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Sequences and series · Check convergence before summing to infinity

Choose between a term and a sum

A term, a finite total and a sum to infinity answer different questions. Name the requested quantity, then choose a formula and check its conditions.

  1. Name the quantity and starting index

    Distinguish a term from the accumulated total. Keep the first index and the number of terms separate.

    Sn=∑j=1nuj.S_n=\displaystyle\sum_{j=1}^n u_j.

    u1=S1.u_1=S_1.

    un=Sn−Sn−1(n≥2).u_n=S_n-S_{n-1}\quad(n\ge2).

    TermRecover one term from partial sums

    Use consecutive partial sums for n≥2n\ge2, then check u1=S1u_1=S_1 separately. A formula supplied only for positive indices must not be extended to index 0 without justification.

    Finite totalCount the included terms

    For integer bounds m≤Nm\le N, the sum from m to N contains N−m+1N-m+1 terms. The upper index is not necessarily the number of included terms. A limit of the terms unu_n is different from a limit of the sums SnS_n.

    Read the final ask again before calculating. A given infinite sum may be a constraint used to find a finite sum, rather than the answer being requested.

  2. Choose a justified finite-sum formula

    When a finite sum is required, identify the model. Here n is a positive integer, a the first term, d the AP common difference and r the GP common ratio.

    AP: Sn=n2(a+un).\text{AP: } S_n=\dfrac n2(a+u_n).

    GP: Sn=a(1−rn)1−r.\text{GP: } S_n=\dfrac{a(1-r^n)}{1-r}.

    GP quotient: r≠1.\text{GP quotient: }r\ne1.

    Arithmetic progressionUse a common difference

    With un=a+(n−1)du_n=a+(n-1)d, use Sn=n(a+un)/2S_n=n(a+u_n)/2. A few matching differences suggest a pattern; a proof or the stated model must establish the AP. One unequal pair can disprove it.

    Geometric progressionUse a common ratio and handle r=1

    For a known geometric model, use the displayed finite formula when r≠1r\ne1. It does not require ∣r∣<1|r|<1 for a fixed finite n. When r=1r=1, every term is a and Sn=naS_n=na; do not divide by zero.

    Reindex later starting points or subtract partial sums first. Constant sequences can fit both models. For a general recurrence or another series, use the supplied relation rather than assuming AP or GP.

  3. Check convergence before an infinite sum

    For a sum to infinity, test the limit of the partial sums. Terms tending to zero alone do not prove convergence of the series.

    S∞=lim⁡n→∞Sn.S_\infty=\displaystyle\lim_{n\to\infty}S_n.

    GP: S∞=a1−r\text{GP: }S_\infty=\dfrac a{1-r}

    when ∣r∣<1.\text{when }|r|<1.

    Geometric modelCheck the ratio before using a/(1−r)

    With nonzero first term, ∣r∣<1|r|<1 makes rn→0r^n\to0 and gives the geometric sum to infinity. At r=−1r=-1 the partial sums alternate between a and 0; at r=1r=1 they equal na. Both boundary cases diverge for a≠0a\ne0, as do the cases ∣r∣>1|r|>1. The all-zero series separately has sum 0.

    Another seriesUse the partial sums actually given

    The geometric ratio condition is not a test for every series. If a finite partial-sum identity is supplied, take its limit with its stated hypotheses, as in Example 26. A finite table or calculator display is not a proof about infinitely many sums.

    A convergent term sequence and a convergent series are different claims. State which quantity has a finite limit; positive or negative infinity is not a finite sum to infinity.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

Example 12(c): check the first term separately

Given Sn=n2+3n+1S_n=n^2+3n+1 for positive integer n, recover the terms. Does the constant disappear from the whole sequence?

Show reasoning

  1. 1. Start with the actual first partial sum

    The supplied formula is for positive indices. The first term is S1S_1, so substituting zero into the quadratic would not supply a justified S0S_0.

    u1=S1=1+3+1=5.\begin{aligned}u_1&=S_1\\&=1+3+1=5.\end{aligned}

  2. 2. Subtract for later terms

    The constant cancels between consecutive partial sums. This calculation applies from n=2; it does not replace the separate first-term check.

    un=Sn−Sn−1=2n+2(n≥2).\begin{aligned}u_n&=S_n-S_{n-1}\\&=2n+2\quad(n\ge2).\end{aligned}

  3. 3. Keep the exception when classifying

    The first three terms are 5, 6 and 8. The consecutive differences 1 and 2 are unequal, and the ratios 6/56/5 and 4/34/3 are unequal. Thus the whole sequence is neither AP nor GP, even though the later terms have a common difference.

    un={5,n=1,2n+2,n≥2.u_n=\begin{cases}5,&n=1,\\2n+2,&n\ge2.\end{cases}
Continue with the lesson question

Example 27: recover the finite sum

A GP has second term 5 and sum to infinity 20. The question asks for the sum of the first eight terms. Which quantity is the final answer?

Show reasoning

  1. 1. Translate the givens into different quantities

    The second term gives ar=5. The stated finite sum to infinity requires |r|<1 and supplies a second equation; 20 is not the requested finite total.

    ar=5,a1−r=20,∣r∣<1.\begin{aligned}ar&=5,\\\frac a{1-r}&=20,\\|r|&<1.\end{aligned}

  2. 2. Find parameters and check the ratio

    Divide the first equation by the second, complete the square, then recover the first term. The resulting ratio satisfies the convergence condition.

    r(1−r)=14,(r−12)2=0,r=12,a=10.\begin{aligned}r(1-r)&=\frac14,\\\left(r-\frac12\right)^2&=0,\\r&=\frac12,\\a&=10.\end{aligned}

  3. 3. Return to the finite quantity requested

    Use eight terms in the finite GP formula. The result is below the given infinite sum, as expected here because all terms are positive.

    S8=20[1−(12)8]=127564.\begin{aligned}S_8&=20\left[1-\left(\frac12\right)^8\right]\\&=\frac{1275}{64}.\end{aligned}
Continue with the lesson question

Example 26: take the given partial-sum limit

The linked question supplies an exact expression for its finite partial sums. Decide what must approach a finite value to establish the series’ sum to infinity.

Show reasoning

  1. 1. Use the supplied finite identity

    The question gives this identity; it does not ask for a partial-fractions derivation. Keep both remainder terms and their signs.

    SN=14−12(N+1)+12(N+2).\begin{aligned}S_N&=\frac14-\frac1{2(N+1)}\\&\quad+\frac1{2(N+2)}.\end{aligned}

  2. 2. Take the limit of the whole partial sum

    Both reciprocal remainder terms tend to zero. The finite limit proves convergence and gives the sum to infinity. No geometric ratio has been assumed.

    lim⁡N→∞SN=14.\lim_{N\to\infty}S_N=\frac14.
Continue with the lesson question