Respuesta :

LammettHash
This is a geometric series with common ratio between terms of [tex]r=\dfrac1{\sqrt2}[/tex].

It should be clear to you that [tex]\dfrac1{\sqrt2}<1[/tex], which means the series will converge.

The sum would be

[tex]\displaystyle\sum_{n\ge1}\left(\frac1{\sqrt2}\right)^n=\frac1{\sqrt2}\sum_{n\ge1}\left(\frac1{\sqrt2}\right)^{n-1}=\frac1{\sqrt2}\times\frac1{1-\frac1{\sqrt2}}=\frac1{\sqrt2-1}=1+\sqrt2[/tex]

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