k is a sequence of positive numbers tending to zero and B_k is any sequence of -1 and 1 such that the series tends to zero.
What about any number of obvious examples such as
[tex]\sum_{k=0}^\infty (2k-1)\left(\frac{1}{3}\right)^k[/tex]


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