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[Петър Евгениев]

[tex]\lim_{n \to \infty}(1+\frac{2}{2^{1}}+\frac{3}{2^{2}}+\frac{4}{2^{3}}+...+\frac{n}{2^{n-1}})[/tex]

[ptj]

[tex]A=\frac{1}{2^0}+\frac{2}{2^{1}}+\frac{3}{2^{2}}+\frac{4}{2^{3}}+...+\frac{n}{2^{n-1}}+...

=\sum_{n=0}^{+\infty }\frac{1}{2^n}+\big(\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{n}{2^{n}}+...\big)=[/tex]

[tex]=\sum_{n=0}^{+\infty }\frac{1}{2^n}+\frac{1}{2}\big(\frac{1}{2^{0}}+\frac{2}{2^{1}}+\frac{3}{2^{2}}+...+\frac{n}{2^{n-1}}+...\big)=
\sum_{n=0}^{+\infty }\frac{1}{2^n}+\frac{1}{2}A[/tex]

[tex]A=2.\sum_{n=0}^{+\infty }\frac{1}{2^n}=4[/tex]

[Петър Евгениев]

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