obliczyć sumy szeregów

[ssskrzat]

a) \(\sum_{ \infty }^{n=1} \frac{n}{2^n}
b) \sum_{ \infty }^{n=1} \frac{n^2}{2^n}
c) \sum_{ \infty }^{n=0} \frac{1}{n!}
d) \sum_{ \infty }^{n=1} \frac{(-1)^n+1}{n}\)

[octahedron]

\(a)
\sum_{n=1}^{\infty}nx^n=x\sum_{n=1}^{\infty}nx^{n-1}=x\sum_{n=1}^{\infty}(x^n)'=x\(\sum_{n=1}^{\infty}x^n\)'=x\(\frac{x}{1-x}\)'=\frac{x}{(1-x)^2}
\sum_{n=1}^{\infty}\frac{n}{2^n}=\sum_{n=1}^{\infty}n\(\frac{1}{2}\)^n=\frac{\frac{1}{2}}{\(1-\frac{1}{2}\)^2}=2\)

[octahedron]

\(b)
\sum_{n=1}^{\infty}n^2x^n=\sum_{n=1}^{\infty}\[n(n-1)x^n+nx^n\]=\sum_{n=1}^{\infty}n(n-1)x^n+\sum_{n=1}^{\infty}nx^n=x^2\sum_{n=1}^{\infty}n(n-1)x^{n-2}+\frac{x}{(1-x)^2}=
=x^2\(\sum_{n=1}^{\infty}x^n\)''+\frac{x}{(1-x)^2}=x^2\(\frac{x}{1-x}\)''+\frac{x}{(1-x)^2}=\frac{2x^2}{(1-x)^3}+\frac{x}{(1-x)^2}
\sum_{n=1}^{\infty}\frac{n^2}{2^n}=\sum_{n=1}^{\infty}n^2\(\frac{1}{2}\)^n=\frac{2\(\frac{1}{2}\)^2}{\(1-\frac{1}{2}\)^3}+2=6\)

[octahedron]

\(c)
e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}
\sum_{n=0}^{\infty}\frac{1}{n!}=e^1=e\)

[octahedron]

\(d)
\sum_{n=1}^\infty\frac{(-1)^n+1}{n}=\frac{2}{2}+\frac{2}{4}+\frac{2}{6}+...=1+\frac{1}{2}+\frac{1}{3}+...=\sum_{n=1}^\infty\frac{1}{n}\)

a ten szereg jest rozbieżny