H2 Math Summation and Series: Study Guide & Examples

JC1 · H2 Mathematics · Study guide

Work through mixed sums, changed bounds, hence replacements, telescoping tails, selected terms and convergence tests with full index working.

Concepts · worked methods · graphs7 min read · practice linked below

Read the general term and both limits

Write the first and last terms before choosing a formula. A constant inside the sum is added once for every index value. Expand a product before splitting a summation: a sum of products is not the product of two sums. Use supplied square- or cube-sum formulae when the question provides them.

Split a mixed geometric and polynomial sum

The index in an exponent gives a GP

Given $\sum_{r=1}^n r^2=n(n+1)(2n+1)/6$ , find $S_n=\sum_{r=1}^n[2^{r+1}+3r-r^2]$ .

\begin{aligned}S_n&=2\sum_{r=1}^n2^r+3\sum_{r=1}^nr\\&\quad-\sum_{r=1}^nr^2,\\&=2\left[\frac{2(2^n-1)}{2-1}\right]\\&\quad+\frac{3n(n+1)}2\\&\quad-\frac{n(n+1)(2n+1)}6,\\&=4(2^n-1)\\&\quad+\frac{n(n+1)}6[9-(2n+1)],\\&=4(2^n-1)+\frac{n(n+1)(4-n)}3.\end{aligned}

Check $n=1$ : the original sum is $4+3-1=6$ . The formula gives $4+2=6$ .

Change the starting point by subtracting earlier terms

Both limits depend on n

Find $\sum_{r=n+1}^{2n}r(r-1)$ , for positive integer $n$ , using the supplied square-sum formula.

First find the sum from 1 to a general upper limit $m$ :

\begin{aligned}F(m)&=\sum_{r=1}^m(r^2-r),\\&=\frac{m(m+1)(2m+1)}6\\&\quad-\frac{m(m+1)}2,\\&=\frac{m(m+1)(m-1)}3.\end{aligned}

To retain terms $n+1$ through $2n$ , subtract $F(n)$ , not $F(n+1)$ :

\begin{aligned}\sum_{r=n+1}^{2n}r(r-1)&=F(2n)-F(n),\\&=\frac{2n(4n^2-1)}3\\&\quad-\frac{n(n^2-1)}3,\\&=\frac{n(7n^2-1)}3.\end{aligned}

For $n=2$ , the terms are $3(2)+4(3)=18$ , agreeing with the formula. There are exactly $n$ terms.

Variation: a constant depends on n but not on r

Find $\sum_{r=n+1}^{2n}(2^r+3^n)$ . The quantity $3^n$ is constant while $r$ varies.

Call this sum $G_n$ . Then

\begin{aligned}G_n&=\frac{2^{n+1}(2^n-1)}{2-1}+n3^n,\\&=2^{n+1}(2^n-1)+n3^n.\end{aligned}

The term $3^n$ occurs $n$ times; it is not a GP in $r$ .

Hence: make the term and both limits match

Shift the index, then remove the unwanted beginning

Given $F(n)=\sum_{r=1}^n(2r^3+5)=n^2(n+1)^2/2+5n$ , find $\sum_{r=2}^n[2(r+2)^3+5]$ , where $n\geq2$ .

Make the general term the same first. Put $s=r+2$ ; the lower limit becomes 4 and the upper limit becomes $n+2$ .

Call the required sum $T_n$ . Thus

\begin{aligned}T_n&=\sum_{s=4}^{n+2}(2s^3+5),\\&=F(n+2)-F(3).\end{aligned}

\begin{aligned}F(3)&=\frac{3^2(4)^2}{2}+5(3)\\&=87,\\F(n+2)&=\frac{(n+2)^2(n+3)^2}{2}\\&\quad+5(n+2).\end{aligned}

Hence the required sum is

\begin{aligned}\frac{(n+2)^2(n+3)^2}{2}+5n-77.\end{aligned}

At $n=2$ , there is only one term: $2(4)^3+5=133$ . The formula also gives 133. This checks both limits, where mistakes most often occur.

