Question

Find whether the following series are convergent or divergent:
$1+\frac{3}{7}x+\frac{3.6}{7.10}{x}^2+\frac{\mathrm{3.6.9}}{\mathrm{7.10.13}}{x}^3+\frac{\mathrm{3.6.9.12}}{\mathrm{7.10.13.16}}{x}^4+....$

Answer 1

$S_n=1+{\Pi }_{i=1}^{\infty }\frac{3n{x}^n}{(3n+4)}{T}_n=\frac{3\cdot 6\cdot 9\cdot .....\cdot \left(3n\right)x^n}{7\cdot 10\cdot 13\cdot ....\cdot (3n+4)}{T}_{n+1}=\frac{3\cdot 6\cdot 9\cdot 12\cdot .....\cdot \left(3n\right)(3n+3)x^{n+1}}{7\cdot 10\cdot 13\cdot ....\cdot (3n+4)(3n+7)}\underset{n\to \infty }{lim}\frac{T_{n+1}}{T_n}=\underset{n\to \infty }{lim}\frac{(3n+3)}{(3n+7)}\frac{x^{n+1}}{x^n}=\underset{n\to \infty }{lim}\left(\frac{3+3/n}{3+7/n}\right)x=x$

For $x>1,{\Pi }_{i=1}^{\infty }\frac{3n{x}^n}{(3n+4)}$ diverges

For $x<1,{\Pi }_{i=1}^{\infty }\frac{3n{x}^n}{(3n+4)}$ converges

Hence, $S_n=1+{\Pi }_{i=1}^{\infty }\frac{3n{x}^n}{(3n+4)}$ converges for $x<1$

$S_n=1+{\Pi }_{i=1}^{\infty }\frac{3n{x}^n}{(3n+4)}$ diverges for $x>1$

For $x=1$, ratio test fails

Then we apply Raabe's Test

$\underset{n\to \infty }{lim}[\frac{T_n}{T_{n+1}}-1]n=\underset{n\to \infty }{lim}n[\frac{3n+7}{3n+3}-1]=\underset{n\to \infty }{lim}n\left[\frac{3n+7-3n+3}{3n+3}\right]=\underset{n\to \infty }{lim}\frac{4n}{n(3+3/n)}=\frac{4}{3}$

Since, $\underset{n\to \infty }{lim}[\frac{T_n}{T_{n+1}}-1]n=\frac{4}{3}>1$

The series is convergent for $x=1$