For what value of 'd' would the sum to infinity of this equation exist?

Question

Sn = 2^a (-2^(dn) +1) / (1-2^d)

I am stuck at this point... Could someone please show, with working the range of values for which the sum to infinity would exist. All values of a, d and n are real numbers and d is not equal to 0. n is the term number for the series.

Thanks

cidyah · Answer

2^a (-2^(dn) +1) / (1-2^d)

Is 2^a a constant?
Did you mean ((-2)^d +1) / (1- 2^d)?

husoski · Answer

The expression 2^a/(1 - 2^d) doesn't depend on n and can be factored out of the sum.  It will be defined for all a and all nonzero d, so you only need to find the convergence of:

Σ (1 - 2^(dn)) .... from n=0 to oo

You need the terms to vanish as n-->oo for that sum to converge:

lim_n-->oo (1 - 2^(dn)) = 0
lim_n-->oo 2^(dn) = 1
lim_n-->oo dn = 0

That's only possible if d=0, so the series never converges for nonzero d.

PS: In the off-chance that you meant S_n = a^n [(-2)^(dn) + 1] / (1 - 2^d) instead, that leads to:

lim_n-->oo [(-2)^(dn) + 1] = 0
lim_n-->oo (-2)^(dn) = -1

...which is impossible for real d.  (Also impossible for complex d, but that's another story.)

Trending News

For what value of 'd' would the sum to infinity of this equation exist?

2 Answers