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? asked in Science & MathematicsMathematics · 9 years ago

Test for convergence of the series.?

Q)

Summation from 1 to infinity

(1+(-1)^i) / (8i+2^i)

This series apparently converges and I can't figure out why.

Update:

Thanks for the reply.

From my understanding, just because the terms approach 0 doesn't necessarily imply convergence, as in the case of harmonic series (1/n).

Also, we can't directly compare the given series to the harmonic series since each of terms of our series is less than the corresponding terms of harmonic series, so we can't establish divergence like that.

But if we compare that to the Geometric series (1/2^i), each of the terms is indeed smaller and would thus imply convergence.

Problem is that I approached the question differently earlier and it gave a different answer. I took 'i' as common from the denominator, so it gave me:

(1+(-1)^i) / i(8+2^i/i)

Now the absolute value of this entire generalization is:

2/i(8+2^i/i)

Rearranging that would give:

2/i * (1/(8+2^i/i)

Now I thought that since 2/i would diverge, the entire series should diverge.

I still don't know what I did wrong with the above method.

2 Answers

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  • CogitoErgoCogitoSum
    Lv 7
    9 years ago
    Favorite Answer

    I dont know what your additional details is saying... but I will have a go at the series itself.

    For each term,

    a_i = (1+(-1)^i) / (8i+2^i)

    Since the numerator is at most a 2, we can directly compare to 2 / (8i + 2^i) and conclude that:

    (1+(-1)^i) / (8i+2^i) ≤ 2 / (8i + 2^i)

    As for 2 / (8i + 2^i), we can directly compare to 2 / 2^i. Since 8i+2^i is always larger than or equal to 2^i. Thus:

    2 / (8i + 2^i) ≤ 2 / 2^i

    Thus,

    (1+(-1)^i) / (8i+2^i) ≤ 2 / 2^i

    And 2/2^i is a convergent geometric series.

  • Johnny
    9 years ago

    I haven't done this for a while but I'll give it a try.

    For the numerator, it will alternating between 0 and 2 since (-1)^i if i is even then 1+1=2 and if i is odd then we have 1-1=0.

    For the denominator, it gonna blow up as i approaches infinity. However, since the numerator is smaller than the denominator, the fraction gonna get smaller (approaching zero). Hence it converges!

    I remember there is a test for it but I don't remember exactly what it is. Alternating maybe?

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