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I was trying to solve he following two questions from a competitive exam paper. Both the questions are linked with following statement.

Let $f_n(x)=\frac{x}{\{(n-1)x+1\}\{nx+1\}}$ and $s_n(x)=\displaystyle \sum_{j=1}^nf_j(x)\forall x\in[0,1]$

First Question:

The sequence $\{s_n\}$

  1. converges uniformly on $[0,1]$
  2. converges pointwise on $[0,1]$ but not uniformly
  3. converges pointwise for $x=0$ but not for $x\in(0,1]$
  4. doesnot converge for $x\in [0,1]$

Second Question:

$\displaystyle\lim_{n\to \infty}\int_0^1s_n(x)dx=1$ is obtained by

  1. dominated convergence theorem
  2. Fatou's Lemma
  3. the fact that $\{s_n\}$ converges uniformly on $[0,1]$
  4. the fact that $\{s_n\}$ converges pointwise on $[0,1]$

What I have done so far for the first question is that $f_n(x)=\frac{1}{\{(n-1)x+1\}}-\frac{1}{\{nx+1\}}$ which shows for a fixed $x$ $f_n(x)\rightarrow 0$ as $n\rightarrow \infty$ So I think pointwise convergence can be assured. But can I claim the pointwise convergence of $\{s_n\}$ from here?I am not sure about this.Also I have no idea about the uniform convergence.

and for the second question I am completely stuck.I only know the statements of the two theorems given in the options. Please help.

I admit that I don't have a great concept about convergence of sequence and series of functions and apologise for not showing much effort from my end.

Any help will be very helpful.Thnx in advance.

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The decomposition

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

shows that the sum telescopes,

$$s_n(x) = 1 - \frac{1}{nx+1}.\tag{1}$$

From that it is pretty easy to see how $(s_n)$ converges and to what. (Note: uniform convergence would imply continuity of the limit.)

$(1)$ shows that $(s_n)$ is uniformly bounded, so the dominated convergence theorem is one way to obtain the integral.

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  • $\begingroup$ Thnx a lot.I missed that telescopic part.Can you please clarify the uniform convergence part of s_n?It will be very helpful then. Thnx again. $\endgroup$ – usermath May 18 '14 at 19:49
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    $\begingroup$ What is the pointwise limit? With the telescoping, that is easily found, and answers the question about uniform convergence. $\endgroup$ – Daniel Fischer May 18 '14 at 19:51
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    $\begingroup$ The pointwise limit is not $1$ everywhere. There is an outsider. $\endgroup$ – Daniel Fischer May 18 '14 at 19:55
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    $\begingroup$ Yes. And that answers the question about uniform convergence. $\endgroup$ – Daniel Fischer May 18 '14 at 19:57
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    $\begingroup$ Since all $s_n$ are continuous, uniform convergence would imply continuity of the limit. The limit is not continuous at $0$, hence the convergence cannot be uniform. $\endgroup$ – Daniel Fischer May 18 '14 at 20:01

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