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1.3 Summation of series

Syllabus
9231–2028–2029
Topic
1.3
Level
AS

Expand first, then apply the three standard power sums

\sum_{r=1}^{n}r=\frac{n(n+1)}2,\qquad \sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}6,\qquad \sum_{r=1}^{n}r^3=\left[\frac{n(n+1)}2\right]^2.

Expand and collect the general term as ar3+br2+cr+da r^3+b r^2+c r+d. Use sigma linearity, replace each power sum by its standard result, and remember ∑r=1nd=dn\sum_{r=1}^{n}d=dn. Factor or simplify only after all four contributions are present.

\sum_{r=1}^{n}(2r-3)(r+4)=\sum_{r=1}^{n}(2r^2+5r-12)=\frac{n(n+1)(2n+1)}3+\frac{5n(n+1)}2-12n=\frac{n(4n^2+21n-55)}6.

If the lower limit is not 1, adjust it explicitly: for example, ∑r=3ng(r)=∑r=1ng(r)−g(1)−g(2)\sum_{r=3}^{n}g(r)=\sum_{r=1}^{n}g(r)-g(1)-g(2). If the upper limit is a number, substitute it only after obtaining the correct finite formula.

The required standard results are for rr, r2r^2 and r3r^3, not the geometric-series formula. Never square or cube ∑r\sum r to obtain ∑r2\sum r^2 or ∑r3\sum r^3; only the displayed identities are valid.

Expose a shifted difference and keep every boundary term

Rewrite the general term as ur−ur+ku_r-u_{r+k} (or the reverse), often by partial fractions. Write at least the first k+1 terms and the last k+1 terms. Cancel identical interior terms with their signs visible. What remains is determined by the initial and final boundaries, not by a guessed pattern.

\frac1{(r+1)(r+3)}=\frac12\left(\frac1{r+1}-\frac1{r+3}\right).

\sum_{r=1}^{n}\frac1{(r+1)(r+3)}=\frac12\left(\frac12+\frac13-\frac1{n+2}-\frac1{n+3}\right).Theshiftis2,sotwoinitialandtwofinalfractionssurvive.The shift is 2, so two initial and two final fractions survive.

Check the formula at n=1n=1: the right side must equal 1/(2⋅4)=1/81/(2\cdot4)=1/8. This catches an incorrect final index or a missing boundary term. If a denominator is zero at an index, the original sum is undefined there; do not telescope across that index.

Cancellation happens only between equal terms with opposite signs after the sum is expanded. A lag-one example leaves one term at each end, but a lag-k difference generally leaves k terms at each end.

Convergence is the finite limit of the exact partial sum

For $S_n=\sum_{r=1}^{n}a_r$, the infinite series converges to $L$ exactly when\lim_{n\to\infty}S_n=Lexists and is finite. Then $\sum_{r=1}^{\infty}a_r=L$.

FromthefiniteresultFrom the finite resultS_n=\frac12\left(\frac12+\frac13-\frac1{n+2}-\frac1{n+3}\right),bothfinalfractionstendto0,soboth final fractions tend to 0, so\sum_{r=1}^{\infty}\frac1{(r+1)(r+3)}=\frac12\left(\frac12+\frac13\right)=\frac5{12}.

If SnS_n contains a parameter power such as xnx^n, first find the values of x for which every n-dependent part has a limit. Usually xn→0x^n\to0 for ∣x∣<1|x|<1; test x=1x=1 and x=−1x=-1 separately in the exact SnS_n, because a cancellation may make an endpoint converge even when the generic rule does not.

Convergence forces an=Sn−Sn−1→0a_n=S_n-S_{n-1}\to0, but an→0a_n\to0 alone does not prove that SnS_n settles. The required argument here is direct: derive SnS_n, take its limit, state the parameter conditions, and only then give the sum to infinity.