First we give a wording the Weierstrass’ M-test:
Suppose that { f n } \left\{f_n\right\} { f n } A A A { M n } \left\{M_n\right\} { M n }
∀ n ≥ 1 , ∀ x ∈ A : ∣ f n ( x ) ∣ ≤ M n ∑ n = 1 ∞ M n < ∞ ⟶ M n converges \forall n\ge 1,\quad \forall x\in A:\quad \left|f_n(x)\right|\le M_n\\[0.5cm]
\sum\limits_{n=1}^{\infty}M_n<\infty\longrightarrow M_n\, \text{converges} ∀ n ≥ 1 , ∀ x ∈ A : ∣ f n ( x ) ∣ ≤ M n n = 1 ∑ ∞ M n < ∞ ⟶ M n converges Then,
∑ n = 1 ∞ f n ( x ) \sum\limits_{n=1}^{\infty}f_n(x) n = 1 ∑ ∞ f n ( x )
converges a bsolutely and uniformly on A .
( More information: https://en.wikipedia.org/wiki/Weierstrass_M-test )
In our case,
∀ x ∈ [ 0 ; k ] : ∣ f n ( x ) ∣ = ∣ x n ( n + 2 ) 2 ∣ ≤ k n ( n + 2 ) 2 \forall x\in[0;k]:\quad\left|f_n(x)\right|=\left|\frac{x}{n(n+2)^2}\right|\le\frac{k}{n(n+2)^2} ∀ x ∈ [ 0 ; k ] : ∣ f n ( x ) ∣ = ∣ ∣ n ( n + 2 ) 2 x ∣ ∣ ≤ n ( n + 2 ) 2 k Then, it remains to prove that the series
∑ n = 1 ∞ k n ( n + 2 ) 2 converges \sum\limits_{n=1}^{\infty}\frac{k}{n(n+2)^2}\,\text{converges} n = 1 ∑ ∞ n ( n + 2 ) 2 k converges For this we use the integral test for convergence.
( More information: https://en.wikipedia.org/wiki/Integral_test_for_convergence )
In our case,
∫ 1 ∞ k d x x ( x + 2 ) 2 = k ⋅ ∫ 1 ∞ d x x ⋅ x 2 ( 1 + 2 / x ) 2 = = [ 2 x = t → d x x 2 = d t − 2 ; x = 1 → t = 2 and x = ∞ → t = 0 ] = = k ⋅ ∫ 2 0 ( t d t − 4 ⋅ ( 1 + t ) 2 ) = k 4 ⋅ ∫ 0 2 ( t + 1 ) − 1 d t ( 1 + t ) 2 = = k 4 ⋅ ∫ 0 2 ( 1 t + 1 − 1 ( t + 1 ) 2 ) d t = k 4 ⋅ ( ln ∣ t + 1 ∣ + 1 t + 1 ) ∣ t = 0 t = 2 = = k 4 ⋅ ( ( ln ∣ 2 + 1 ∣ + 1 2 + 1 ) − ( ln ∣ 0 + 1 ∣ + 1 0 + 1 ) ) = = k 4 ⋅ ( ( ln 3 + 1 3 ) − ( ln ∣ 1 ∣ + 1 ) ) = k 4 ⋅ ( ln 3 − 2 3 ) < ∞ \int_1^\infty\frac{kdx}{x(x+2)^2}=k\cdot\int_1^\infty\frac{dx}{x\cdot x^2(1+2/x)^2}=\\[0.3cm]
=\left[\frac{2}{x}=t\to\frac{dx}{x^2}=\frac{dt}{-2};\,x=1\to t=2\,\,\text{and}\,x=\infty\to t=0\right]=\\[0.3cm]
=k\cdot\int_2^0\left(\frac{tdt}{-4\cdot (1+t)^2}\right)=\frac{k}{4}\cdot\int_0^2\frac{(t+1)-1dt}{ (1+t)^2}=\\[0.3cm]
=\frac{k}{4}\cdot\int_0^2\left(\frac{1}{t+1}-\frac{1}{(t+1)^2}\right)dt=\frac{k}{4}\cdot\left.\left(\ln|t+1|+\frac{1}{t+1}\right)\right|_{t=0}^{t=2}=\\[0.3cm]
=\frac{k}{4}\cdot\left(\left(\ln|2+1|+\frac{1}{2+1}\right)-\left(\ln|0+1|+\frac{1}{0+1}\right)\right)=\\[0.3cm]
=\frac{k}{4}\cdot\left(\left(\ln3+\frac{1}{3}\right)-\left(\ln|1|+1\right)\right)=\frac{k}{4}\cdot\left(\ln3-\frac{2}{3}\right)<\infty ∫ 1 ∞ x ( x + 2 ) 2 k d x = k ⋅ ∫ 1 ∞ x ⋅ x 2 ( 1 + 2/ x ) 2 d x = = [ x 2 = t → x 2 d x = − 2 d t ; x = 1 → t = 2 and x = ∞ → t = 0 ] = = k ⋅ ∫ 2 0 ( − 4 ⋅ ( 1 + t ) 2 t d t ) = 4 k ⋅ ∫ 0 2 ( 1 + t ) 2 ( t + 1 ) − 1 d t = = 4 k ⋅ ∫ 0 2 ( t + 1 1 − ( t + 1 ) 2 1 ) d t = 4 k ⋅ ( ln ∣ t + 1∣ + t + 1 1 ) ∣ ∣ t = 0 t = 2 = = 4 k ⋅ ( ( ln ∣2 + 1∣ + 2 + 1 1 ) − ( ln ∣0 + 1∣ + 0 + 1 1 ) ) = = 4 k ⋅ ( ( ln 3 + 3 1 ) − ( ln ∣1∣ + 1 ) ) = 4 k ⋅ ( ln 3 − 3 2 ) < ∞ ∫ 1 ∞ k d x x ( x + 2 ) 2 = k 4 ⋅ ( ln 3 − 2 3 ) ≈ k ⋅ 0.10799 \boxed{\int_1^\infty\frac{kdx}{x(x+2)^2}=\frac{k}{4}\cdot\left(\ln3-\frac{2}{3}\right)\approx k\cdot0.10799} ∫ 1 ∞ x ( x + 2 ) 2 k d x = 4 k ⋅ ( ln 3 − 3 2 ) ≈ k ⋅ 0.10799
Conclusion,
Since ∑ n = 1 ∞ k n ( n + 2 ) 2 converges → ∑ n = 1 ∞ x n ( n + 2 ) 2 uniformly converges \text{Since}\quad\sum\limits_{n=1}^{\infty}\frac{k}{n(n+2)^2}\,\text{converges}\to\\[0.3cm]
\sum\limits_{n=1}^{\infty}\frac{x}{n(n+2)^2}\,\text{ uniformly converges} Since n = 1 ∑ ∞ n ( n + 2 ) 2 k converges → n = 1 ∑ ∞ n ( n + 2 ) 2 x uniformly converges
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