vc quer calcular limite quando n vai pro infinito de:
\frac{ \sum_{k=0}^n \frac{1}{\sqrt{2k+1}} }{ \sum_{k=1}^n
\frac{1}{\sqrt{2k}} } + 1 =
\frac{ \sum_{k=1}^n \frac{1}{\sqrt{k}} }{ \sum_{k=1}^n \frac{1}{\sqrt{2k}}
} =
\sqrt{2} \frac{ \sum_{k=1}^2n \frac{1}{\sqrt{k}} }{ \sum_{k=1}^n
\frac{1}{\sqrt{k}} } =
\sqrt{2} [1 + \frac{ \sum_{k=n}^2n \frac{1}{\sqrt{k + n}} }{ \sum_{k=1}^n
\frac{1}{\sqrt{k}} } ] =
\sqrt{2} + \sqrt{2} \frac{ \sum_{k=n}^2n \frac{1}{\sqrt{k + n}} }{
\sum_{k=1}^n \frac{1}{\sqrt{k}} }
Mas \frac{ \sum_{k=n}^2n \frac{1}{\sqrt{k + n}} }{ \sum_{k=1}^n
\frac{1}{\sqrt{k}} } vai pra zero com n, para ver basta usar que a média
dos quadrados é maior ou igual a média aritmética, assim:
\sum_{k=n}^2n \frac{1}{\sqrt{k}} \leq \frac{1}{n}\sqrt{\sum_{k=n}^2n
\frac{1}{k + n}} <
\frac{1}{n}\sqrt{\sum_{k=n}^2n \frac{1}{n}} = \frac{1}{n}.
Enquanto \sum_{k=1}^n \frac{1}{\sqrt{k}} > 1.
