We now turn our attention to convergent series. There are two cases.

Case A: The partial sums grow but the rate of growth slows down fast enough for the sequence to approach a finite limit. Examples are various geometric series satisfying $0 < |r| < 1$ such as:

$$S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots=\frac{1}{1\mathbin{-}\frac{1}{2}}=2$$

$$S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\cdots=\frac{1}{1\mathbin{-}\frac{1}{3}}=\frac{3}{2}$$

$$S=1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}+\cdots=\frac{1}{1\mathbin{-}\frac{1}{4}}=\frac{4}{3}$$

and in general

$$S=1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\frac{1}{n^4}+\cdots=\frac{1}{1\mathbin{-}\frac{1}{n}}=\frac{n}{n\mathbin{-}1}$$

where n is a non-zero integer (positive or negative)

Note how as the common ratio goes from $\frac{1}{2}$ to $\frac{1}{3}$ to $\frac{1}{4}$ eventually tending to zero, the sum of the series goes from 2.0 to 1.5 to 1.33333. . . eventually tending to 1.

Case B: The partial sums oscillate eventually settling down to a finite value.

Consider the alternating harmonic series (alternating because its terms are alternately positive and negative)

$$S = 1\mathbin{-} \frac{1}{2} + \frac{1}{3} \mathbin{-}\frac{1}{4} +\frac{1}{5} + \cdots$$

Let us dissect what is happening here and try to find an estimate for the sum of the this series. Starting at 1 we subtract $\frac{1}{2}$ and end up at $\frac{1}{2}$. Next we add $\frac{1}{3}$ which takes us to$\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ or approximately 0.83. Next we subtract $\frac{1}{4}=0.25$ and drop to 0.58. Next we add $\frac{1}{5}=0.20$ and jump to 0.78. Now the exact value of the series must lie between any two consecutive partial sums. Do you see why? So taking the average of 0.58 and 0.78 we arrive at a final estimate of 0.68 for the sum of the alternating harmonic series.

Next we go one level higher and calculate the exact sum using calculus.

We start with the geometric series

$$ \frac{1}{1 + r} = 1\mathbin{-}r + r^2\mathbin{-}r^3 + r^4\mathbin{-}\cdots$$

We integrate with respect to $dr$ between zero and $x$

$$\int_0^x\frac{dr}{1 + r} = \int_0^x(1\mathbin{-} r + r^2\mathbin{-}r^3 + r^4\mathbin{-}\cdots)dr$$

$$ \left[\ln(1 + r)\right]_0^x = \left[r \mathbin{-}\frac{r^2}{2} + \frac{r^3}{3}\mathbin{-}\frac{r^4}{4} + \cdots\right]_0^x$$

$$\ln(1 + x) = x \mathbin{-} \frac{x^2}{2} + \frac{x^3}{3}\mathbin{-}\frac{x^4}{4} + \cdots$$

The geometric series we started with is valid for 0 < |r| < 1 so we might expect the above expression for ln(1 + x) to be valid only for 0 < |x| < 1. Actually is can be shown that this expression is valid for $0\leq |x| \le 1$. Its radius of convergence is said to be 1. Substituting $x=\mathbin{-}1$, $x = 0$ and $x = 1$ yields the following 

$$\ln0 = \mathbin{-}1\mathbin{-} \frac{1}{2}\mathbin{-}\frac{1}{3}\mathbin{-}\frac{1}{4}\mathbin{-}\frac{1}{5} +\cdots = -\infty$$

$$\ln1 = 0\mathbin{-}\frac{0^2}{2} + \frac{0^3}{3}\mathbin{-}\frac{0^4}{4} + \cdots = 0$$

$$\ln2 = 1\mathbin{-}\frac{1}{2} + \frac{1}{3}\mathbin{-} \frac{1}{4} + \cdots = 0.693147$$

The last result vindicates our previous estimate of 0.68 for the sum of the alternating harmonic series. There are some rich ideas lurking here so let’s take some time and uncover them. In the geometric series we started with above make the substitution $r=y^n$ where $n=1,2,3. . .$ is a positive integer to get

$$\frac{1}{1+y^n}=1\mathbin{-}y^n+y^{2n}\mathbin{-}y^{3n}+y^{4n}\mathbin{-}\cdots$$ 

Next we integrate both sides with respect to $dy$ from 0 to $x$ yielding

$$\int_0^x\frac{dy}{1+y^n}=\int_0^x\left(1\mathbin{-}y^n+y^{2n}\mathbin{-}y^{3n}+y^{4n}\mathbin{-}\cdots\right)$$

$$\int_0^x\frac{dy}{1+y^n}=x\mathbin{-}\frac{x^{n+1}}{n+1}+\frac{x^{2n+1}}{2n+1}\mathbin{-}\frac{x^{3n+1}}{3n+1}+\frac{x^{4n+1}}{4n+1}\mathbin{-}\cdots$$

Note on the right hand side the exponents and denominators 1, $n+1$, $2n+1$, $3n+1$, $4n+1. . .$ are in arithmetic progression with first term $a=1$ and common difference $d=n$. Substituting $n = 2$ gives:

$$\int_0^x\frac{dy}{1+y^2}=\arctan(x) = x\mathbin{-} \frac{x^3}{3} + \frac{x^5}{5}\mathbin{-}\frac{x^7}{7} + \frac{x^9}{9}\mathbin{-} \cdots$$

