Y=f(n)=axrn, if r=1/2 what is the horizontal asymptote for this function?

y=?

First term(a1)=

Common Ratio(r) =

Number of terms (n) =

Choose what to compute

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## 1 Answer

My guess is that rn should be rⁿ, and, since r=½, rⁿ→0 as n increases so the asymptote is y=0.

It’s not clear what x is in this context, nor what a1 represents. Since f is a function of n, not of x, then x appears to be a constant in this context, unless it is supposed to be a fixed subscript of a.

The domain of n has not been given. I assume n is an integer >0, in other words, a natural number.

a1 or a₁ is not defined in this context.

My guess is that f(n)=a_nxⁿrⁿ may be the intended function, where the underscore indicates that n is a subscript of a. If this is so, f(1)=a₁x/2, f(2)=a₂x²/4, f(3)=a₃x³/8, etc. However, this does not give any clues about the value of a₁. If n is an integer ≥0, then f(0)=a₀.

A better statement of this question is needed to resolve ambiguities and explain the context more clearly. A full answer can’t be given without more explanation. For example, a series would consist of a sum of consecutive values of f(n).

by Top Rated User (660k points)

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