If $\frac{u_{n}}{u_{n + 1}} = \frac{n^{k} + A n^{k - 1} + B n^{k - 2} + C n^{k - 3} + . . . . .}{n^{k} + a n^{k - 1} + b n^{k - 2} + c n^{k - 3} + . . . . .}$ , where $k$ is positive integer, show that the series $u_{1} + u_{2} + u_{3} + . . . . . .$ is convergent if $A - a - 1$ is positive, and divergent if $A - a - 1$ is negative or zero.

Open in App

Solution

We apply Raabe's test
$lim n \to \infty n [\frac{u_{n}}{u_{n + 1}} - 1] = lim n \to \infty n [\frac{[n^{k} + A n^{k - 1} + B n^{k - 2} + . . .]}{[n^{k} + a n^{k - 1} + b n^{k - 2} + . . .]} - 1] = lim n \to \infty n \frac{[n^{k} + A n^{k - 1} + B n^{k - 2} + . . .] - [n^{k} + a n^{k - 1} + b n^{k - 2} + . . .]}{[n^{k} + a n^{k - 1} + b n^{k - 2} + . . .]} = lim n \to \infty n [\frac{[(A - a) n^{k - 1} + (B - b) n^{k - 2} + . . .]}{[n^{k} + a n^{k - 1} + b n^{k - 2} + . . .]}] = lim n \to \infty \frac{n^{k} [(A - a) + \frac{(B - b)}{n} + . . .]}{n^{k} [1 + \frac{a}{n} + \frac{b}{n^{2}} + . . . .]} = (A - a)$
If $A - a > 1 A - a - 1 > 0$
The series is divergent
If $A - a < 1 A - a - 1 < 0$
The series is convergent

Suggest Corrections

Similar questions

u_{1} + u_{2} + u_{3} + u_{4} + . . . . . .

lim n \to \infty \sqrt[n]{u_{n}}

< 1

> 1

\frac{1}{u_{0}} - \frac{1}{u_{1}} + \frac{1}{u_{2}} - \frac{1}{u_{3}} + \dots + {(- 1)}^{n} \frac{1}{u_{n}} = \frac{1}{u_{0} +} \frac{u_{0}^{2}}{u_{1} - u_{0} +} \frac{u_{1}^{2}}{u_{2} - u_{1} +} \dots \frac{u_{n - 1}^{2}}{u_{n} - u_{n - 1}}

u_{n} = \sqrt[3]{n^{3} + 1} - n

n_{1}, n_{2}, \dots, n_{k}

n

1

n .

n_{1} + n_{2} + \dots + n_{k} = 91,

\frac{1}{n_{1}} + \frac{1}{n_{2}} + \dots + \frac{1}{n_{k}}

\frac{1}{x y} - \frac{1}{(x + 1) (y + 1)} + \frac{1}{(x + 2) (y + 2)} - \frac{1}{(x + 3) (y + 3)} + . . . .,

x

y

Join BYJU'S Learning Program