It is given in CRM book, by MAA (A Radical Approach to Real Analysis By David M. Bressoud) in exercise of Q. 2.2.9 Q. 2.2.9

It is tempting to integrate each side of eqn. (2.11) wrt $x$ and to assert that

The eqn. (2.11) is for sum of infinite geometric series as given by:
$1+x+x^2+x^3+\cdots=\frac1{1-x}$, if $|x|\lt 1$.

$x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\cdot=-\ln(1-x).$ (2.13)

Following Cauchy's advice (on pg. 19), we know we need to be careful, but now we run into trouble.

This means that need take a finite lower bound on the number of terms ($n$). E.g., the eqn. (2.10) is for sum of geometric series, for a lower bound $n$ as given by:
$1+x+x^2+x^3+\cdots+x^{n-1}=\frac1{1-x} - \frac {x^n}{1-x}$, if $|x|\lt 1$.

What happens when we try to integrate $\frac{x^n}{(1-x)}$?

Doubt 1: Not clear about it, as need to find integration using usual methods. But, if take it as a stand-alone problem, then how to pursue/ attack it.

Fortunately, we do not have to find the exact value of the difference between $ x+\frac{x^2}2+\frac{x^3}3+\cdots+\frac{x^n}n$ and $-\ln(1-x)$. All we have to show is that we can make this difference as small as we wish by taking enough terms. We can do this by bounding the integral of $\frac{x^n}{(1-x)}$.

Need find bounds $a,b$ s.t. form limits for sum of enough terms, denoted by the difference $|b-a|$
$\int_a^b \frac{x^n}{(1-x)}dx$

We use the fact that if
$|f(x)|\lt g(x) $ for all x, then $|\int_a^b f(x)dx|\lt \int_a^b g(x)dx$.
If $0\lt x \lt 1$, then we can find a number $a$ so that $0\lt x\lt a\lt 1$ and
$|\frac{x^n}{1-x}| \lt x^n(1-a)^{-1}$

Doubt 2: Unable to see the significance of absolute value of limits of integrand obtained for $f(x)$.
The inequality $|f(x)|\lt g(x)$ shows that for function $f(x)$ the negative values taken are also summed up.
Is this a well-known result, if so what is its name?
Also, what is the significance of it here.

Integrate this bounding function w.r.t. $x$, & show that if $0\lt x \lt 1$, then the partial sums of the series in eqn. (2.13) approach the target value of $-\ln(1-x)$ as $x$ increases.

Doubt 3: Unable to pursue this computation. Need some clue.

Explain what happens for $-1\lt x\lt 0$. Justify your answer.

Doubt 4: Unable to find the significance of such bounds.

  • 1
    $\begingroup$ Judging from this question alone, it appears that your book is good and you are lucky to have it. Most calculus textbooks don't deal with this issue. $\endgroup$
    – Paramanand Singh
    Apr 29, 2019 at 18:02
  • 1
    $\begingroup$ The key is to analyze the integral $\int_{0}^{x}\frac{t^n}{1-t}\,dt$. If $0\leq x<1$ then we have $1/(1-t)<1/(1-x)$ for $0<t<x$. Thus the integral lies between $0$ and $x^{n+1}/((n+1)(1-x)) $ and this is less than $1/(n+1)(1-x)$ and so tends to $0$. You should do a similar analysis when $-1\leq x<0$ by putting $t=-u$ in integral. $\endgroup$
    – Paramanand Singh
    Apr 29, 2019 at 18:06
  • $\begingroup$ I am in process of following your advice for Doubt #4, as based on your clue for Doubt #3. But, am confused as how the term $\frac{x^n}{((n+1)(1-x))}$ came into picture from $|\frac{x^n}{1-x}| \lt x^n(1-a)^{-1}$. If could explain that, then understanding your clue & its appln. to Doubt #4 is easy. Also, request to supply answer (preferably) for Doubt #2. Also, as asked in there, is this a well known result? $\endgroup$
    – jiten
    Apr 30, 2019 at 0:18
  • 1
    $\begingroup$ First the absolute value signs are not necessarily as $x$ as well as $1-x$ are positive. The desired inequality as per my notation is $t^n/(1-t)<t^n(1-x)^{-1}$ and now integrate this inequality wrt $t$ and get desired result. $\endgroup$
    – Paramanand Singh
    Apr 30, 2019 at 2:48
  • $\begingroup$ Please give some hint as directly integrating the below is not helping : $|\frac{t^n}{1-t}| \lt t^n(1-x)^{-1} \implies |\frac{t^n}{(1-t)(1-x)}| \lt t^n$. Taking the lhs, get: $\implies |\int_a^b \frac{x^n}{(1-x)(1-a)}|$, but this is not leading further. I am also scared of the comptn. I am feeling that here answer will help a lot. $\endgroup$
    – jiten
    Apr 30, 2019 at 4:16

