Efter videon följer ett problem som du kan lösa för att testa att du tillgodogjort dig innehållet.
Problem:
Vad konvergerar talföljderna \[a_n=\sqrt{n^2+n}-\sqrt{n^2-n}, \qquad n=1,2,3,\ldots\]och\[b_n=\sqrt{n^4+n}-\sqrt{n^4-n}, \qquad n=1,2,3,\ldots\] mot?Svar:
Vi förlänger med konjugat-uttrycken: \[a_n=\sqrt{n^2+n}-\sqrt{n^2-n}\]\[=\dfrac{\left(\sqrt{n^2+n}-\sqrt{n^2-n}\right)\left(\sqrt{n^2+n}+\sqrt{n^2-n}\right)}{\sqrt{n^2+n}+\sqrt{n^2-n}}\]\[=\dfrac{(n^2+n)-(n^2-n)}{\sqrt{n^2+n}+\sqrt{n^2-n}}=\dfrac{2n}{\sqrt{n^2+n}+\sqrt{n^2-n}}\]\[=\dfrac{2}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1-\dfrac{1}{n}}}\to \dfrac{2}{1+1}=1.\]
\[b_n=\sqrt{n^4+n}-\sqrt{n^4-n}\]\[=\dfrac{\left(\sqrt{n^4+n}-\sqrt{n^4-n}\right)\left(\sqrt{n^4+n}+\sqrt{n^4-n}\right)}{\sqrt{n^4+n}+\sqrt{n^4-n}}\]\[=\dfrac{(n^4+n)-(n^4-n)}{\sqrt{n^4+n}+\sqrt{n^4-n}}=\dfrac{2n}{\sqrt{n^4+n}+\sqrt{n^4-n}}\]\[=\dfrac{\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n^3}}+\sqrt{1-\dfrac{1}{n^3}}}\to \dfrac{0}{1+1}=0.\]
\[b_n=\sqrt{n^4+n}-\sqrt{n^4-n}\]\[=\dfrac{\left(\sqrt{n^4+n}-\sqrt{n^4-n}\right)\left(\sqrt{n^4+n}+\sqrt{n^4-n}\right)}{\sqrt{n^4+n}+\sqrt{n^4-n}}\]\[=\dfrac{(n^4+n)-(n^4-n)}{\sqrt{n^4+n}+\sqrt{n^4-n}}=\dfrac{2n}{\sqrt{n^4+n}+\sqrt{n^4-n}}\]\[=\dfrac{\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n^3}}+\sqrt{1-\dfrac{1}{n^3}}}\to \dfrac{0}{1+1}=0.\]
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