Find ((ln 2)/(2!))+((ln 3)/(3!))+((ln 4)/(4!))+...+((ln n)/(n!))+...=?


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Question Number 79947 by mr W last updated on 29/Jan/20

$F i n d$ $\frac{\ln 2}{2!} + \frac{\ln 3}{3!} + \frac{\ln 4}{4!} + . . . + \frac{\ln n}{n!} + . . . = ?$
Commented by mr W last updated on 29/Jan/20

$t h a n k s a g a i n!$ $i n d e e d t o f i n d t h e l i m i t n u m e r i c a l l y$ $w e o n l y n e e d t o l o o k t h e f i r s t f e w$ $t e r m s . t a k i n g t h e f i r s t 20 t e r m s i$ $g e t P \approx 1.829 024 679 563 571 8 .$
Commented by jagoll last updated on 29/Jan/20

$\underset{n = 2}{\sum^{\infty}} \frac{\ln (n)}{n!}$ $tobe continue$
Commented by MJS last updated on 29/Jan/20

$I stay with \underset{k = 2}{\prod^{n}} k^{\frac{1}{k!}}$ $the {primes}^{'} potencies sum up as follows$ $(for n = 25)$ $2 : \frac{1}{2!} + \frac{2}{4!} + \frac{1}{6!} + \frac{3}{8!} + \frac{1}{10!} + \frac{2}{12!} + \frac{1}{14!} + \frac{4}{16!} + \frac{1}{18!} + \frac{2}{20!} + \frac{1}{22!} + \frac{3}{24!} \approx . 584797$ $3 : \frac{1}{3!} + \frac{1}{6!} + \frac{2}{9!} + \frac{1}{12!} + \frac{1}{15!} + \frac{2}{18!} + \frac{1}{21!} + \frac{1}{24!} \approx . 168061$ $5 : \frac{1}{5!} + \frac{1}{10!} + \frac{1}{15!} + \frac{1}{20!} + \frac{2}{25!} \approx . 00833361$ $7 : \frac{1}{7!} + \frac{1}{14!} + \frac{1}{21!} \approx . 000198413$ $11 : \frac{1}{11!} + \frac{1}{22!} \approx 2.5 \times 10^{- 8}$ $13 : \frac{1}{13!} \approx 1.6 \times 10^{- 10}$ $17 : \frac{1}{17!} \approx 2.8 \times 10^{- 15}$ $19 : \frac{1}{19!} \approx 8.2 \times 10^{- 18}$ $23 : \frac{1}{23!} \approx 3.9 \times 10^{- 23}$ $maybe we can find the limits of the first few$ $of these$
Commented by abdomathmax last updated on 29/Jan/20

$y o u a r e r i g h t s i r i h a v e c o m m i t e d a e r r o r b e c a u s e$ $i h s v e u s e d t h i s i n e q u a l i t y w i t h o u t p r o o f . . .$
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