1 Answer 1


We could have used the definition of incomplete Beta function

$$B(z; a,b) = \int_0^z u^{a-1}(1-u)^{b-1}\, du$$

to express

$$\int_0^{z}\frac{u^n}{1-u}\, du = \int_0^z u^{(n+1)-1}(1-u)^{-1}\, du = B(z; n+1, 0).$$

but if you don't understand the property of incomplete Beta function, it doesn't give you good control.

Also, for $|x| < 1$,

$$\sum_{i=0}^{n-1}x^i = \frac1{1-x}- \frac{x^n}{1-x}$$

$$\int \frac{x^n}{1-x}\, dx = -\ln (1-x) - \sum_{i=1}^{n}\frac{x^i}{i}+C$$

which is exactly the error term.

The message that the book is trying to send is you don't have to work with this quantity directly.

$$\int_{0}^x \frac{t^n}{1-t}\, dt = \int_0^x\left(\frac{1}{1-t}-\sum_{i=0}^{n-1}t^i \right)\, dt=-\ln (1-x) - \sum_{i=1}^n \frac{x^i}{i}$$

The error term is obtained by taking the absolute value,

$$\left| \int_{0}^x \frac{t^n}{1-t}\, dt\right|.$$

So our goal is to show that as $n$ is big enough, the error term shrink to zero.

Fixing a value of $x \in (0,1)$.

$$\left| \int_{0}^x \frac{t^n}{1-t}\, dt\right| \le \int_0^x \frac{t^n}{1-x} \, dt = \frac{x^{n+1}}{(n+1)(1-x)} \to 0$$

Fixing a value of $x \in (-1, 0)$,

$$\left| \int_{0}^x \frac{t^n}{1-t}\, dt\right| \le \int_x^0 |t|^n\, dt = \frac{|x|^{n+1}}{n+1}\to 0$$

  • $\begingroup$ Thanks a lot. Have never read about these special functions before. Please elaborate your answer for the last two lines as you have not shown values for $x,n, n+1, x^{n+1}, |x|^{n+1}$. $\endgroup$
    – jiten
    May 1, 2019 at 11:53
  • 1
    $\begingroup$ They are supposed to be general. There is no values to show. $\endgroup$ May 1, 2019 at 12:01
  • $\begingroup$ For me it is impossible to understand the meaning of that (your last comment), might be unable to grasp the answer. Request chat. $\endgroup$
    – jiten
    May 1, 2019 at 12:11
  • $\begingroup$ If you prefer, let $x=0.5$ for the case when $x \in (0,1)$ and also let $x=-0.5$ for the case when $x \in (-1,0)$. $\endgroup$ May 1, 2019 at 12:14
  • $\begingroup$ please refer to some literature or googling keywords for details on this problem. I tried for a lot of hours, but is like a dead end. Request to kindly accede to it, as you can by better knowledge. Without more literature, this question goes nowhere, by itself. $\endgroup$
    – jiten
    May 1, 2019 at 12:35

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