Sum converging to 0

Bipolarity

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

Vargo

1+-1+.5+-5+.25+-.25+.125+-.125+......

Bipolarity

Can you find an explicit representation for that seqence (i.e. with sigma notation) ?

BiP

Bipolarity

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

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

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

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

The problem requires it.

BiP

haruspex

How about (1-λ)λⁿ-(1-μ)μⁿ for n >= 0, where λ is algebraic and μ is not?

Millennial

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

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]