I was wondering if this proof and my later assertions were correct.

$f_n(x) = x/(1+x)^n$

Let $M_n = x/n(1+x)$. Here, $|f_n(x)| \le M_n$ since $n(1+x) \le (1+x)^n$ for $x \in [1, 2]$ and for $n>0$.

$\sum_{n=1}^{\infty} M_n$ converges by the ratio test.

Thus, $\sum_{n=1}^{\infty} x/(1+x)^n$ is uniformly convergent for $ x \in [1,2]$ by the Weierstrass $M$ test.

Also, since it is uniformly convergent, it is convergent.

Furthermore, $\int_1^2 (\sum_{1}^{\infty} f_n(x)) dx = \sum_{1}^{\infty} \int_1^2 f_n(x)dx$ since the series converges uniformly.

Thanks ahead of time.


Your application of $M_n$ test is wrong

Insted show $f_n(x)=x/(1+x)^n$ is decreasing function by derivative test


$f_n'(x)<0$ for $n>1$ as $x\in [1,2]$

SO $f_n(x)=x/(1+x)^n\leq f_n(1)$

i.e $M_n =1/2^n$

$\sum_1^{\infty}1/2^n =1$

SO given series is uniformly convergent by Weierstrass $M_n$ Test

  • $\begingroup$ Ahh ok, I understand now, thank you. So then if I prove uniform convergence like this, are the other two assertions I made valid? $\endgroup$ – user591271 Dec 1 '18 at 17:28
  • 1
    $\begingroup$ Yes . If you have uniformly convergent series then you can interchange summation and integration over compact set $\endgroup$ – MathLover Dec 1 '18 at 17:57

No, that is not a correct application of the M-test. First, and most importantly, $M_n$ must be a constant for the whole interval, not dependent on $x$. Without that, we wouldn't be talking about uniform convergence. Claiming "$M_n=\frac{x}{n(1+x)}$" can't possibly work.
Second, that choice doesn't work because the sum doesn't converge; $\sum_{n=1}^{\infty} \frac{x}{n(1+x)}=\frac{x}{1+x}\sum_{n=1}^{\infty}\frac1n$ is a multiple of the harmonic series, which diverges.


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