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Question Number 79947 by mr W last updated on 29/Jan/20
Find ((ln 2)/(2!))+((ln 3)/(3!))+((ln 4)/(4!))+...+((ln n)/(n!))+...=?
$${Find} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}!}+\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}!}+\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{4}!}+…+\frac{\mathrm{ln}\:{n}}{{n}!}+…=? \\ $$
Commented by mr W last updated on 29/Jan/20
thanks again! in deed to find the limit numerically we only need to look the first few terms. taking the first 20 terms i get P≈1.829 024 679 563 571 8.
$${thanks}\:{again}! \\ $$$${in}\:{deed}\:{to}\:{find}\:{the}\:{limit}\:{numerically} \\ $$$${we}\:{only}\:{need}\:{to}\:{look}\:{the}\:{first}\:{few} \\ $$$${terms}.\:{taking}\:{the}\:{first}\:\mathrm{20}\:{terms}\:{i} \\ $$$${get}\:{P}\approx\mathrm{1}.\mathrm{829}\:\mathrm{024}\:\mathrm{679}\:\mathrm{563}\:\mathrm{571}\:\mathrm{8}. \\ $$
Commented by jagoll last updated on 29/Jan/20
Σ_(n=2) ^∞ ((ln(n))/(n!)) tobe continue
$$\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{ln}\left(\mathrm{n}\right)}{\mathrm{n}!}\: \\ $$$$\mathrm{tobe}\:\mathrm{continue} \\ $$
Commented by MJS last updated on 29/Jan/20
I stay with Π_(k=2) ^n k^(1/(k!)) the primes′ potencies sum up as follows (for n=25) 2: (1/(2!))+(2/(4!))+(1/(6!))+(3/(8!))+(1/(10!))+(2/(12!))+(1/(14!))+(4/(16!))+(1/(18!))+(2/(20!))+(1/(22!))+(3/(24!))≈.584797 3: (1/(3!))+(1/(6!))+(2/(9!))+(1/(12!))+(1/(15!))+(2/(18!))+(1/(21!))+(1/(24!))≈.168061 5: (1/(5!))+(1/(10!))+(1/(15!))+(1/(20!))+(2/(25!))≈.00833361 7: (1/(7!))+(1/(14!))+(1/(21!))≈.000198413 11: (1/(11!))+(1/(22!))≈2.5×10^(−8) 13: (1/(13!))≈1.6×10^(−10) 17: (1/(17!))≈2.8×10^(−15) 19: (1/(19!))≈8.2×10^(−18) 23: (1/(23!))≈3.9×10^(−23) maybe we can find the limits of the first few of these
$$\mathrm{I}\:\mathrm{stay}\:\mathrm{with}\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}{k}^{\frac{\mathrm{1}}{{k}!}} \\ $$$$\mathrm{the}\:\mathrm{primes}'\:\mathrm{potencies}\:\mathrm{sum}\:\mathrm{up}\:\mathrm{as}\:\mathrm{follows} \\ $$$$\left(\mathrm{for}\:{n}=\mathrm{25}\right) \\ $$$$ \\ $$$$\mathrm{2}:\:\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{2}}{\mathrm{4}!}+\frac{\mathrm{1}}{\mathrm{6}!}+\frac{\mathrm{3}}{\mathrm{8}!}+\frac{\mathrm{1}}{\mathrm{10}!}+\frac{\mathrm{2}}{\mathrm{12}!}+\frac{\mathrm{1}}{\mathrm{14}!}+\frac{\mathrm{4}}{\mathrm{16}!}+\frac{\mathrm{1}}{\mathrm{18}!}+\frac{\mathrm{2}}{\mathrm{20}!}+\frac{\mathrm{1}}{\mathrm{22}!}+\frac{\mathrm{3}}{\mathrm{24}!}\approx.\mathrm{584797} \\ $$$$\mathrm{3}:\:\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{6}!}+\frac{\mathrm{2}}{\mathrm{9}!}+\frac{\mathrm{1}}{\mathrm{12}!}+\frac{\mathrm{1}}{\mathrm{15}!}+\frac{\mathrm{2}}{\mathrm{18}!}+\frac{\mathrm{1}}{\mathrm{21}!}+\frac{\mathrm{1}}{\mathrm{24}!}\approx.\mathrm{168061} \\ $$$$\mathrm{5}:\:\frac{\mathrm{1}}{\mathrm{5}!}+\frac{\mathrm{1}}{\mathrm{10}!}+\frac{\mathrm{1}}{\mathrm{15}!}+\frac{\mathrm{1}}{\mathrm{20}!}+\frac{\mathrm{2}}{\mathrm{25}!}\approx.\mathrm{00833361} \\ $$$$\mathrm{7}:\:\frac{\mathrm{1}}{\mathrm{7}!}+\frac{\mathrm{1}}{\mathrm{14}!}+\frac{\mathrm{1}}{\mathrm{21}!}\approx.\mathrm{000198413} \\ $$$$\mathrm{11}:\:\frac{\mathrm{1}}{\mathrm{11}!}+\frac{\mathrm{1}}{\mathrm{22}!}\approx\mathrm{2}.\mathrm{5}×\mathrm{10}^{−\mathrm{8}} \\ $$$$\mathrm{13}:\:\frac{\mathrm{1}}{\mathrm{13}!}\approx\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{10}} \\ $$$$\mathrm{17}:\:\frac{\mathrm{1}}{\mathrm{17}!}\approx\mathrm{2}.\mathrm{8}×\mathrm{10}^{−\mathrm{15}} \\ $$$$\mathrm{19}:\:\frac{\mathrm{1}}{\mathrm{19}!}\approx\mathrm{8}.\mathrm{2}×\mathrm{10}^{−\mathrm{18}} \\ $$$$\mathrm{23}:\:\frac{\mathrm{1}}{\mathrm{23}!}\approx\mathrm{3}.\mathrm{9}×\mathrm{10}^{−\mathrm{23}} \\ $$$$\mathrm{maybe}\:\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{the}\:\mathrm{limits}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{few} \\ $$$$\mathrm{of}\:\mathrm{these} \\ $$
Commented by abdomathmax last updated on 29/Jan/20
you are right sir i have commited a error because i hsve used this inequality without proof...
$${you}\:{are}\:{right}\:{sir}\:{i}\:{have}\:{commited}\:{a}\:{error}\:{because} \\ $$$${i}\:{hsve}\:{used}\:{this}\:{inequality}\:{without}\:{proof}… \\ $$

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