This little gem is known as Gregory’s Series after James Gregory (1638$\mathbin{-}$1675) who found series expansions of several trigonometrical functions. It is valid for a radius of convergence of $|x|\le1$. If we substitute $x = 1$ then since arctan(1) is $\frac{\pi}{4}$ radians (or $45^{\circ}$) we have:

$$\frac{\pi}{4} = 1\mathbin{-}\frac{1}{3} + \frac{1}{5}\mathbin{-}\frac{1}{7} + \frac{1}{9}\mathbin{-} \cdots$$

This is Leibniz’s alternating series for $\pi$, named after the great German mathematician and philosopher Gottfried Wilhelm Leibniz (1646$\mathbin{-}$1716) who was one of the co-inventors of calculus with Sir Isaac Newton. The series is extremely slow in converging and is not very useful as a means to estimate $\pi$ (a task for which other relations are more suitable).

Setting $x =1$ in the equation for $\int_0^x\frac{dy}{1+y^n}$ above gives us

$$I_n = \int_0^1\frac{dy}{1 + y^n} = 1\mathbin{-} \frac{1}{n + 1} + \frac{1}{2n + 1}\mathbin{-}\frac{1}{3n + 1}+ \cdots$$

From this equation we can see that as $n$ tends to infinity, the integral tends to one. We now give specific instances of this formula for $n = $1, 2, 3, 4 and 5

$$I_1 = \int_0^1\frac{dy}{1 + y} = \ln2 = 0.693147. . . = 1\mathbin{-} \frac{1}{2} +\frac{1}{3}\mathbin{-}\frac{1}{4} + \cdots$$

$$I_2=\int_0^1\frac{dy}{1+y^2}=\frac{\pi}{4}=0.785398. . . =1\mathbin{-}\frac{1}{3}+\frac{1}{5}\mathbin{-}\frac{1}{7}+\cdots$$

$$I_3 =\int_0^1\frac{dy}{1 + y^3} = \frac{\ln2}{3}+\frac{\pi\sqrt{3}}{9} = 0.835649. . . = 1\mathbin{-}\frac{1}{4} + \frac{1}{7}\mathbin{-}\frac{1}{10} + \cdots$$

$$I_4 =\int_0^1\frac{dy}{1 + y^4} = \frac{\sqrt{2}}{8}(\pi + \ln(3 + 2\sqrt{2})) = 0.866973. . . = 1\mathbin{-} \frac{1}{5} + \frac{1}{9}\mathbin{-} \frac{1}{13} + \cdots $$

$$I_5 =\int_0^1\frac{dy}{1 + y^5} =\frac{\ln2}{5}+\frac{\sqrt{5}}{20}\ln\frac{3+\sqrt{5}}{3\mathbin{-}\sqrt{5}}+\frac{\pi}{50}\left(2\sqrt{10\mathbin{-}2\sqrt{5}}+\sqrt{10+2\sqrt{5}}\right)= 0.888313. . . = 1\mathbin{-} \frac{1}{6} + \frac{1}{11} \mathbin{-}\frac{1}{16} + \cdots $$

Looking at the decimal value of the integrals, we can somewhat appreciate that $\lim_{n\to\infty}I_n = 1$. In case you are wondering how we evaluated the integrals $I_3$, $I_4$ and $I_5$ see the problems below for an illustration of the general technique.

Exercises:

1. Show that 

$$\int_0^1\frac{2x\mathbin{-}1}{x^2\mathbin{-}x+1}dx=0$$

Hint: Use $$\int\frac{f^{\prime}(x)}{f(x)}dx=\ln(f(x))+C$$

2. Show that

$$\int_0^1\frac{dx}{x^2\mathbin{-}x+1} = \frac{2\pi}{3\sqrt{3}}$$

Hint: complete the square in the denominator then make a simple substitution to put the integrand in the following form

$$\int\frac{du}{u^2+a^2}=\frac{1}{a}\arctan\left(\frac{u}{a}\right)+C$$

3. Show that

$$\int_0^1\frac{x\mathbin{-}2}{x^2\mathbin{-}x+1}dx =\mathbin{-}\frac{\pi}{\sqrt{3}}$$

Hint: write $x\mathbin{-}2$ as $\frac{1}{2}(2x\mathbin{-}1\mathbin{-}3)$ and break the integrand into two parts that you can integrate separately using results established from questions (1) and (2)

4. We can factorize $1+x^3$ as

$$1+x^3=(1+x)(x^2\mathbin{-}x+1)$$

Thus using the method of partial fractions we can write

$$\frac{1}{1+x^3}=\frac{A}{1+x}+\frac{Bx+C}{x^2\mathbin{-}x+1}$$

Set $x=0, 1, 2$ in turn to find a system of three equations and solve it to show that $A=\frac{1}{3}$, $B=\mathbin{-}\frac{1}{3}$, $C=\frac{2}{3}$

5. Finally use the partial fraction decomposition and the result from question 4 to show that

$$\int_0^1\frac{dx}{1+x^3}=\frac{\ln2}{3}+\frac{\pi\sqrt{3}}{9}$$