$a\text{)}1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\dots$  

zauważmy, że

$q=\frac{1}{3}$  

zatem:  

$\left|\frac{1}{3}\right|<1$  

więc:

$\underset{n\to \infty }{\lim }{S}_n=\frac{a_1}{1-q}$  

$S_n=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}$  

---

$b\text{)}1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\dots$  

zauważmy, że

$q=-\frac{1}{3}$  

zatem:

$\left|-\frac{1}{3}\right|<1$  

więc:

$\underset{n\to \infty }{\lim }{S}_n=\frac{a_1}{1-q}$  

$S_n=\frac{1}{1-\left(-\frac{1}{3}\right)}=\frac{1}{\frac{4}{3}}=\frac{3}{4}$  

---

$c\text{)}\frac{1}{\sqrt{2}}+\frac{1}{2}+\frac{1}{2\sqrt{2}}+\dots$  

zauważmy, że

$q=\frac{\sqrt{2}}{2}$  

zatem:

$\left|\frac{\sqrt{2}}{2}\right|<1$  

więc:

$\underset{n\to \infty }{\lim }{S}_n=\frac{a_1}{1-q}$  

$S_n=\frac{\frac{1}{\sqrt{2}}}{1-\frac{\sqrt{2}}{2}}=\frac{1}{\sqrt{2}}\cdot \frac{1}{1-\frac{\sqrt{2}}{2}}=\frac{1}{\sqrt{2}-\frac{2}{2}}=\frac{1}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{2-1}=\sqrt{2}+1$  

---

$d\text{)}5+2+0,8+\dots$  

zauważmy, że

$q=\frac{2}{5}$  

zatem:

$\left|\frac{2}{5}\right|<1$  

więc:

$\underset{n\to \infty }{\lim }{S}_n=\frac{a_1}{1-q}$  

$S_n=\frac{5}{1-\frac{2}{5}}=\frac{5}{\frac{3}{5}}=\frac{25}{3}=8\frac{1}{3}$  

---

$e\text{)}-100-60-36+\dots$  

zauważmy, że

$q=\frac{6}{10}=\frac{3}{5}$  

zatem:

$\left|\frac{3}{5}\right|<1$  

więc:

$\underset{n\to \infty }{\lim }{S}_n=\frac{a_1}{1-q}$  

$S_n=\frac{-100}{1-\frac{3}{5}}=\frac{-100}{\frac{2}{5}}=-100\cdot \frac{5}{2}=-250$  

---

$f\text{)}-4+\frac{5}{3}-\frac{25}{36}+\dots$  

zauważmy, że

$q=\frac{\frac{5}{3}}{-4}=\frac{5}{3}\cdot \left(-\frac{1}{4}\right)=\frac{5}{-12}$  

zatem:

$\left|-\frac{5}{12}\right|<1$  

więc:

$\underset{n\to \infty }{\lim }{S}_n=\frac{a_1}{1-q}$  

$S_n=\frac{-4}{1-\left(-\frac{5}{12}\right)}=\frac{-4}{\frac{17}{12}}=\frac{-48}{17}=-2\frac{14}{17}$