{
  "icon" : "sum",
  "id" : "51B1395A-3F4D-5D02-A810-6038C387A70D",
  "lines" : [
    {
      "id" : "7722BCD4-99B7-5F7D-A485-4E81CAA56C17",
      "source" : "# Geometric series"
    },
    {
      "id" : "5FE5FB37-DC15-593D-8406-5DF70FFD5961",
      "source" : "\/\/ The partial sums of 1\/2 + 1\/4 + 1\/8 + … close in on 1."
    },
    {
      "id" : "62A59C32-485C-5FD4-82B8-19C3C4EC1132",
      "source" : "sum(1\/2^n, n, 1, 10)"
    },
    {
      "id" : "977B7E45-8E6D-58E5-9CA8-CF26159E3EE3",
      "source" : "sum(1\/2^n, n, 1, 50)"
    },
    {
      "id" : "48B1D9C3-63BB-55BD-B31C-32E621906D15",
      "source" : ""
    },
    {
      "id" : "771CE978-DD28-5352-90CE-D016851E7590",
      "source" : "# The p-series"
    },
    {
      "id" : "48E2C179-6E7D-5D12-AE97-A1A12F336C45",
      "source" : "\/\/ 1 + 1\/4 + 1\/9 + … converges to π²\/6."
    },
    {
      "id" : "88A12FFF-562A-562C-B94F-C095D4204574",
      "source" : "sum(1\/n², n, 1, 1000)"
    },
    {
      "id" : "B9CB4549-8B5D-5BDD-80C6-6DCC16FE6EDB",
      "source" : "pi²\/6"
    },
    {
      "id" : "3C942A8D-C919-511D-8D9D-25AEFD434B82",
      "source" : "sum(1\/n², n, 1, 1000) - pi²\/6"
    },
    {
      "id" : "F4F723A9-6E8D-5C54-9732-F0DEE1DD186B",
      "source" : ""
    },
    {
      "id" : "A70BAC9B-C8DE-57CD-AB7B-F1A670452E17",
      "source" : "# The harmonic surprise"
    },
    {
      "id" : "27456558-14E8-51A7-BDD6-EF615EED7701",
      "source" : "\/\/ The terms 1\/n shrink to 0, yet the sum grows without bound, about as fast as ln(n)."
    },
    {
      "id" : "3F308A1A-ACF9-55DF-A757-325EE43BF6D3",
      "source" : "sum(1\/n, n, 1, 100)"
    },
    {
      "id" : "DF9CD3AC-6D63-59F2-B8AE-8AD434958334",
      "source" : "sum(1\/n, n, 1, 10000)"
    },
    {
      "id" : "AD1462A3-2B42-597B-9AD3-9AFACCFB36BB",
      "source" : "ln(10000)"
    },
    {
      "id" : "891E6DF9-E7B3-5BCD-A66F-2C1ED3D89B9F",
      "source" : ""
    },
    {
      "id" : "16F8B6BB-1A7D-587C-BE8B-B8B1B8237783",
      "source" : "# Telescoping"
    },
    {
      "id" : "F06F17ED-67A1-5903-9E15-8584EB0C5B36",
      "source" : "\/\/ 1\/(n(n+1)) = 1\/n - 1\/(n+1), so the sum collapses to 1 - 1\/101."
    },
    {
      "id" : "B30FFD9A-076F-5D64-8FB1-9E62A8993A21",
      "source" : "sum(1\/(n*(n+1)), n, 1, 100)"
    },
    {
      "id" : "DEB93F4B-39A8-59B2-A4FC-77C23A161BE8",
      "source" : "1 - 1\/101"
    }
  ],
  "numberStyle" : {
    "decimal" : ".",
    "grouping" : ",",
    "secondaryGroupSize" : 3
  },
  "title" : "Infinite series"
}