{"id":10710,"date":"2017-09-08t12:29:59","date_gmt":"2017-09-08t19:29:59","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=10710"},"modified":"2017-09-08t12:29:59","modified_gmt":"2017-09-08t19:29:59","slug":"ap-calculus-bc-review-series-convergence","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/ap-calculus-bc-review-series-convergence\/","title":{"rendered":"ap calculus bc review: series convergence"},"content":{"rendered":"

a series<\/strong> is the sum of infinitely many terms. but how can you add up an infinite<\/em> number of things? well, it turns out that sometimes we can do exactly that! but first we’ll need to understand the concepts of series convergence and divergence.<\/p>\n

\"series
infinite series convergence may seem mysterious at first, but it’s not<\/em> the work of aliens…. so far as i know.<\/figcaption><\/figure>\n

working with series<\/h2>\n

typically, a series is expressed either by writing out a few terms in order to establish a pattern, or by using sigma notation<\/strong>. <\/p>\n

\"series<\/p>\n

here, an<\/sub><\/em> is the general term<\/strong> for the series.<\/p>\n

check out the following article for more explanation as well as examples: ap calculus bc review: series fundamentals<\/a>.<\/p>\n

partial sums<\/h3>\n

any time that infinity is involved, we have to be extra careful with our definitions. so how do you add up infinitely many items? one item at a time! this is a job for partial sums<\/em>.<\/p>\n

a partial sum is something like a running total<\/em> for a series.<\/p>\n

let   \"series\"   be a series.<\/p>\n

the k<\/em>th partial sum<\/strong> of the series is:<\/p>\n

\"partial<\/p>\n

in other words, the k<\/em>th partial sum is the sum of the first k<\/em> terms of the series.<\/p>\n

the first four partial sums of a series are:<\/p>\n

\"first<\/p>\n

now, the useful thing about partial sums is that they form a sequence<\/em>,<\/p>\n

\"sequence<\/p>\n

and then we define series convergence in terms of the convergence of this sequence of partial sums.<\/p>\n

series convergence and divergence — definitions<\/h3>\n

a series σan<\/sub><\/em> converges<\/strong> to a sum s<\/em> if and only if the sequence of partial sums converges to s<\/em>. that is, a series converges if the following limit exists:<\/p>\n

\"definition<\/p>\n

otherwise, if the limit of sk<\/sub><\/em> (as k<\/em> → ∞) is infinite or fails to exist, then the series diverges<\/strong>.<\/p>\n

ways that a series could diverge<\/h3>\n

a series may diverge in three different ways:<\/p>\n

    \n
  1. the series diverges to infinity<\/strong> (∞) if the partial sums increase without bound. that is, sk<\/sub><\/em> → ∞.<\/li>\n
  2. the series diverges to negative infinity<\/strong> (-∞) if the partial sums decrease without bound. that is, sk<\/sub><\/em> → -∞.<\/li>\n
  3. if the limit of partial sums does not exists and is not ∞ or -∞, then we just say that the series diverges. this case occurs whenever the sequence of partial sums oscillates among multiple values or jumps around randomly.<\/li>\n<\/ol>\n

    limit test for divergence<\/h2>\n

    there is a quick and easy test that can be used to show that a series diverges. (however, it can’t be used to show a series converges.)<\/p>\n

    limit divergence test:<\/strong> if the limit of the general term of a series is not equal to 0, then the series diverges.<\/p>\n

    that is, if \"limit, then the series σan<\/sub><\/em> diverges.<\/p>\n

    unfortunately, if the limit does turn out to be zero, then the test is inconclusive<\/em>. then you’d have to use additional convergence tests to figure out series convergence or divergence.<\/p>\n

    let’s take a look at a few examples.<\/p>\n

    using the limit divergence test<\/h3>\n

    which of the following series must diverge according to the divergence test?<\/p>\n

    \"four<\/p>\n

    solutions<\/h4>\n

    remember, for this question, we’re only allowed to use the divergence test.<\/p>\n

    (a) as n<\/em> → ∞, the values of 2n<\/sup><\/em> also approach ∞. the bottom line is that the limit of 2n<\/sup><\/em> is certainly not equal to 0.<\/p>\n

    therefore, series (a) diverges.<\/p>\n

    (b) \"limit<\/p>\n

    here, the answer is not clear. a limit of 0 does not automatically mean that the series will not diverge. so the limit divergence test alone cannot say anything more about series convergence or divergence here.<\/p>\n

    in fact, we will prove that this series does converge by other methods.<\/p>\n

    (c) again, the divergence test is inconclusive, because 1\/n<\/em> → 0 as n<\/em> → ∞.<\/p>\n

    now this series actually diverges<\/em>. it’s called the harmonic series<\/strong>, and it’s just the sum of all of the reciprocals of the natural numbers.<\/p>\n

    \"harmonic<\/p>\n

    we’ll need another method to decide for sure that the harmonic series diverges. it’s far from obvious!<\/p>\n

    (d) \"limit<\/p>\n

    with a nonzero limit, the divergence test conclusively states that this series must diverge.<\/p>\n

    geometric series<\/h2>\n

    let’s take a closer look at examples (a) and (b). both of these are geometric<\/strong> series. <\/p>\n

    a geometric series is the sum of the powers of a constant base. <\/p>\n

    \"geometric<\/p>\n

    there is a straightforward test to decide whether any geometric series converges or diverges.<\/p>\n