thomas49th said:Find the sum of
e^(-nt) as n = 0,1,2,3,...
Apparently it's 1/(1-e^-t)
Can't use normal geometric series formula though as if a =1, r = e?
Any ideas
Thanks :)
The formula for calculating the infinite sum of infinite negative series e^(-nt) is:
Σ e^(-nt) = 1/(1-e^(-t))
This formula is also known as the geometric series formula.
The infinite sum of infinite negative series e^(-nt) has many applications in mathematics, physics, and engineering. It is used to calculate the total energy of a system in quantum mechanics, to model the decay of radioactive materials, and to solve differential equations, among other things.
Yes, the infinite sum of infinite negative series e^(-nt) can converge to a finite value under certain conditions. This occurs when the absolute value of the ratio between consecutive terms (r) is less than 1, meaning that the series is convergent. When r is greater than or equal to 1, the series is divergent and does not have a finite sum.
The infinite sum of infinite negative series e^(-nt) is closely related to the exponential function. In fact, when r is less than 1, the infinite sum of infinite negative series can be rewritten as 1/(1-r), which is the same as the power series representation of the exponential function. This shows that the infinite sum of infinite negative series converges to the exponential function under certain conditions.
Yes, there are many real-world applications of the infinite sum of infinite negative series e^(-nt). One example is in finance, where it is used to calculate the present value of a stream of future cash flows. It is also used in computer science to approximate infinite series, and in chemistry and biology to model chemical reactions and biological processes, respectively.