Question

$L\ \equiv\ \lim_{n\to\infty}\ \left[\sum_{k=0}^n\ \log_2\ \left(1-2^{2^k}+4^{2^k}\right)-2^{n+2}\right] .$ .
Calculati

Answer 1

$\lim_{n\to\infty}\ \left[\sum_{k=0}^n\ \log_2\ \left(1-2^{2^k}+4^{2^k}\right)-2^{n+2}\right]=$

$\lim_{n\to \infty}\ \left[\log_2\left(\prod_{k=0}^n\left(1-2^{2^k}+4^{2^k}\right)\right)-2^{n+2}\right]=$

$=\lim_{n\to \infty}\ \left[\log_2\left(\frac{1+2^{2^{n+1}}+2^{2^{n+2}}}{1+2+2^2}\right)-2^{n+2}\right]=$

$\lim_{n\to \infty}\ \left[\log_2{2^{2^{n+2}}\left(\frac{1}{2^{2^{n+2}}}+\frac{1}{2^{2^{n+1}}}+1\right)}-\log_2{7}-2^{n+2}\right]=$

[tex]\log_2\left(\lim_{n\to \infty}\left[\frac{1}{2^{2^{n+2}}}+\frac{1}{2^{2^{n+1}}}+1\right]\right)-\log_2{7}=\log_2{1}-\log_2{7}=-\log_2{7} [/tex]

[tex]Identitatea (1-x+x^2)(1-x^2+x^4)\cdot ...\cdot (1-x^{2^{n-1}}+x^{2^n})=\frac{1+x^{2^n}+x^{2^{n+1}}}{1+x+x^2} [/tex]


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