# values of $(1+1/n)^{n}$ for $0

The following table gives the numerator and denominator of $\left(1+{1\over n}\right)^{n}$ as well as the decimal expansion to 20 places.

 $n$ Numerator of $\left(1+{1\over n}\right)^{n}$ Denominator of $\left(1+{1\over n}\right)^{n}$ Decimal value of $\left(1+{1\over n}\right)^{n}$ 1 2 1 2.0000000000000000000 2 9 4 2.2500000000000000000 3 64 27 2.3703703703703703704 4 625 256 2.4414062500000000000 5 7776 3125 2.4883200000000000000 6 117649 46656 2.5216263717421124829 7 2097152 823543 2.5464996970407131139 8 43046721 16777216 2.5657845139503479004 9 1000000000 387420489 2.5811747917131971820 10 25937424601 10000000000 2.5937424601000000000 11 743008370688 285311670611 2.6041990118975308782 12 23298085122481 8916100448256 2.6130352902246781603 13 793714773254144 302875106592253 2.6206008878857322211 14 29192926025390625 11112006825558016 2.6271515563008693884 15 1152921504606846976 437893890380859375 2.6328787177279190470 16 48661191875666868481 18446744073709551616 2.6379284973665998588 17 2185911559738696531968 827240261886336764177 2.6424143751831096203 18 104127350297911241532841 39346408075296537575424 2.6464258210976854673 19 5242880000000000000000000 1978419655660313589123979 2.6500343266404449073 20 278218429446951548637196401 104857600000000000000000000 2.6532977051444201339 21 15519448971100888972574851072 5842587018385982521381124421 2.6562632139261049855 22 907846434775996175406740561329 341427877364219557396646723584 2.6589698585377882029 23 55572324035428505185378394701824 20880467999847912034355032910567 2.6614501186387814545 24 3552713678800500929355621337890625 1333735776850284124449081472843776 2.6637312580685940367 25 236773830007967588876795164938469376 88817841970012523233890533447265625 2.6658363314874199930

With a large enough value of $n$, this formula approximates the natural log base $e$. For example, with $n$ set to ten million, we get 2.7182816925449662712, which is 0.000000135914078964161737245574 short of 2.7182818284590452354 (this calculation took almost four minutes with Mathematica 4.2). It is less computationally intensive to use

 $\sum_{i=0}^{n}\frac{1}{i!},$

which with $n$ set to 100 gives in less than a second a result to 20 places that is indistinguishable from $e$.

Title values of $(1+1/n)^{n}$ for $0 ValuesOf11nnFor0N26 2013-03-22 17:02:26 2013-03-22 17:02:26 PrimeFan (13766) PrimeFan (13766) 5 PrimeFan (13766) Data Structure msc 33B99