Method of differences: show what cancels

A finite sum, an infinite sum and an accuracy requirement

Find $S_n=\sum_{r=1}^n1/[r(r+1)]$ . Hence find $\sum_{r=5}^{n+1}1/[r(r+1)]$ , its sum to infinity, and the smallest $n$ for which that finite sum is within 0.01 of its sum to infinity.

\begin{aligned}\frac1{r(r+1)}&=\frac1r-\frac1{r+1},\\S_n&=\left(1-\frac12\right)\\&\quad+\left(\frac12-\frac13\right)\\&\quad+\cdots+\left(\frac1n-\frac1{n+1}\right),\\&=1-\frac1{n+1}.\end{aligned}

All the middle terms cancel. For $n\geq4$ ,

\begin{aligned}T_n&=S_{n+1}-S_4,\\&=\left(1-\frac1{n+2}\right)-\left(1-\frac15\right),\\&=\frac15-\frac1{n+2}.\end{aligned}

As $n\to\infty$ , the last fraction tends to zero, so $T_n\to1/5$ . Thus the series converges to $1/5$ . Its finite sum is always below this value.

\begin{aligned}\frac15-T_n&=\frac1{n+2},\\\frac1{n+2}&<0.01,\\n+2&>100,\\n&>98.\end{aligned}

The smallest integer is 99. At $n=98$ the difference is exactly 0.01; at $n=99$ it is $1/101<0.01$ .

See the partial sum and the remaining tail

Upper limit is n + 1; choose n:8

n = 8: the sum from 5 to 9 is 0.100000; the remaining tail is 0.100000.

Select odd terms by removing the even terms

Find a sum of odd cubes

Given $\sum_{r=1}^n r^3=n^2(n+1)^2/4$ , find $1^3+3^3+\cdots+(2k-1)^3$ .

The last odd term is below $2k$ . Sum all cubes up to $2k$ , then subtract $2^3,4^3,\ldots,(2k)^3$ .

Let $O_k$ denote the required odd-cube sum.

\begin{aligned}O_k&=\sum_{r=1}^{2k}r^3-8\sum_{r=1}^kr^3,\\&=\frac{(2k)^2(2k+1)^2}{4}\\&\quad-8\frac{k^2(k+1)^2}{4},\\&=k^2[(2k+1)^2-2(k+1)^2],\\&=k^2(2k^2-1).\end{aligned}

For $k=2$ , $1^3+3^3=28=2^2(2\cdot2^2-1)$ .

Variation: symmetric limits

For $\sum_{r=-n}^n(r^2+3r+5)$ , the $3r$ terms cancel in pairs; the square terms occur twice; the constant occurs $2n+1$ times.

Calling the symmetric sum $H_n$ ,

\begin{aligned}H_n&=2\sum_{r=1}^nr^2+5(2n+1),\\&=\frac{n(n+1)(2n+1)}3\\&\quad+5(2n+1).\end{aligned}

The zero index contributes the constant 5, so do not count only $2n$ terms.

Convergence concerns the sum, not only its terms

Combine two convergent series from their correct starting terms

Find $\sum_{r=2}^{\infty}[2^{-r}+1/(r(r-1))]$ .

The GP begins at $2^{-2}=1/4$ . Replacing $s=r-1$ in the second sum gives the telescoping form already used.

\begin{aligned}\sum_{r=2}^n\frac1{r(r-1)}&=\sum_{s=1}^{n-1}\frac1{s(s+1)},\\&=1-\frac1n,\\\sum_{r=2}^{\infty}2^{-r}&=\frac{1/4}{1-1/2}\\&=\frac12,\\\sum_{r=2}^{\infty}\frac1{r(r-1)}&=\lim_{n\to\infty}\left(1-\frac1n\right)\\&=1.\end{aligned}

The required sum is therefore $1/2+1=3/2$ . Each partial-sum limit is finite, so adding the limits is justified. Terms tending to zero are necessary for convergence, but are not by themselves sufficient.

Use an unfamiliar test exactly as supplied

Apply a given root test

A question supplies the root test: $\sum a_r$ converges if $\lim_{r\to\infty}\sqrt[r]{|a_r|}<1$ , and the test is inconclusive if that limit equals 1. Use it for $a_r=[(-2r^3+rx)/(5r^3+7)]^r$ , where $x$ is fixed.

\begin{aligned}\sqrt[r]{|a_r|}&=\left|\frac{-2r^3+rx}{5r^3+7}\right|,\\&=\left|\frac{-2+x/r^2}{5+7/r^3}\right|,\\\lim_{r\to\infty}\sqrt[r]{|a_r|}&=\left|-\frac25\right|\\&=\frac25<1.\end{aligned}

Thus the series converges for every fixed real $x$ . The question supplied the test; the task is to simplify its expression and compare the limit with 1. If the limit had equalled 1, this test would not decide convergence.

Before you finish

Does the formula use the right first term and number of terms?
Have both limits changed when the index changed?
Did you subtract through the term immediately before the new lower limit?
For a sum to infinity, have you taken a finite partial-sum limit or used the supplied test correctly?

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