study-tbe-convergence-of-n-1-nln-1-n-

Question Number 127778 by Bird last updated on 02/Jan/21

s t u d y t b e c o n v e r g e n c e o f

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

Answered by mnjuly1970 last updated on 02/Jan/21

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

= \underset{n = 2}{\sum^{\infty}} \frac{1}{n l n (n)} \underset{t h e o r e m}{\overset{c a u c h y d e n s i t y}{\approx}} \underset{k = 1}{\sum^{\infty}} \frac{2^{k}}{2^{k} l n (2^{k})}

= \frac{1}{l o g (2)} \underset{k = 1}{\sum^{\infty}} \frac{1}{k} \to \infty

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

t h e n t h e o r g i n a l s e i e s i s d i v e r g e n t \dots

c o m p a r i s o n t e s t \dots

Answered by mindispower last updated on 02/Jan/21

f : x \to x l n (1 + x) p o s i t i v i n c r e a s e f u n c t i o n

\int_{1}^{\infty} f (x) d x a n d Σ \frac{1}{n l n (1 + n)} s a m n a t u r e

\int_{1}^{\infty} \frac{1}{x l n (1 + x)} d x = B y p a r t {[\frac{l n (x)}{l n (1 + x)}]}_{1}^{\infty} - \int_{1}^{\infty} \frac{l n (x)}{(1 + x) {l n}^{2} (1 + x)}

w h e n x \to \infty

\frac{l n (x)}{(1 + x) {l n}^{2} (1 + x)} \sim \frac{1}{(1 + x) l n (1 + x)}

x \to \frac{1}{(1 + x) l n (1 + x)} n o t c v i n + \infty

\int \frac{d x}{(1 + x) l n (1 + x)} = l n (l n (1 + x) \to \infty

\Rightarrow Σ \frac{1}{n l n (1 + n)} D v