Sumu szeregów

[NieRozumiem85]

Obliczyć sumy szeregów:

\(\sum_{n=1}^{ \infty } \frac{(-1)^{n+1}}{n(n+1)}\)

\(\sum_{1}^{ \infty } \frac{n^2}{9^n}\)

[kerajs]

1)
\(\sum_{n=1}^{ \infty } \frac{(-1)^{n+1}}{n(n+1)}=
\frac{1}{1 \cdot 2} -\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}-\frac{1}{4 \cdot 5}+...=
\frac{2-1}{1 \cdot 2} -\frac{3-2}{2 \cdot 3}+\frac{4-3}{3 \cdot 4}-\frac{5-4}{4 \cdot 5}+...=\\=
1- \frac{1}{2}- \frac{1}{2}+ \frac{1}{3} + \frac{1}{3}- \frac{1}{4}- \frac{1}{4}+ \frac{1}{5} + ...=
-1+2(1- \frac{1}{2}+ \frac{1}{3} - \frac{1}{4}+ \frac{1}{5}-... )=\\=-1+2\ln 2\)


2)
\(1+x+x^2+x^3+x^4+...= \frac{1}{1-x} \\
(1+x+x^2+x^3+x^4+...)'= (\frac{1}{1-x})' \\
1+2x+3x^2+4x^3+5x^4+...= \frac{1}{(x-1)^2}\\
x+2x^2+3x^3+4x^4+5x^5+...=\frac{x}{(x-1)^2} \\
(x+2x^2+3x^3+4x^4+5x^5+...)'=(\frac{x}{(x-1)^2})' \\
1+4x+9x^2+16x^3+25^4+...=\frac{-1-x}{(x-1)^3}\\
x+4x^2+9x^3+16x^4+25^5+...=x\frac{-1-x}{(x-1)^3}\)
\(\sum_{n=1}^{ \infty }n^2x^{n}=- \frac{x^2+x}{(x-1)^3}\)
wystarczy wstawić \(x= \frac{1}{9}\)

[NieRozumiem85]

1)
\(\sum_{n=1}^{ \infty } \frac{(-1)^{n+1}}{n(n+1)}=
\frac{1}{1 \cdot 2} -\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}-\frac{1}{4 \cdot 5}+...=
\frac{2-1}{1 \cdot 2} -\frac{3-2}{2 \cdot 3}+\frac{4-3}{3 \cdot 4}-\frac{5-4}{4 \cdot 5}+...=\\=
1- \frac{1}{2}- \frac{1}{2}+ \frac{1}{3} + \frac{1}{3}- \frac{1}{4}- \frac{1}{4}+ \frac{1}{5} + ...=
-1+2(1- \frac{1}{2}+ \frac{1}{3} - \frac{1}{4}+ \frac{1}{5}-... )=\\=-1+2\ln 2\)


2)
\(1+x+x^2+x^3+x^4+...= \frac{1}{1-x} \\
(1+x+x^2+x^3+x^4+...)'= (\frac{1}{1-x})' \\
1+2x+3x^2+4x^3+5x^4+...= \frac{1}{(x-1)^2}\\
x+2x^2+3x^3+4x^4+5x^5+...=\frac{x}{(x-1)^2} \\
(x+2x^2+3x^3+4x^4+5x^5+...)'=(\frac{x}{(x-1)^2})' \\
1+4x+9x^2+16x^3+25^4+...=\frac{-1-x}{(x-1)^3}\\
x+4x^2+9x^3+16x^4+25^5+...=x\frac{-1-x}{(x-1)^3}\)
\(\sum_{n=1}^{ \infty }n^2x^{n}=- \frac{x^2+x}{(x-1)^3}\)
wystarczy wstawić \(x= \frac{1}{9}\)

A czy mogłabym poprosić o wytłumaczenie?

[kerajs]

A konkretnie, to co mam wytłumaczyć?

To samo dotyczy: viewtopic.php?f=37&t=84159