Why is the lim [(-1)^(n+1)]/n (n->infinty) =0
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Why is the lim [(-1)^(n+1)]/n (n->infinty) =0

[From: ] [author: ] [Date: 11-12-18] [Hit: ]
it is obvious that the denominator will become larger, hence the number will come closer and closer to 0.-you have n in denominator whose value is infinity.any thing divided by infinity is zero.......
If we list out the first few terms it might become apparent:

For n = 1 to 6,

1, -1/2, 1/3, -1/4, 1/5, -1/6 ...

As you can see, the denominator increases with 1 for each n, whilst the numerator remains of the same absolute value (albeit with a different sign). The larger n becomes, it is obvious that the denominator will become larger, hence the number will come closer and closer to 0.

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you have n in denominator whose value is infinity.
any thing divided by infinity is zero.
since anythin divided by zero is infinite
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keywords: lim,Why,infinty,the,is,gt,Why is the lim [(-1)^(n+1)]/n (n->infinty) =0
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