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 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 with $n$ being a positive or negative integer (but not zero) we have:

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

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 decreases from 2 to 1.5 to 1.33333… eventually tending to 1.

Case B: The partial sums oscillate but eventually settle down at 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’s dissect what happening here and see if we can obtain an approximate value of this series. We start at one and then subtract a half ending up at $\frac{1}{2}$. Next we add one third which takes us to $\frac{1}{2} + \frac{1}{3} =\frac{5}{6}$ or approximately 0.83. Next we subtract a quarter or 0.25 so we drop to 0.58. Next we add a fifth or 0.20 and we get to 0.78. The exact value of the series must lie between any two consecutive partial sums. Do you see why? Thus the series can be approximated by the average of any two consecutive partial sums. We take the average of 0.58 and 0.78 to get a final estimate of approximately 0.68 for the sum of the alternating harmonic series. Next we show how to calculate the sum exactly with a little integration.

Start with the geometric series

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

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 $\mathbin{-}1 \le x \le 1$. 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 uncover them. In Equation 1 we make the substitution $r = y^n$ where $n$ is a positive integer $n = 1, 2, 3 . . .$

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

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

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

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

Note on the right hand side the exponents and denominators form the sequence 1, $n +1$, $2n +1$, $3n + 1$, $4n + 1\cdots$ which is an A.P. 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 and it is valid for a radius of convergence of $|x|\le1$. If we substitute $x = 1$ then because arctan(1) is 45 degrees or $\frac{\pi}{4}$ radians 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 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 Equation (2) above gives us

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

We now give specific instances this formula for $n =$ 1, 2, 3, 4, 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 $\lim_{n\to\infty}I_n = 1$ as should be evident from equation (3). In case you are wondering how we evaluated the integrals $I_3$, $I_4$ and $I_5$ see the exercise 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 to show that

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