Question

Find whether the following series are convergent or divergent:
$1+\frac{\alpha .\beta }{1.\gamma }x+\frac{a(a+1)\beta (\beta +1)}{\mathrm{1.2.}\gamma (\gamma +1)}{x}^2+\frac{a(a+1)(a+2)\beta (\beta +1)(\beta +)}{\mathrm{1.2.3.}\gamma (\gamma +1)(\gamma +2)}{x}^3+...$

Answer 1

Let $T_n$ be the general term of the series

$T_n=\frac{\alpha (\alpha +1)\beta (\beta +1)...(\alpha +(n-1\left)\right)(\beta +(n-1\left)\right)x^n}{(n!)\gamma (\gamma +1)(\gamma +2)....(\gamma +(n-1\left)\right)}{T}_{n+1}=\frac{\alpha (\alpha +1)\beta (\beta +1)...(\alpha +(n-1\left)\right)(\beta +(n-1\left)\right)(\alpha +n)(\beta +n)x^{n+1}}{(n+1)!\gamma (\gamma +1)(\gamma +2)....(\gamma +(n-1\left)\right)(\gamma +n)}\underset{n\to \infty }{lim}\frac{T_{n+1}}{T_n}=\underset{n\to \infty }{lim}\frac{(\alpha +n)(\beta +n)(n!)x}{(n+1)!(\gamma +n)}=\frac{n^2(1+\alpha /n)(1+\beta /n)x}{n^2(1+1/n)(1+\gamma /n)}=x$

If $x<1$, the series is convergent

If $x>1$, the series is divergent

For $x=1$, ratio test fails

We apply Raabe's test

$\underset{n\to \infty }{lim}n[\frac{T_n}{T_{n+1}}-1]=\underset{n\to \infty }{lim}n[\frac{(\gamma +n)(n+1)}{(\alpha +n)(\beta +n)}-1]=\underset{n\to \infty }{lim}n\left[\frac{(1+\gamma )n+n^2+\gamma -n^2-\alpha \beta -(\alpha +\beta )n}{(\alpha +n)(\beta +n)}\right].=\underset{n\to \infty }{lim}\frac{(1+\gamma -\alpha -\beta )n^2+\gamma n-\alpha \beta n}{(\alpha +n)(\beta +n)}=\underset{n\to \infty }{lim}\frac{(1+\gamma -\alpha -\beta )+\frac{\gamma }{n}-\frac{\alpha \beta }{n}}{(1+\frac{\alpha }{n})(1+\frac{\beta }{n})}=(1+\gamma -\alpha -\beta )$
If

$1+\gamma -\alpha -\beta <1$, the series is divergent

$\gamma -\alpha -\beta <0$, the series is divergent

If

$1+\gamma -\alpha -\beta >1$, the series is convergent

$\gamma -\alpha -\beta >0$, the series is convergent