Look at the following series: 1 + 2x + 3x^2 + 4x^3 + 5x^4 + ..... You can say by using any method that the series is divergent. It indeed diverges but we use this as a series expansion for 1/(1-x)^2. I think it is wrong to expand functions like that by using Maclaurin series expansion method. According to me calculating the sum of an alternating series is also incorrect and this misconception is also due to the expansion of a function by Maclaurin series expansion method. Let's put 2 instead of x in the above function(i.e 1/(1-x)^2) We get 1 as a sum of the infinite series obtained by expanding the function by Maclaurin series expansion method and then inserting 2 instead of x. We will get a divergent series as evident from the expansion of the function. I came up with this idea when i studied a research paper published by Leonhard Euler on Serieses of that type. The main point is that a method does not work for all situations-it fails somewhere. So if you agree then say yes, you are true and if there is a mistake then please correct me.


The sum is not divergent for $|x|<1$: in this case the exponential decay of the $x^n$ factor is fast enough to mitigate the linear growth of the $n$ factor*. We can't use the sum to represent $1/(1-x)^2$ on the entire domain of this function; we can only use the sum for the sub-domain $(-1,1)$.

However, there do exist infinitely differentiable functions whose Taylor series diverge except at the point of expansion. Similarly, there exist infinitely differentiable functions whose Taylor series converge but to the wrong function, again except at the point of expansion. The latter has a classic example, given by Taylor expanding the function

$$f(x)=\begin{cases} e^{-1/x^2} & x \neq 0 \\ 0 & x=0 \end{cases}$$

at $x=0$.

* Clarification: if the limit given by the ratio test is $r<1$, then there exists $N$ such that for $n \geq N$, $|a_n| \leq \left ( \frac{1+r}{2} \right )^n$. That is, the summands are eventually dominated by the summands of a convergent geometric series with a slightly larger base than the limit. The reverse happens if $r>1$.

  • $\begingroup$ If we put x<1, then shall we get the same answer from both sides? $\endgroup$ – user258250 Aug 10 '15 at 9:34
  • $\begingroup$ @user258250 With $|x|<1$ they are indeed equal, yes. $\endgroup$ – Ian Aug 10 '15 at 10:25

Let's put 2 instead of x in the above function(i.e 1/(1-x)^2) We get 1/4

That's not right: $\frac{1}{(1-2)^2}=1$.

Perhaps you have in mind the series $1-2+3-4+\cdots$. See the Wikipedia article for full details. Briefly, Euler reasoned that $1-2x+3x^2-4x^4+\cdots=\frac{1}{(1+x)^2}$ and substituted $x=1$. Today, we would say that this is not a valid way to find the ordinary sum of the series, which diverges at $x=1$. On the other hand, since the series converges for $|x|<1$, it demonstrates that the Abel sum is $\frac14$.

If you're studying Euler's research papers on divergent series, keep in mind that they were written two and a half centuries ago. To avoid confusion, you should read them alongside a modern commentary using modern definitions. In the case of this series, check on the Euler Archive on E352 and read the synopses, which explain the connection to Abel summation.

The main point is that a method does not work for all situations-it fails somewhere.

It is true that regularized sums of divergent series are trickier to work with than sums of convergent series. Not all of the theorems you're used to will apply in more general settings. This is inconvenient, but hardly fatal to the theory.

  • $\begingroup$ O i see, i have made a mistake in the value substitution. That is indeed 1 if we put 2 into the function. $\endgroup$ – user258250 Aug 10 '15 at 9:31

As already pointed out by @Ian, the series $\sum_{n=0}^\infty (n+1)x^n$ does converge for $|x|<1$ and diverges elsewhere.

I am not sure if this is where you were going with all of this, but it seems instructive to discuss another topic that might be lurking here, and that is the topic of an Asymptotic Series. Asymptotic series are series that can actually diverge, but whose partial sums can prove very useful and powerful as approximations.

For example, the the Exponential Integral $\text{Ei}(x)$ has Asymptotic Series

$$\text{Ei}(x)\sim e^{-x}\sum_{k=0}^{\infty}\frac{(-1)^n\,n!}{x^{n+1}}$$

Clearly this series diverges for all $x$! The ratio test yields $\left|\frac{a_{n+1}}{a_n}\right|=\frac{n+2}{x}\to \infty$. But, a truncated series provides a better and better approximation for the Exponential Integral as $x$ gets larger and larger. We see that if we use just the first term of the series, then the approximation of the Exponential Integral is

$$\text{Ei}(x)\approx \frac{e^{-x}}{x}$$

which has an error of order $O\left(\frac{1}{x}\right)$.

  • $\begingroup$ What do you think about the sum of alternating series? I think that should be a wrong idea. $\endgroup$ – user258250 Aug 13 '15 at 9:35
  • $\begingroup$ An example of alternating series is 1-2+3-4+5-6+.. A paper of Euler rewritten in 2006 has the sum of that series as 1/4. The sum of the series 1-1+1-1+1-1+... has been given to be 1/2. The sum of the series 1-2^2+3^2-4^2+... has been given to be 0. Strange results!! How he found these results is by knowing that 1-x^2+x^3-x^4+... = 1/1+x and without thinking about convergence and divergence he put 1 in the formula(1/1+x) and its series and found what i have stated(1-1+1-..). $\endgroup$ – user258250 Aug 15 '15 at 11:09
  • $\begingroup$ Another man who is said to have proved the existence of God stated that the sum of that series is 1. The point is that summing alternating serieses should be prohibited. $\endgroup$ – user258250 Aug 15 '15 at 11:10
  • $\begingroup$ Please let me know how I can improve my answer. I really just want to give you the best answer I can. $\endgroup$ – Mark Viola Aug 17 '15 at 15:13

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