Can I find the sum of a series by the use of differentiation and limits?

Question

Find the sum function of the series $$\sum f_n$$ where

$f_n(x)=\frac{nx}{(1+n^2x^2)}-\frac{(n-1)x}{[1+(n-1)^2x^2]}$, and $x∈[0,1].$

Solution
I am new in the topic of Power series and its sum or convergence etc. My approach is:-

Let $u_n = \frac{nx}{(1+n^2x^2)}$ and $u_{n-1}=\frac{(n-1)x}{(1+n^2x^2)}$

We have

$f_n(x)=\frac{nx}{(1+n^2x^2)}-\frac{(n-1)x}{(1+n^2x^2)}$

$f_n'(x)=\frac{d}{dx}[\frac{nx}{(1+n^2x^2)}]-\frac{d}{dx}[\frac{(n-1)x}{(1+n^2x^2)}]$

$=\frac{n(1-n^2x^2)}{(1+n^2x^2)^2} - \frac{(n-1)[1-(n-1)^2x^2]}{[1+(n-1)^2x^2]^2}$

Now setting $f_n'(x)=0$

$→ \frac{n(1-n^2x^2)}{(1+n^2x^2)^2} - \frac{(n-1)[1-(n-1)^2x^2]}{[1+(n-1)^2x^2]^2}=0$

$→ \frac{n(1-n^2x^2)}{(1+n^2x^2)^2} = \frac{(n-1)[1-(n-1)^2x^2]}{[1+(n-1)^2x^2]^2}$

$→ u_n'(x)=u_{n-1}'(x)$

Now for $n→∞$, we have

$\lim_{n\to\infty}|\frac{\frac{n(1-n^2x^2)}{(1+n^2x^2)^2}}{\frac{(n-1)[1-(n-1)^2x^2]}{[1+(n-1)^2x^2]}}|$ approaches to $1$

Since this series is not uniformly convergent it becomes pointwise convergent and the sum of the series is $\lim_{x\to 1}\frac{\frac{n(1-n^2x^2)}{(1+n^2x^2)^2}}{\frac{(n-1)[1-(n-1)^2x^2]}{[1+(n-1)^2x^2]}} -\lim_{x\to 0}\frac{\frac{n(1-n^2x^2)}{(1+n^2x^2)^2}}{\frac{(n-1)[1-(n-1)^2x^2]}{[1+(n-1)^2x^2]}}$

$= \frac{(1+n)(n^2-2n+1)^2}{(2-n)(1+n^2)^2} - \frac{n}{(n-1)}$

In my logic Since $x∈[0,1]$ then the difference between $lim{f_n'(x)}$ at $x=1$ and $lim{f_n'(x)}$ at $x=0$ should be the sum of the series.

If it is wrong then please suggest me a correct approach.Thank you.

Answer 1

Note that: $$\begin{align} f(1) = \frac{x}{1+x^2}- 0\\ f(2) = \frac{2x}{1+4x^2}-\frac{x}{1+x^2}\\ f(3) = \frac{3x}{1+9x^2}-\frac{2x}{1+4x^2}\\ \ldots \\ f(n) = \frac{nx}{1+n^2x^2} - \frac{(n-1)x}{1+(n-1)^2x^2} \\ \implies\sum_{1}^{n} \,f(k) = \frac{nx}{1+n^2x^2} \end{align}$$

We can also compute : $$\lim_{n \to \infty} \sum_{1}^{n} \, f(k) = \lim_{n \to \infty} \frac{nx}{1+n^2x^2} = 0$$ using L’Hôpital’s rule.

Answer 2

Let $G(n)=nx/(1+n^2x^2).$ Then $f_n=G(n)-G(n-1).$ So $$\sum_{n=1}^Mf_n=\sum_{n=1}^MG(n)-G(n-1)=G(M)-G(0).$$ For example $f_3+f_2+f_1=(G(3)-G(2))+(G(2)-G(1))+(G(1)-G(0))=G(3)-G(0).$

This is an example of a telescoping series, as said in the comment by Robert Israel.

Since $G(M)\to 0$ as $M\to \infty,$ regardless of $x,$ we have $$\sum_{n=1}^{\infty}f_n=\lim_{M\to \infty}\sum_{n=1}^Mf_n=\lim_{M\to \infty}(G(M)-G(0))=-G(0)=0.$$ Footnote: If $x\ne 0$ and $M> 0$ then $|G(M)|=|Mx|/(1+M^2x^2)<|Mx|/(M^2x^2)=1/(M|x|)\to 0$ as $M\to \infty.$ If $x=0$ then $G(M)=0$ for all $M.$ So $G(M)\to 0$ as $M\to \infty$, for any $x$.