Determine if the following geometric series converges or diverges. If it converges, what is its sum? $$\sum_{i=0}^n \frac{1}{(\sqrt[]{2})^n}$$
So I know that it converges because $$r =\frac{1}{(\sqrt[]{2})}$$
The answer is $$ 2 + \sqrt[]{2}$$
I have no idea how they got there. Can anybody help?
$\endgroup$ 41 Answer
$\begingroup$Formula for sum of G.P. with infinite terms ,
$$S_{\infty}=\frac{a}{1-r}$$
Where $a$ is the first term of series and $r$ is common ratio. ($|r|<1$)
In your case :
$$S_{\infty}=\frac{1}{(\sqrt2)^0}+\frac{1}{(\sqrt2)^1}+\frac{1}{(\sqrt2)^2}+\frac{1}{(\sqrt2)^3} \dots \infty$$
Since $(\sqrt2)^0=1$
$$S_{\infty}=1+\frac{1}{\sqrt2}+\frac{1}{(\sqrt2)^2}+\frac{1}{(\sqrt2)^3} \dots \infty$$ $$a=1 ~~~~\text{and}~~~~ r=\frac{1}{\sqrt2}$$
$$S_{\infty}=\frac{1}{1-\frac{1}{\sqrt2}} =\frac{\sqrt2}{{\sqrt2}-1} \times \frac{{\sqrt2}+1}{{\sqrt2}+1}=2+{\sqrt2}$$
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