Which-of-the-series-converge-and-which-diverge-Check-by-the-limit-comparison-test-1-n-2-1-n-ln-n-n-2-5-2-n-1-ln-n-n-3-2-3-n-3-1-ln-lnn-4-n-1-

Question Number 67244 by Learner-123 last updated on 24/Aug/19

W h i c h o f t h e s e r i e s c o n v e r g e a n d

w h i c h d i v e r g e ? C h e c k b y t h e l i m i t

c o m p a r i s o n t e s t .

1) \underset{n = 2}{\sum^{\infty}} \frac{1 + n l n (n)}{n^{2} + 5}

2) \underset{n = 1}{\sum^{\infty}} \frac{l n (n)}{n^{\frac{3}{2}}}

3) \underset{n = 3}{\sum^{\infty}} \frac{1}{l n (l n n)}

4) \underset{n = 1}{\sum^{\infty}} \frac{1}{n {(n)}^{\frac{1}{n}}}

? ?

Commented by mathmax by abdo last updated on 24/Aug/19

1) \sum_{n = 2}^{\infty} \frac{1 + n l n (n)}{n^{2} + 5} = \sum_{n = 2}^{\infty} \frac{1}{n^{2} + 5} + \sum_{n = 2}^{\infty} \frac{n l n (n)}{n^{2} + 5}

\frac{1}{n^{2} + 5} = \frac{1}{n^{2} (1 + 5 n^{- 2})} \sim \frac{1}{n^{2}} s o t h i s s e r i e c o n v e r g e s

\frac{n l n (n)}{n^{2} + 5} = \frac{n l n (n)}{n^{2} (1 + 5 n^{- 2})} \sim \frac{l n (n)}{n} l e t d e t e r m i n e n a t u r e o f Σ \frac{l n (n)}{n}

f (t) = \frac{l n (t)}{t} w i t h t ⩾ 2 \Rightarrow f^{^{'}} (t) = \frac{1 - l n t}{t^{2}} < 0 \Rightarrow f d e c r e a s e s \Rightarrow

Σ \frac{l n (n)}{n} a n d \int_{2}^{+ \infty} \frac{l n (t)}{t} d t a r e t h e s a m e n a t u r e

c h a n g e m e n t l n (t) = u g i v e \int_{2}^{+ \infty} \frac{l n (t)}{t} d t = \int_{l n (2)}^{+ \infty} \frac{u}{e^{u}} e^{u} d u

= \int_{l n (2)}^{+ \infty} u d u = + \infty s o t h i s s e r i e d i v e r g e s .

Commented by mathmax by abdo last updated on 24/Aug/19

2) l e t u_{n} = \frac{l n (n)}{n^{\frac{3}{2}}} a n d f (x) = \frac{l n x}{x^{\frac{3}{2}}} w i t h x > 1 w e h a v e

f^{^{'}} (x) = \frac{\frac{1}{x} x^{\frac{3}{2}} - \frac{3}{2} x^{\frac{1}{2}} l n x}{x^{3}} = \frac{\sqrt{x} - \frac{3}{2} \sqrt{x} l n (x)}{x^{3}} = \frac{1}{x^{\frac{5}{2}}} \times (1 - l n x) < 0 \Rightarrow

f d e c r e a s e s o (u_{n}) d e c r e a s e s \Rightarrow t h e s e r i e h a s t h e n a t u r e o f

I \int_{1}^{+ \infty} \frac{l n (x)}{x^{\frac{3}{2}}} d x c h a n g e m e n t l n (x) = t g i v e

I = \int_{0}^{\infty} \frac{t}{e^{\frac{3}{2} t}} d t = \int_{0}^{\infty} t e^{- \frac{3}{2} t} d t i t s c l e a r t h a t I c o n v e r g e s \Rightarrow

t h i s s e r i e c o n v e r g e s .

Commented by mathmax by abdo last updated on 24/Aug/19

f^{^{'}} (x) = \frac{1}{x^{\frac{5}{2}}} \times (1 - \frac{3}{2} l n (x)) < 0

Commented by Learner-123 last updated on 25/Aug/19

S i r, i n 1) \to f^{'} (t) = \frac{1 - l n t}{t^{2}} < 0

\forall t ϵ (e, \infty) . B u t ⩾ 0 \forall t ϵ [2, e] .

Commented by mathmax by abdo last updated on 25/Aug/19

s i r t h e n a t u r e o f a s e r i e d o n t c h a n g e s f o r a f e w t e r m s \dots

Commented by Learner-123 last updated on 25/Aug/19

o k, t h a n k y o u p r o f .

Commented by mathmax by abdo last updated on 26/Aug/19

3) t h e s e q u e n c e u_{n} = \frac{1}{l n (l n n)} i s d e c r e a s i n g o n [3, + \infty [s o

Σ u_{n} a n d \int_{3}^{+ \infty} \frac{d x}{l n (l n x)} h a v e t h e s a m e n a t u r e

c h a n g e m e n t l n x = t g i v e \int_{3}^{+ \infty} \frac{d x}{l n (l n x)} = \int_{l n (3)}^{+ \infty} \frac{e^{t} d t}{l n (t)}

w e h a v e {l i m}_{t \to + \infty} \frac{e^{t}}{l n t} = + \infty \Rightarrow t h e i n t e g r a l d i v e r g e s \Rightarrow Σ u_{n}

d i v e r g e s .