Question

Prove that a series of complex terms diverge if $\mathit{\rho }\mathbf{>}\mathbf{1}$( = ratio test limit). Hint: The${\mathit{n}}^{\mathbf{t}\mathbf{h}}$term of a convergent series tends to zero.

Short answer

The series is divergent for $\rho >1$.

Step-by-step solution

Given Information.

The value of $\rho >1$ in ratio test.

Definition of Convergent and Divergent series.

A convergent series is one in which the partial sums all gravitate to the same finite number, also known as a limit. Divergent refers to any series that is not converging.

Calculate the value of ρn.

Assume the series.

$S=\sum _{}{C}_n=C_1+C_2+C_3+...$ ...(1)

Use the ratio test and find the value of

${\mathrm{\rho }}_{\mathrm{n}}=\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}}$

Test the series for convergence or divergence.

Calculate the value of $\mathrm{\rho }$, i.e.,

role="math" localid="1658733057990" $$\begin{gathered} \mathrm{\rho }=\underset{\mathrm{n}\to \infty }{\lim }\left|{\mathrm{\rho }}_{\mathrm{n}}\right| \\ \mathrm{\rho }=\underset{\mathrm{n}\to \infty }{\lim }\left|\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}}\right|...\left(2\right) \end{gathered}$$

Possibilities for $\mathrm{\rho }$are mentioned below:

If $\mathrm{\rho }<1$each term is smaller than the previous term, as the series progresses, the coefficients become smaller, and the series approaches a specific number.

If $\mathrm{\rho }=1$each term equals the preceding term, and the coefficient remains constant as the series progresses, we will have no idea how the series will turn out at infinity, thus we must find another way.

If $\mathrm{\rho }>1$ each term is larger than the previous term, as the series progresses, the coefficients will have a very large value that we can't estimate, and the series will approach infinity, which is undetermined, indicating that the series is divergent.

For $\mathrm{\rho }>1$, series diverges.

Hence, the series is divergent for $\mathrm{\rho }>1$.