znaleźć sumę

[basia]

\(\displaystyle{ \sum ^{\infty}_{n=1}n\cdot x^{n}}\)

[przemk20]

mozna tez bez calek
\(\displaystyle{ S = x + 2x^2 + 3x^3 + .... | x \\
Sx = x^2 + 2x^3 + 3x^4 + ....... \\
S - Sx = x + x^2 + x^3 + x^4 + . ... = \frac{x}{1-x} \\
S = \frac{x}{(1-x)^2}}\)
kuch2r ale
\(\displaystyle{ \int n x^n = \frac{n}{n+1} x^{n+1}}\)

[kuch2r]

mozna tez bez calek
\(\displaystyle{ S = x + 2x^2 + 3x^3 + .... | x \\
Sx = x^2 + 2x^3 + 3x^4 + ....... \\
S - Sx = x + x^2 + x^3 + x^4 + . ... = \frac{x}{1-x} \\
S = \frac{x}{(1-x)^2}}\)
kuch2r ale
\(\displaystyle{ \int n x^n = \frac{n}{n+1} x^{n+1}}\)

??:
\(\displaystyle{ \sum_{n=0}^{\infty}n\cdot x^n=\sum_{n=0}^{\infty}(n+1)\cdot x^n - x^n=\sum_{n=0}^{\infty}(n+1)\cdot x^n- \sum_{n=0}^{\infty}x^n}\)
Niech:
\(\displaystyle{ S_1=\sum_{n=0}^{\infty}(n+1)\cdot x^n\\
S_2=\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}}\)

\(\displaystyle{ \int (n+1)\cdot x^n dx=x^{n+1}=\frac{x}{1-x}\\
(\frac{x}{1-x})'=\frac{1}{(1-x)^2}\\
S=S_1-S_2=\frac{1}{(1-x)^2}-\frac{1}{1-x}}\)

[Kostek]

Mozna tez tak :
\(\displaystyle{ S=S_{1}+S_{2}+.......}\)
\(\displaystyle{ S_{1}=x+x^{2}+x^{3}....=\frac{x}{1-x}}\)
\(\displaystyle{ S_{2}=x^{2}+x^{3}+x^{4}....=\frac{x^{2}}{1-x}}\)
\(\displaystyle{ S_{3}=x^3+x^{4}+x^{5}....=\frac{x^{3}}{1-x}}\)
\(\displaystyle{ S_{4}=x^{4}+x^{5}+x^{6}....=\frac{x^{4}}{1-x}}\)
.
.
.

\(\displaystyle{ S=\frac{1}{1-x}(x+x^{2}+x^{3}....)=\frac{x}{(1-x)^{2}}}\)

[basia]

to znaczy że kuch2r zrobił źle??
ps. Kostek-fajny sposób

[max]

kuch2r podał ten sam wynik, tylko w postaci sumy ułamków prostych.