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Math Help - Generic Summation questions

  1. #1
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    Generic Summation questions

    I need help with these. Decide whether or not each of the following statements is true all of the time. If it's true, write an explanation or a proof. If it's false, give a counterexample. Assume (a sub n) and (b sub n) > 0 for all n.


    If summation 1 to infinity of (a sub n) = s, then summation 1 to infinity of (1/ a sub n) = 1/s.

    If summation (a sub n) is a convergent series, then so is summation ((a sub n)^2).

    If summation (a sub n) and summation (b sub n) are both divergent series where (a sub n) doesn't equal (b sub n), then so is the summation (a sub n - b sub n).

    If summation [(a sub n)/(b sub n)] is a divergent series, then summation (b sub n) must be a divergent series.

    If summation 1 to infinity of (a sub n) = A and summation 1 to infinity of (b sub n) = B, then summation 1 to infinity of [(a sub n)*(b sub n)] = AB.

    I thought the first one was false and I gave the example (a sub n) = square root of n.

    I guessed that the last one is false because you'd be multiplying the numbers of the series with summation [(a sub n)*(b sub n)] and with AB, you be multiplying the overall sum. I don't know if I'm right though.

    Anyways, I did 15 of them and I'm now stuck on these 5. Thanks a lot guys and gals!
    Last edited by thegame189; April 17th 2008 at 02:54 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by thegame189 View Post
    I need help with these. Decide whether or not each of the following statements is true all of the time. If it's true, write an explanation or a proof. If it's false, give a counterexample. Assume (a sub n) and (b sub n) > 0 for all n.


    If summation 1 to infinity of (a sub n) = s, then summation 1 to infinity of (1/ a sub n) = 1/s.
    This is not true as to converge the terms in an infinite sum must go to zero, and so their reciprocals must go to infinity, and so the series of reciprocals cannot converge.

    RonL
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    Quote Originally Posted by thegame189 View Post
    If summation (a sub n) is a convergent series, then so is summation ((a sub n)^2).
    If the series \sum_1^{\infty}a_n converges \lim_{n \to \infty}a_n=0 , so there exists n_0 such that for all n>n_0,\ |a_n|<1, and so |a_n^2|<|a_n| .

    Hence \sum_1^{\infty}a_n^2 converges.

    RonL
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    Quote Originally Posted by thegame189 View Post
    If summation (a sub n) and summation (b sub n) are both divergent series where (a sub n) doesn't equal (b sub n), then so is the summation (a sub n - b sub n).
    Consider a_n=1/n, b_n=1/(n+1)

    RonL
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    Quote Originally Posted by thegame189 View Post

    If summation [(a sub n)/(b sub n)] is a divergent series, then summation (b sub n) must be a divergent series.
    Consider a_n=1/n, b_n=1/n^2

    RonL
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    Quote Originally Posted by thegame189 View Post
    If summation 1 to infinity of (a sub n) = A and summation 1 to infinity of (b sub n) = B, then summation 1 to infinity of [(a sub n)*(b sub n)] = AB.
    If this were true it would be true for finite sums. In particular it would be true that:

     <br />
a_1b_1+a_2b_2=(a_1+a_2)(b_1+b_2)<br />

    RonL
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    Quote Originally Posted by thegame189 View Post
    I need help with these. Decide whether or not each of the following statements is true all of the time. If it's true, write an explanation or a proof. If it's false, give a counterexample. Assume (a sub n) and (b sub n) > 0 for all n.
    [snip]

    If summation (a sub n) is a convergent series, then so is summation ((a sub n)^2).

    [snip]
    Abel's test is a simple (and the laziest) way of proving that this statement is true.
    Last edited by mr fantastic; April 18th 2008 at 05:08 AM.
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