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Base of Natural Logarithms

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E == Limit[(1/n + 1)^n, n -> Infinity] ==
Limit[Sum[(1/n)^i*Binomial[n, i], {i, 0, n}], n -> Infinity] ==
Limit[Sum[(n!*(1/n)^i)/(i!*(n - i)!), {i, 0, n}], n -> Infinity] ==
Limit[Sum[Product[(n - j)/n, {j, 0, i}]/i!, {i, 0, n}], n -> Infinity] ==
Limit[Sum[Product[1 - j/n, {j, 0, i}]/i!, {i, 0, n}], n -> Infinity] ==
Limit[Sum[1/i!, {i, 0, n}], n -> Infinity] ==
Sum[1/i!, {i, 0, Infinity}]


IP属地:广东1楼2013-07-25 23:10回复
    E^x == Limit[(1/m + 1)^m, m -> Infinity]^x ==
    Limit[(x/(m*x) + 1)^(m*x), m -> Infinity] ==
    Limit[(x/n + 1)^n, n -> Infinity] ==
    Limit[Sum[(x/n)^i*Binomial[n, i], {i, 0, n}], n -> Infinity] ==
    Limit[Sum[(x^i*(n!*(1/n)^i))/(i!*(n - i)!), {i, 0, n}], n -> Infinity] ==
    Limit[Sum[(x^i*Product[(n - j)/n, {j, 0, i}])/i!, {i, 0, n}], n -> Infinity] ==
    Limit[Sum[(x^i*Product[1 - j/n, {j, 0, i}])/i!, {i, 0, n}], n -> Infinity] ==
    Limit[Sum[x^i/i!, {i, 0, n}], n -> Infinity] ==
    Sum[x^i/i!, {i, 0, Infinity}]


    IP属地:广东4楼2013-07-26 00:28
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