The ratio and root tests

Question

Use the ratio tests to determine the a_k of this series below.

∞

∑ (n𝜋^n)/(-4)^(n-1)

n = 1 

I know that it is convergent but I was not able to figure out the a_k specifically.

Answer 1

If you are, in fact, looking for the sum of this series, note that the series is of the form ∑kr^k which can be obtained by several methods, e.g. so-called partial summation.

Also note that it is a series which meets the criteria for Leibnitz's Theorem so it converges...but you were asked to prove convergence by other means.

Answer 2

This is a_n= (n𝜋^n)/(-4)^(n-1).

since there is (-4)^(n-1), you can separate odd and even members.

a_{2k}=2k𝜋^(2k))/(-4)^(2k-1).=-2k𝜋^(2k))/4^(2k-1).

and

a_{2k-1}=(2k-1)𝜋^(2k-1))/(-4)^(2k-2).=(2k-1)𝜋^(2k-1))/(4)^(2(k-1)).

Do you know how to figure out convergence?

Answer 3

∑_n=1∞(n𝜋ⁿ)/(-4)^n-1···

lim n→∞ I(n + 1)πⁿ⁺¹/(-4)ⁿ·(-4)^n-1/(πⁿnⁿ)I = lim n→∞I(1 + 1/n)π/(-4)I = π/4 < 1.

So, series is convergent·