# Prove $\sum_{n=1}^{\infty} x/(1+x)^n$ is uniformly convergent for $x \in [1,2]$

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.

Your application of $$M_n$$ test is wrong

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

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

$$f_n'(x)<0$$ 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

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